# CBSE Class 12 Maths Revision Notes Chapter 10

## Class 12 Mathematics Chapter 10 notes

Mathematics is an essential subject in Class 12. Students must  understand the fundamentals of  this subject to crack the national level entrance exams  to get admission to various colleges and universities for  higher studies. The CBSE syllabus has been formulated perfectly and creates a strong base of all the mathematical concepts.

To master the subject, students must consider using the Class 12 Mathematics Chapter 10 notes. Students can  enhance  their knowledge and concept to score high marks in the board examinations. Along with several NCERT books, students must consider referring to the Class 12 Mathematics Chapter 10 notes and CBSE sample papers.

### NCERT  Class 12 Mathematics Chapter 10: Key Topics

The Class 12 Mathematics Chapter 10 notes provide detailed information about the following topics:

• Introduction
• Some basic concepts
• Type of Vectors
• Product of vectors (dot product and cross product)
• Product of Two Vectors

### Introduction

In the Class 12 Mathematics Chapter 10 notes, some interesting facts about vectors and their applications in our day-to-day life are discussed in detail.

Following are some examples:

• In the motion of an aeroplane, both speed and direction are taken into account.
• Suppose when a plastic ball is thrown with some speed, and  suppose the direction in which it is thrown  is northeast. Therefore, the ball has speed as well as direction.
• Fluid mechanics, static, Electrical Engineering etc., use vectors.
• In the game of carrom board, we hit the striker with speed and direction.

### Vectors

The physical quantities which have directions and magnitude are known as vectors. They are represented using an arrow.

Consider a vector AB. It will start from point A and end at point B. Therefore, point A is the initial point, and B is the terminal point.

The distance between points A and B (initial and terminal points) is termed the magnitude of the vector. It is denoted by AB or a, where the arrow indicates the direction.

Consider the following examples:

 Measures Scalar Vector Criteria 10 kg Yes No It has only magnitude. 21 m north-west No Yes It has magnitude and direction. 50° Yes No It has only magnitude. Force No Yes It has magnitude and direction. Speed Yes No It has only magnitude.

### The right-handed rectangular coordinate system

The right-hand thumb rule is applied to understand the conventions for vectors in three dimensions.

If we fold our four fingers and move from X-axis to Y-axis, then the thumb will indicate the direction of the Z-axis.

#### Position Vector

Let a point P in space have coordinates (x, y, z) with the origin O (0, 0, 0).

Then, the vector OP with O and P as its initial and terminal points, respectively, is the position vector of point P with point O.

The value of OP= x2+y2+z2

The Class 12 Mathematics Chapter 10 notes include several practice questions based on this concept.

### Direction Angle & Direction Cosines

Consider a position vector OP of a point P(x,y, z). The angles α, β and are made by the vector r with the positive  x,y and z-axis.

Cosine values, i.e., cos α, cos β and cos , are termed as the direction cosines of the vector r

They are denoted by l, m and n. Hence, cos α = l, cos β= m and cos = n.

Consider a right-angled OAP,

We know that cos α=(xr) where (r= |r|). Similarly, for right-angled triangles OBP and OCP, we get cos β = (yr) and cos =(zr).

The coordinates of point P are written as (lr, mr, nr) are the direction ratios of vector r. They are denoted as a, b and c, respectively. In general l2+m2+n2 =1.

Different Types of Vectors

#### Zero Vector:

Consider a vector r. If the starting and final points are equal, then the vector r is known as a zero null vector and is denoted by 0.

The vector does not have a definite direction because it is zero in magnitude.

#### Unit Vector:

A vector r, whose magnitude is 1 unit, is known as a unit vector. The unit vector in the direction is denoted by a.

#### Coinitial Vectors:

Consider vectors r and s. If the two vectors have the same initial point are known as Coinitial vectors.

#### Collinear Vectors:

The two or more vectors are known as collinear vectors if they are parallel to the same line, independent of the magnitude and direction.

For example: Consider three vectors, they all are parallel to each other, but their magnitudes and directions are different.

#### Equal Vectors:

The equal vectors have the same magnitude and direction irrespective of the initial points.

#### Negative of a Vector:

A negative vector that has the same magnitude and opposite direction as that of a given vector is called the negative of a given vector.

#### Free Vectors:

The vectors that will not change when it is displaced in a parallel direction, keeping their magnitude and direction the same, are called free vectors.

Consider a man moving from point A to point B and then from point B to point C. We can say that the total displacement the man makes from point A to point C is denoted by the vector AC and is expressed as AC = AB + BC. This is the triangle law of addition.

Refer to the Class 12 Mathematics Chapter 10 notes to get in-depth knowledge about Vectors

For the addition of two vectors, they are positioned so that the starting point of a given vector coincides with the final point of the other.

Consider two vectors, a and b. We move them parallel to each other in such a way that the tail of b touches the head of b.

The resultant vector is given as vector (a + b).

Also we know that AC = AB + BC

Therefore, AC = – CA .

Hence, AA= AB + BC + CA =0.

This implies that when the sides of the triangle are considered in order, it results in zero resultant because the initial and terminal points are coinciding with each other.

Constructing a vector where the magnitude of BC’ is the same as the vector BC, but the direction is opposite to the vector BC. This implies BC’= – BC.

Apply the triangle law of addition we get, AC’= AB+ BC= AB+ (-BC)

AC’ = (ab).

Therefore, vector AC’ represents the difference between vectors a and b.

### Parallelogram law of vector addition:

Consider two vectors (a and b). Let them be represented using the two adjacent sides of a parallelogram in magnitude and direction. Then their sum (a + b) will be given in magnitude and direction by the diagonal of the parallelogram through a common point.

Applying the parallelogram law of vector, we will get

OC = OA + AC

In triangle OAC, we apply the triangle law of addition.

Therefore, OC = OA + AC

or OC = a + b

Using the triangle law of addition in triangle OCB, we get

OC = OB + CB

OC = a + b

This is termed the Parallelogram law for vector addition.

The parallelogram law of vector addition is very crucial. Students are advised to refer to the Class 12 Chapter 10 Mathematics notes to get an in-depth understanding of this concept.

#### Commutative property:

Consider any two vectors a and b, then vector addition (a + b) or (b + a) will be equal i.e. (a + b) =(b + a).

#### Associative property:

Consider any three vectors a, b, and c, then (a + b) + c = a+ (b + c)

The addition of vectors a and b, when added to c will be the same if we first add vectors b and c and then add the sum to the vector a.

Additive Identity:- Consider 0 vector. If we add 0 to vector a, then we will get vector a

(a + 0) =(0 + a). = a.

### Multiplication of any Vector by any Scalar:-

Let a be any vector and λ be a scalar.

The product of vector a and scalar λ is denoted as (λ a), which has the same or opposite to that of vector a.

The magnitude of the product is given as |λ a| = | λ|| a|

When a vector is added with itself but in the opposite direction, then the resultant is 0.

If λ =-1 is a scalar, then (λ a) = – a, the negative vector has the same magnitude but is opposite in direction. The vector – a is known as the additive inverse of the vector a and is given as a + (–a) = (–a) + a = 0.

#### Unit Vector in a direction

A unit vector is a vector of magnitude 1.

It is denoted by a.

Let λ be a scalar, then λ = (1a ), then | λ a | =| λ|a .

=> (1a ) a = 1.

Therefore λ a represents the unit vector in the direction of vector a.

Unit vector can be written as a = (1a ) a.

The Class 12 Mathematics Chapter 10 notes explain all the properties of vectors in a detailed manner.

### Components of a vector:

#### Unit Vector:

Let points A =(1,0,0), B=(0,1,0) and C=(0,0,1) be on the x-axis y-axis and z-axis, respectively.

|OA|=1, |OB|=1 and |OC |= 1 shows all three vectors have the same magnitude of 1.

The vectors |OA|, |OB|and |OC | are termed as unit vectors along OX, OY and OZ, respectively. They are denoted by î, ĵ and k̂, respectively.

#### Position vector:

Let P(x, y, z) be the position vector, and P1 be the perpendicular from point P onto the plane XoY. This implies that P1P is parallel to the z-axis.

Let î be the unit vectors along the x-axis, ĵ be the unit vectors along the y-axis and k̂ be the unit vector along the z-axis. Then the coordinates of P, P1P = OR = z k̂.

and QP1= OS= yĵ. Also, OQ = x î .

Hence, OP1 = OQ + QP1 = x î + yĵ and

OP = OP1+ P1P= x î + yĵ + z k̂

Therefore, the position vector of P at O is given as OP or r =x î +yĵ+ z k̂, where r is known as the position vector at point P.

This form of a vector is called the component form.

x, y and z are known as the scalar components and x î + yĵ + z k̂ are called vector components of vector r

The length of r= x î + yĵ + z k̂ is |r| =x2+y2+z2.

### Vectors Operations

Consider two vectors given by a= a1î + a2ĵ + a3k̂ and b= b1î + b2ĵ + b3

Sum of the vectors is (a +b) =(a1+ b1) î + (a2+ b2)ĵ + (a3+ b3) k̂

Difference of the vectors is (ab) =(a1b1) î + (a2b2)ĵ + (a3b3) k̂

Vectors a & b are equal iff a1= b1, a2= b2 and a3= b3.

Multiplication of vector by scalar (λa )=( λ a1) î +( λ a2) ĵ + (λ a3) k̂

#### Distributive law

Consider vectors, a and b and scalars, k and m. Then

ka + m a =(k+m)a

k (ma ) =(km) a

k(a + b ) = k a +k b

#### Collinear Vectors

Let λ be a non-zero scalar. Then, the vector (λ a ) is said to be collinear to the vector a if there exists λ such that b = λ a.

Therefore (a1b1) = (a2b2) = (a3b3) = λ

Vectors joining two points

Let P1(x1, y1, z1) be the initial point and P2(x2, y2, z2) be the terminal point. Then the vector joining P1 and P2is P1P2=( (x2i+ y2j+ z2k) − (x1i+ y1j+ z1k)) =(x2x1)i+ (y2y1)j+ (z2z1) k

The magnitude of P1P2 is given as:-

P1P2= (x2x1)2+(y2y1)2+ (z2z1) 2

The vector operations are very important. Students can refer to the Class 12 Mathematics Chapter 10 notes for a step-wise explanation.

### Section Formula

Consider two points, P and Q, represented by the position vectors OP and OQ, respectively. The line segment joining P and Q is divided by  third point R, internally and externally.

Case 1: R divides PQ internally, such that m (RQ) = n (PR), where m and n are scalar quantities.

Then, the position vector of the R, which divides points P and Q internally in the ratio (m:n), is

OR = mb + nam+n

Case 2:-When R divides PQ externally, such that (PRQR) = (mn), where m and n are scalar quantities.

Then, the position vector of the R, which divides points P and Q externally in the ratio (m:n), is

OR = mb – nam-n

Students will have to use the section formula to solve several sums included in the Chapter 10 Mathematics Class 12 notes.

### Product of Two Vectors

For cross product, the resultant will be a vector quantity.

For dot product, the resultant will be a scalar quantity.

1. Dot Product of two vectors

The dot product of any two nonzero vectors a and b is given by a .b = | a | |b | cos θ

Where θ = angle between the two vectors and | a |, | b |is the magnitude of vectors a and b, respectively.

The dot or scalar product is commutative, i.e., a . b = b. a

The Angle between Vectors: The scalar product is given as cos θ = (a . b| a | |b |).

#### Properties of scalar product:

Consider any two vectors a and b and a scalar λ. Then,

a). b = λ (a. b) = a (λ. b)

Also, we know that, a = a1 î +a2 ĵ +a3 k̂ and b = b1î +b2 ĵ +b3k̂, then

(a . b) = (a1b1) + (a2b2) + (a3b3

#### Projection of a vector on a line

Consider a vector AB. Suppose it makes an angle θ with a directed line l. Then the projection of AB on line l is a vector p having magnitude |AB| cosθ. The direction of p can be the same or opposite to the line l, depending on the value of cos θ.

The vector p is known as the projection vector, and its magnitude p is termed as the projection of the vector AB on line l.

1. Vector (Cross) Product

The vector product of two nonzero vectors, a and b, is given as (a x b ) =|a ||b | n sin θ, where n is the unit vector perpendicular to vectors a and b. The value of the unit vector (n ) is given by the right-hand thumb rule.

### Right-handed rectangular coordinate systems

To find the product of 2 vectors, a and b, the direction of a x b is given by the thumb.

Consider two nonzero vectors, a and b. Then a x b = 0 if a and b are parallel or collinear to each other, i.e., if a x b = 0, then |a ||b|.

The vector product is not commutative, i.e, (a x b) = – (b x a). The magnitudes will be the same, but the directions are different.

Students can find several problems based on the area of a triangle and the area of a parallelogram in the Class 12 Mathematics Chapter 10 notes.

#### Area of triangle

Let a and b be the adjacent sides, then the area of a triangle is given as (12) a X b

Area of Parallelogram

If a and b are the sides, then the area of a parallelogram is given as a X b.

Distributive Property of Vector Cross Product:

If a , b and c are vectors and λ be a scalar, then a x (b + c) = a x b + a x c.

λ(a x b) = (λ a) x (b) = a x (λ b

The Class 12 Mathematics notes Chapter 10 is an essential study material that covers all important concepts.

## Class 12 Mathematics Chapter 10 notes:Exercise & Solutions

To help students acquire knowledge, Extramarks provides the Class 12 Mathematics Chapter 10 notes. It enables students to access themselves based on their preparation level. Solutions to all exercise questions and solved examples are given in the CBSE revision notes. The CBSE extra questions are formulated concept-wise for easy practice. Extramarks provides the best Class 12 Mathematics Chapter 10 notes to help students boost confidence and develop problem-solving skills to easily tackle any question asked in the exams. All important questions, concepts and formulas are highlighted so that students can focus on them.

Students can find the Class 12 Mathematics Chapter 10 notes and CBSE  past years’ question papers to judge their preparation. Extramarks, an online learning platform, aims to provide a holistic learning experience and enable each student to focus and improve their score gradually.

Click on the following links mentioned below to  access  the solutions to all questions in the Class 12 Mathematics Chapter 10 notes:

### Class 12 Mathematics Chapter 10 Notes : Key Features

• The notes are segregated, well-organised and cover the entire CBSE syllabus.
• Using the notes will help to develop  conceptual clarity and  knowledge  to clarify all your doubts.
• The Class 12 Mathematics Chapter 10 notes strictly follow the guideline issued by the CBSE board because most of the CBSE Board questions asked are from NCERT Books.
• It is curated by  subject matter experts  at Extramarks.
• With the help of the Class 12 Mathematics Chapter 10 notes, students can understand the various types of questions asked in the exams. It gives them an in-depth understanding of subjects.
• These notes prepare the students for Class 12 exams and National level competitive examinations such as JEE, NDA, etc.
• The solution to every question is explained in a systematic  topic-wise manner in the Class 12 Mathematics Chapter 10 notes.

Q.1

$\mathrm{Find}\text{}\mathrm{angle}\text{}\mathrm{\theta }\text{}\mathrm{between}\text{}\mathrm{the}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}\text{}=\text{}\stackrel{^}{\mathrm{i}}\text{}+\stackrel{^}{\mathrm{j}}\text{}-\text{}\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\stackrel{^}{\mathrm{i}}\text{}-\stackrel{^}{\mathrm{j}}\text{}+\text{}\stackrel{^}{\mathrm{k}}.$

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\mathrm{the}\text{}\mathrm{vecotrs}\text{}\stackrel{\to }{\mathrm{a}}\text{}=\text{}\stackrel{^}{\mathrm{i}}\text{}+\stackrel{^}{\mathrm{j}}\text{}-\text{}\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\stackrel{^}{\mathrm{i}}\text{}-\stackrel{^}{\mathrm{j}}\text{}+\text{}\stackrel{^}{\mathrm{k}}\text{}\mathrm{is}\\ \mathrm{cos\theta }\text{}=\text{}\frac{\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}}{\left|\stackrel{\to }{\mathrm{a}}\right|\left|\stackrel{\to }{\mathrm{b}}\right|}\\ =\text{}\frac{\left(\stackrel{^}{\mathrm{i}}\text{}+\stackrel{^}{\mathrm{j}}\text{}-\text{}\stackrel{^}{\mathrm{k}}\right)\left(\stackrel{^}{\mathrm{i}}\text{}-\stackrel{^}{\mathrm{j}}\text{}+\text{}\stackrel{^}{\mathrm{k}}\right)}{\sqrt{{1}^{2}\text{}+{1}^{2}\text{}+\text{}{1}^{2}}\sqrt{{1}^{2}\text{}+{1}^{2}\text{}+\text{}{1}^{2}}}\\ =\text{}\frac{1-1-1}{\sqrt{3}\sqrt{3}}\text{}=\text{}-\frac{1}{3}\\ \mathrm{cos\theta }\text{}=\text{}-\frac{1}{3}\text{}⇒\text{}\mathrm{\theta }\text{}={\mathrm{cos}}^{-1}\text{}\left(-\frac{1}{3}\right)\end{array}$

Q.2

$\begin{array}{l}\mathrm{Express}\text{}\mathrm{the}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{a}}\text{}=\text{}5\stackrel{^}{\mathrm{i}}\text{}-2\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}\text{}\mathrm{as}\text{}\mathrm{sum}\text{}\mathrm{of}\text{}\mathrm{two}\text{}\mathrm{vectors}\text{}\mathrm{such}\\ \mathrm{that}\text{}\mathrm{one}\text{}\mathrm{is}\text{}\mathrm{parallel}\text{}\mathrm{to}\text{}\mathrm{the}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}3\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\mathrm{other}\text{}\mathrm{is}\text{}\\ \mathrm{perpendicular}\text{}\mathrm{to}\text{}\stackrel{\to }{\mathrm{b}}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{}\stackrel{\to }{\mathrm{a}}\text{}=\stackrel{\to }{\mathrm{p}}\text{}+\text{}\stackrel{\to }{\mathrm{d}}\text{}\mathrm{where}\text{}\stackrel{\to }{\mathrm{p}}\text{}\mathrm{is}\text{}\mathrm{parallel}\text{}\mathrm{to}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{d}}\text{}\mathrm{is}\text{}\mathrm{perpendicular}\text{}\stackrel{\to }{\mathrm{b}}.\\ \stackrel{\to }{\mathrm{p}}\text{}=\text{}\mathrm{\lambda }\stackrel{\to }{\mathrm{b}}\text{}\left[\text{}\stackrel{\to }{\mathrm{p}}\text{}||\text{}\stackrel{\to }{\mathrm{b}}\right]\\ ⇒\stackrel{\to }{\mathrm{p}}\text{}=\text{}\mathrm{\lambda }\left(3\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{k}}\right)\text{}=3\mathrm{\lambda }\stackrel{^}{\mathrm{i}}\text{}+\text{}\mathrm{\lambda }\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{d}}\text{}=\stackrel{\to }{\mathrm{a}}\text{}-\text{}\stackrel{\to }{\mathrm{p}}\\ =\text{}5\stackrel{^}{\mathrm{i}}\text{}-2\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}\text{}-3\mathrm{\lambda }\stackrel{^}{\mathrm{i}}\text{}+\text{}\mathrm{\lambda }\stackrel{^}{\mathrm{k}}\\ =\text{}\left(5\text{}-3\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-2\stackrel{^}{\mathrm{j}}\text{}+\text{}\left(5\text{}-\mathrm{\lambda }\right)\text{}\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{d}}.\stackrel{\to }{\mathrm{b}}\text{}=\text{}0\text{}\left[\text{}\stackrel{\to }{\mathrm{d}}\text{}\perp \text{}\stackrel{\to }{\mathrm{b}}\right]\\ ⇒\text{}\left[\left(5\text{}-3\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-2\stackrel{^}{\mathrm{j}}\text{}+\text{}\left(5\text{}-\mathrm{\lambda }\right)\text{}\stackrel{^}{\mathrm{k}}\right].\left(3\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{k}}\right)\text{}=\text{}0\\ ⇒\text{}\left(5\text{}-3\mathrm{\lambda }\right)3+\left(5\text{}-\mathrm{\lambda }\right).1\text{}=\text{}0\\ ⇒\text{}15\text{}-9\mathrm{\lambda }+5\text{\hspace{0.17em}}-\mathrm{\lambda }\text{}=\text{}0\\ ⇒\text{}20\text{}-10\mathrm{\lambda }\text{}=\text{}0\\ ⇒\text{}\mathrm{\lambda }\text{}=\text{}2\\ \mathrm{Hence},\\ \stackrel{\to }{\mathrm{p}}\text{}=\text{}3\mathrm{\lambda }\stackrel{^}{\mathrm{i}}\text{}+\text{}\mathrm{\lambda }\stackrel{^}{\mathrm{k}}=6\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{d}}\text{}=\text{}\left(5\text{}-3\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-2\stackrel{^}{\mathrm{j}}\text{}+\text{}\left(5\text{}-\mathrm{\lambda }\right)\text{}\stackrel{^}{\mathrm{k}}\\ =\left(5\text{}-6\right)\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-2\stackrel{^}{\mathrm{j}}\text{}+\text{}\left(5\text{}-2\right)\text{}\stackrel{^}{\mathrm{k}}\\ =\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-2\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\end{array}$

Q.3

$\mathrm{Find}\text{}\mathrm{unit}\text{}\mathrm{vector}\text{}\mathrm{in}\text{}\mathrm{the}\text{}\mathrm{direction}\text{}\mathrm{of}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{a}}\text{}=\stackrel{^}{\mathrm{i}}\text{}+2\stackrel{^}{\mathrm{j}}\text{}+\text{}\stackrel{^}{\mathrm{k}}.$

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{unit}\text{}\mathrm{vector}\text{}\mathrm{in}\text{}\mathrm{the}\text{}\mathrm{direction}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{a}}\text{}=\frac{1}{\left|\stackrel{\to }{\mathrm{a}}\right|}\stackrel{\to }{\mathrm{a}}.\\ \mathrm{Now},\text{}\left|\stackrel{\to }{\mathrm{a}}\right|\text{}=\text{}\sqrt{{1}^{2}\text{}+\text{}{2}^{2\text{}}+\text{}{1}^{2}}\text{}=\text{}\sqrt{6}\\ \mathrm{Therefore},\text{}\stackrel{^}{\mathrm{a}}\text{}=\text{}\frac{1}{\sqrt{6}}\text{}\left(\stackrel{^}{\mathrm{i}}\text{}+2\stackrel{^}{\mathrm{j}}\text{}+\text{}\stackrel{^}{\mathrm{k}}\right)\end{array}$

Q.4 Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are collinear.

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{given}\text{}\mathrm{points}\text{}\mathrm{are}\text{}\mathrm{A}\left(1,\text{}2,\text{}7\right),\text{}\mathrm{B}\left(2,\text{}6,\text{}3\right)\text{}\mathrm{and}\text{}\mathrm{C}\text{}\left(3,\text{}10,\text{}-1\right).\\ \therefore \text{}\stackrel{\to }{\mathrm{AB}}\text{}=\text{}\left(2-1\right)\stackrel{^}{\mathrm{i}}\text{}+\text{}\left(6-2\right)\stackrel{^}{\mathrm{j}}\text{}+\text{}\left(3-7\right)\stackrel{^}{\mathrm{k}}\text{}=\text{}\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}-\text{\hspace{0.17em}}4\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{BC}}\text{}=\text{}\left(3-2\right)\stackrel{^}{\mathrm{i}}\text{}+\text{}\left(10-6\right)\stackrel{^}{\mathrm{j}}\text{}+\text{}\left(-1-3\right)\stackrel{^}{\mathrm{k}}\text{}=\text{}\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}-\text{\hspace{0.17em}}4\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{AC}}\text{}=\text{}\left(3-1\right)\stackrel{^}{\mathrm{i}}\text{}+\text{}\left(10-2\right)\stackrel{^}{\mathrm{j}}\text{}+\text{}\left(-1-7\right)\stackrel{^}{\mathrm{k}}\text{}=\text{}2\stackrel{^}{\mathrm{i}}\text{}+\text{}8\stackrel{^}{\mathrm{j}}\text{}-\text{\hspace{0.17em}}8\stackrel{^}{\mathrm{k}}\\ \left|\stackrel{\to }{\mathrm{AB}}\right|\text{}=\text{}\sqrt{{1}^{2}\text{}+\text{}{4}^{2}\text{}+\text{}{\left(-4\right)}^{2}}\text{}=\text{}\sqrt{1+16+16}\text{}=\text{}\sqrt{33}\\ \left|\stackrel{\to }{\mathrm{BC}}\right|\text{}=\text{}\sqrt{{1}^{2}\text{}+\text{}{4}^{2}\text{}+\text{}{\left(-4\right)}^{2}}\text{}=\text{}\sqrt{1+16+16}\text{}=\text{}\sqrt{33}\\ \left|\stackrel{\to }{\mathrm{AC}}\right|\text{}=\text{}\sqrt{{2}^{2}\text{}+\text{}{8}^{2}\text{}+\text{}{\left(-8\right)}^{2}}\text{}=\text{}\sqrt{4+64+64}\text{}=\text{}\sqrt{132}\text{}=\text{}2\sqrt{33}\\ \therefore \text{}\left|\stackrel{\to }{\mathrm{AC}}\right|\text{}=\text{}\left|\stackrel{\to }{\mathrm{AB}}\right|\text{}+\text{}\left|\stackrel{\to }{\mathrm{BC}}\right|\text{}\\ \mathrm{Hence},\text{}\mathrm{the}\text{}\mathrm{given}\text{}\mathrm{points}\text{}\mathrm{A},\text{}\mathrm{B}\text{}\mathrm{and}\text{}\mathrm{C}\text{}\mathrm{are}\text{}\mathrm{collinear}.\end{array}$

Q.5 If A, B, C have position vectors (2, 0, 0), (0, 1, 0) and (0, 0, 2), show that ΔABC is isosceles.

Ans

$\begin{array}{l}\stackrel{\to }{\mathrm{AB}}\text{}=\text{}\mathrm{Position}\text{}\mathrm{vector}\text{}\mathrm{of}\text{}\mathrm{B}\text{}-\text{}\mathrm{Position}\text{}\mathrm{vector}\text{}\mathrm{of}\text{}\mathrm{A}\\ ⇒\text{}\stackrel{\to }{\mathrm{AB}}\text{}=\text{}\left(0\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{\hspace{0.17em}\hspace{0.17em}}0\stackrel{^}{\mathrm{k}}\right)-\text{}\left(2\stackrel{^}{\mathrm{i}}\text{}+\text{}0\stackrel{^}{\mathrm{j}}\text{}+\text{}0\stackrel{^}{\mathrm{k}}\right)\text{}=\text{}-2\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{\hspace{0.17em}}0\stackrel{^}{\mathrm{k}}\\ ⇒\text{}\mathrm{AB}\text{}=\text{}\left|\stackrel{\to }{\mathrm{AB}}\right|\text{}=\text{}\sqrt{{\left(-2\right)}^{2}\text{}+\text{}{1}^{2}\text{}+\text{}{0}^{2}}\text{}=\text{}\sqrt{5}\\ \stackrel{\to }{\mathrm{BC}}\text{}=\text{}\mathrm{Position}\text{}\mathrm{vector}\text{}\mathrm{of}\text{}\mathrm{C}\text{}-\text{}\mathrm{Position}\text{}\mathrm{vector}\text{}\mathrm{of}\text{}\mathrm{B}\\ ⇒\text{}\stackrel{\to }{\mathrm{BC}}\text{}=\text{}\left(0\stackrel{^}{\mathrm{i}}\text{}+\text{}0\stackrel{^}{\mathrm{j}}\text{}+\text{\hspace{0.17em}\hspace{0.17em}}2\stackrel{^}{\mathrm{k}}\right)-\text{}\left(0\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{}0\stackrel{^}{\mathrm{k}}\right)\text{}=\text{}0\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{\hspace{0.17em}}2\stackrel{^}{\mathrm{k}}\\ ⇒\text{}\mathrm{BC}\text{}=\text{}\left|\stackrel{\to }{\mathrm{BC}}\right|\text{}=\text{}\sqrt{{\left(0\right)}^{2}\text{}+\text{}{\left(-1\right)}^{2}\text{}+\text{}{2}^{2}}\text{}=\text{}\sqrt{5}\\ \mathrm{So},\text{}\mathrm{AB}\text{}=\text{}\mathrm{BC}\\ \mathrm{Hence},\text{}\mathrm{\Delta ABC}\text{}\mathrm{is}\text{}\mathrm{an}\text{}\mathrm{isosceles}\text{}\mathrm{triangle}.\end{array}$

Q.6

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{scalar}\text{}\mathrm{projection}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{a}}\text{}=\stackrel{^}{\mathrm{i}}\text{}-\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\text{}\mathrm{on}\text{}\mathrm{the}\\ \mathrm{vector}\text{}\stackrel{\to }{\mathrm{b}}\text{}=4\stackrel{^}{\mathrm{i}}\text{}-\text{}4\stackrel{^}{\mathrm{j}}\text{}+\text{\hspace{0.17em}}7\stackrel{^}{\mathrm{k}}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Scalar}\text{}\mathrm{projection}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{on}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\text{}\mathrm{given}\text{}\mathrm{by}\\ \frac{\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}}{\left|\stackrel{\to }{\mathrm{b}}\right|}=\text{}\frac{\left(\stackrel{^}{\mathrm{i}}\text{}-\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right).\left(4\stackrel{^}{\mathrm{i}}\text{}-\text{}4\stackrel{^}{\mathrm{j}}\text{}+\text{\hspace{0.17em}}7\stackrel{^}{\mathrm{k}}\right)}{\sqrt{{4}^{2}\text{}+\text{}{4}^{2}\text{}+\text{}{7}^{2}}}\\ =\text{}\frac{4+8+7}{\sqrt{16+16+49}}\\ =\text{}\frac{19}{\sqrt{81}}\text{}=\text{}\frac{19}{9}\end{array}$

Q.7 Prove that the angle in a semi-circle is a right angle.

Ans

Let O be the centre of the semi-circle and r be its radius.
Let C be any point on the semi-circle. $\begin{array}{l}\mathrm{Now},\text{}\\ \stackrel{\to }{\mathrm{CA}}.\text{}\stackrel{\to }{\mathrm{CB}}\text{}=\text{}\left(\stackrel{\to }{\mathrm{CO}}+\text{}\stackrel{\to }{\mathrm{OA}}\right).\text{}\left(\stackrel{\to }{\mathrm{CO}}+\text{}\stackrel{\to }{\mathrm{OB}}\right)\\ =\text{}\left(-\stackrel{\to }{\mathrm{c}}+\text{}\stackrel{\to }{\mathrm{a}}\right).\text{}\left(-\stackrel{\to }{\mathrm{c}}+\text{}\stackrel{\to }{\mathrm{b}}\right)\\ =\text{}\left(-\stackrel{\to }{\mathrm{c}}+\text{}\stackrel{\to }{\mathrm{a}}\right).\text{}\left(-\stackrel{\to }{\mathrm{c}}-\text{}\stackrel{\to }{\mathrm{a}}\right)\text{}\left[\stackrel{\to }{\mathrm{b}}=\text{}-\stackrel{\to }{\mathrm{a}}\text{}\right]\\ =\text{}{\left|-\stackrel{\to }{\mathrm{c}}\right|}^{2}\text{}-\text{}{\left|\stackrel{\to }{\mathrm{a}}\right|}^{2}\\ =\text{}{\mathrm{c}}^{2}\text{}-\text{}{\mathrm{a}}^{2}\\ =\text{}{\mathrm{r}}^{2}\text{}-\text{}{\mathrm{r}}^{2}\text{\hspace{0.17em}}\left[\mathrm{Radius}\text{}\mathrm{of}\text{}\mathrm{same}\text{}\mathrm{circle}\right]\\ =\text{}0\\ \therefore \text{}\stackrel{\to }{\mathrm{CA}}.\text{}\stackrel{\to }{\mathrm{CB}}\text{}=\text{}0\\ ⇒\text{}\stackrel{\to }{\mathrm{CA}}\text{}\perp \text{}\stackrel{\to }{\mathrm{CB}}\text{}\end{array}$

Q.8

$\begin{array}{l}\mathrm{If}\text{}\stackrel{\to }{\mathrm{a}},\text{}\stackrel{\to }{\mathrm{b}},\text{}\stackrel{\to }{\mathrm{c}}\text{}\mathrm{are}\text{}\mathrm{unit}\text{}\mathrm{vectors}\text{}\mathrm{such}\text{}\mathrm{that}\text{}\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}\text{}=\stackrel{\to }{0},\\ \mathrm{then}\text{}\mathrm{find}\text{}\mathrm{the}\text{}\mathrm{value}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}.\end{array}$

Ans

$\begin{array}{l}\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}\text{}=\stackrel{\to }{0}\\ \mathrm{or}\\ \left(\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}\right).\left(\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}\right)\text{}=\text{}0\\ ⇒\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{b}}\text{}+\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{c}}\text{}+\text{}2\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\text{}2\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\text{}2\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}\text{}=\text{}0\\ ⇒{\left|\stackrel{\to }{\mathrm{a}}\right|}^{2}+\text{}{\left|\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}+{\left|\stackrel{\to }{\mathrm{c}}\right|}^{2}\text{}+\text{}2\left(\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}\right)\text{}=\text{}0\\ ⇒1+\text{}1\text{}+1\text{}+\text{}2\left(\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}\right)\text{}=\text{}0\\ ⇒\text{}2\left(\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}\right)\text{}=\text{}-3\\ ⇒\text{}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}\text{}=\text{}-\frac{3}{2}\end{array}$

Q.9

$\mathrm{Find}\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|,\text{}\mathrm{if}\text{}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}\text{}+\text{}5\stackrel{^}{\mathrm{j}}\text{}-\text{}2\stackrel{^}{\mathrm{k}}.\text{}$

Ans

$\begin{array}{l}\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ 2\text{\hspace{0.17em}}1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\\ 3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\text{}-2\end{array}\right|\\ =\left(-2-15\right)\stackrel{^}{\mathrm{i}}\text{}-\left(-4-9\right)\stackrel{^}{\mathrm{j}}\text{}+\text{}\left(10-3\right)\stackrel{^}{\mathrm{k}}\\ =-17\stackrel{^}{\mathrm{i}}\text{}+13\stackrel{^}{\mathrm{j}}\text{}+\text{}7\stackrel{^}{\mathrm{k}}\\ \left|\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}\sqrt{{17}^{2}\text{}+\text{}{13}^{2}\text{}+\text{}{7}^{2}}=\text{}\sqrt{507}\end{array}$

Q.10

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{parallelogram}\text{}\mathrm{whose}\text{}\mathrm{adjacent}\text{}\mathrm{sides}\text{}\mathrm{are}\text{}\mathrm{given}\\ \mathrm{by}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{}\stackrel{^}{\mathrm{k}}.\text{}\end{array}$

Ans

$\begin{array}{l}\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{parallelogram}\\ \stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ 3\text{\hspace{0.17em}}1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\\ 1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-1\text{\hspace{0.17em}}1\end{array}\right|\\ =\left(1+4\right)\stackrel{^}{\mathrm{i}}\text{}-\left(3-4\right)\stackrel{^}{\mathrm{j}}\text{}+\text{}\left(-3-1\right)\stackrel{^}{\mathrm{k}}\\ =5\stackrel{^}{\mathrm{i}}\text{}+\stackrel{^}{\mathrm{j}}\text{}-\text{}4\stackrel{^}{\mathrm{k}}\\ \left|\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}\sqrt{{5}^{2}\text{}+\text{}{1}^{2}\text{}+\text{}{4}^{2}}=\text{}\sqrt{42}\end{array}$

Q.11

$\mathrm{If}\text{}\stackrel{^}{\mathrm{a}}\text{}\mathrm{is}\text{}\mathrm{a}\text{}\mathrm{unit}\text{}\mathrm{vector}\text{}\mathrm{and}\text{}\left(\stackrel{\to }{\mathrm{x}}\text{}-\text{}\stackrel{^}{\mathrm{a}}\right).\text{}\left(\stackrel{\to }{\mathrm{x}}\text{}+\text{}\stackrel{^}{\mathrm{a}}\right)=8,\text{}\mathrm{find}\text{}\left|\stackrel{\to }{\mathrm{x}}\right|.$

Ans

$\begin{array}{l}\mathrm{Since}\text{}\stackrel{^}{\mathrm{a}}\text{}\mathrm{is}\text{}\mathrm{a}\text{}\mathrm{unit}\text{}\mathrm{vector}\text{}\mathrm{so}\text{}\left|\stackrel{^}{\mathrm{a}}\right|=1\\ \left(\stackrel{\to }{\mathrm{x}}\text{}-\text{}\stackrel{^}{\mathrm{a}}\right).\text{}\left(\stackrel{\to }{\mathrm{x}}\text{}+\text{}\stackrel{^}{\mathrm{a}}\right)=8\\ \stackrel{\to }{\mathrm{x}}.\stackrel{\to }{\mathrm{x}}\text{}+\text{}\stackrel{\to }{\mathrm{x}}.\stackrel{^}{\mathrm{a}}\text{}-\text{}\stackrel{^}{\mathrm{a}}.\stackrel{\to }{\mathrm{x}}\text{}-\text{}\stackrel{^}{\mathrm{a}}.\stackrel{^}{\mathrm{a}}\text{}=\text{}8\\ {\left|\stackrel{\to }{\mathrm{x}}\right|}^{2}\text{}-1\text{}=8\\ ⇒\text{}\left|\stackrel{\to }{\mathrm{x}}\right|\text{}=3\end{array}$

Q.12

$\mathrm{Find}\text{}\left|\stackrel{\to }{\mathrm{a}}\right|\text{}\mathrm{and}\text{}\left|\stackrel{\to }{\mathrm{b}}\right|,\text{}\mathrm{if}\text{}\left(\stackrel{\to }{\mathrm{a}}\text{}-\text{}\stackrel{\to }{\mathrm{b}}\right).\text{}\left(\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right)=27\text{}\mathrm{and}\text{}\left|\stackrel{\to }{\mathrm{a}}\right|=\text{}2\left|\stackrel{\to }{\mathrm{b}}\right|.$

Ans

$\begin{array}{l}\left(\stackrel{\to }{\mathrm{a}}\text{}-\text{}\stackrel{\to }{\mathrm{b}}\right).\text{}\left(\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right)=27\\ ⇒\text{}{\left|\stackrel{\to }{\mathrm{a}}\right|}^{2}-\text{}{\left|\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}=\text{}27\\ ⇒\text{}4{\left|\stackrel{\to }{\mathrm{b}}\right|}^{2}-\text{}{\left|\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}=\text{}27\\ ⇒\text{}3\text{}{\left|\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}=\text{}27\\ ⇒\text{}{\left|\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}=\text{}9\\ ⇒\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}3\\ \mathrm{Now},\text{}\left|\stackrel{\to }{\mathrm{a}}\right|\text{}=\text{}2\left|\stackrel{\to }{\mathrm{b}}\right|\\ ⇒\text{}\left|\stackrel{\to }{\mathrm{a}}\right|\text{}=\text{}6\\ \mathrm{Thus},\text{}\left|\stackrel{\to }{\mathrm{a}}\right|\text{}=\text{}6\text{}\mathrm{and}\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}3\end{array}$

Q.13

$\begin{array}{l}\mathrm{The}\text{}\mathrm{two}\text{}\mathrm{adjacent}\text{}\mathrm{sides}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{parallelogram}\text{}\mathrm{are}\text{}2\stackrel{^}{\mathrm{i}}\text{}-4\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}\\ \mathrm{and}\text{}\stackrel{^}{\mathrm{i}}\text{}-2\stackrel{^}{\mathrm{j}}\text{}-\text{}3\stackrel{^}{\mathrm{k}}.\text{}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{unit}\text{}\mathrm{vector}\text{}\mathrm{parallel}\text{}\mathrm{to}\text{}\mathrm{its}\text{}\mathrm{diagonal}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Adjacent}\text{}\mathrm{sides}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{parallelogram}\text{}\mathrm{are}\text{}\mathrm{given}\text{}\mathrm{as}:\\ \stackrel{\to }{\mathrm{a}}\text{}=\text{}2\stackrel{^}{\mathrm{i}}\text{}-4\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\stackrel{^}{\mathrm{i}}\text{}-2\stackrel{^}{\mathrm{j}}\text{}-\text{}3\stackrel{^}{\mathrm{k}}\\ \mathrm{Then},\text{}\mathrm{the}\text{}\mathrm{diagonal}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{parallelogram}\text{}\mathrm{is}\text{}\mathrm{given}\text{}\mathrm{by}\text{}\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\\ \stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left(2+1\right)\stackrel{^}{\mathrm{i}}\text{}+\left(-4-2\right)\stackrel{^}{\mathrm{j}}\text{}+\text{}\left(5-3\right)\stackrel{^}{\mathrm{k}}\text{}=\text{}3\stackrel{^}{\mathrm{i}}\text{}-6\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}\\ \mathrm{Thus},\text{}\mathrm{the}\text{}\mathrm{unit}\text{}\mathrm{vector}\text{}\mathrm{parallel}\text{}\mathrm{to}\text{}\mathrm{the}\text{}\mathrm{diagonal}\text{}\mathrm{is}\\ \frac{\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}}{\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|}\text{}=\text{}\frac{3\stackrel{^}{\mathrm{i}}\text{}-6\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}}{\sqrt{{3}^{2}\text{}+{\left(-6\right)}^{2}\text{}+\text{}{2}^{2}}}\\ =\text{}\frac{3\stackrel{^}{\mathrm{i}}\text{}-6\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}}{\sqrt{9+36+4}}\\ =\text{}\frac{3\stackrel{^}{\mathrm{i}}\text{}-6\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}}{7}\\ =\text{}\frac{3}{7}\stackrel{^}{\mathrm{i}}\text{}-\frac{6}{7}\stackrel{^}{\mathrm{j}}\text{}+\text{}\frac{2}{7}\stackrel{^}{\mathrm{k}}\end{array}$

Q.14

$\mathrm{If}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{are}\text{}\mathrm{any}\text{}\mathrm{vectors},\text{}\mathrm{then}\text{}\mathrm{prove}\text{}\mathrm{that}\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|\le \text{}\left|\stackrel{\to }{\mathrm{a}}\right|+\left|\stackrel{\to }{\mathrm{b}}\right|.$

Ans

$\begin{array}{l}{\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}=\text{}\left(\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right).\text{}\left(\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right)=\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{b}}\\ =\text{}{\left|\stackrel{\to }{\mathrm{a}}\right|}^{2}\text{}+\text{}2\left|\stackrel{\to }{\mathrm{a}}\right|\left|\stackrel{\to }{\mathrm{b}}\right|\text{}\mathrm{cos\theta }\text{}+\text{}{\left|\stackrel{\to }{\mathrm{b}}\right|}^{2}\\ \le \text{}{\left|\stackrel{\to }{\mathrm{a}}\right|}^{2}\text{}+\text{}2\left|\stackrel{\to }{\mathrm{a}}\right|\left|\stackrel{\to }{\mathrm{b}}\right|\text{}+\text{}{\left|\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}=\text{}{\left(\left|\stackrel{\to }{\mathrm{a}}\right|\text{}+\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\right)}^{2}\\ ⇒\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|\text{}\le \text{}\left|\stackrel{\to }{\mathrm{a}}\right|\text{}+\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\end{array}$

Q.15

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{scalar}\text{}\mathrm{product}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{vectors}\text{}\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}-8\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\\ -3\stackrel{^}{\mathrm{i}}\text{}-\text{}5\stackrel{^}{\mathrm{j}}\text{}+4\stackrel{^}{\mathrm{k}}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{}\stackrel{\to }{\mathrm{a}}\text{}=\text{}\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}-8\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}=-3\stackrel{^}{\mathrm{i}}\text{}-\text{}5\stackrel{^}{\mathrm{j}}\text{}+4\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}=\left(\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}-8\stackrel{^}{\mathrm{k}}\right).\left(-3\stackrel{^}{\mathrm{i}}\text{}-\text{}5\stackrel{^}{\mathrm{j}}\text{}+4\stackrel{^}{\mathrm{k}}\right)\\ =\text{}-3-15-32\text{}=\text{}-50\end{array}$

Q.16

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{value}\text{}\mathrm{of}\text{}\mathrm{\lambda }\text{}\mathrm{such}\text{}\mathrm{that}\text{}\mathrm{the}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{are}\text{}\\ \mathrm{penpendicular}\text{}\mathrm{where}\text{}\stackrel{\to }{\mathrm{a}}=\mathrm{\lambda }\text{}\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}\text{}+\stackrel{^}{\mathrm{k}},\text{}\stackrel{\to }{\mathrm{b}}=\text{}4\stackrel{^}{\mathrm{i}}\text{}-\text{}9\stackrel{^}{\mathrm{j}}\text{}+2\stackrel{^}{\mathrm{k}}.\end{array}$

Ans

$\begin{array}{l}\stackrel{\to }{\mathrm{a}}\text{}\perp \text{}\stackrel{\to }{\mathrm{b}}\text{}⇒\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}=\text{}0\\ ⇒\left(\mathrm{\lambda }\text{}\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}\text{}+\stackrel{^}{\mathrm{k}}\right).\left(4\stackrel{^}{\mathrm{i}}\text{}-\text{}9\stackrel{^}{\mathrm{j}}\text{}+2\stackrel{^}{\mathrm{k}}\right)\text{}=\text{}0\\ ⇒\text{}4\mathrm{\lambda }\text{}-\text{}18\text{}+\text{}2\text{}=\text{}0\\ ⇒\text{}4\mathrm{\lambda }\text{}-\text{}16=\text{}0\\ ⇒\text{}4\mathrm{\lambda }\text{}=\text{}16\\ ⇒\text{}\mathrm{\lambda }\text{}=\text{}4\end{array}$

Q.17 Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) as its vertices.

Ans

$\begin{array}{l}\mathrm{Here}\text{}\stackrel{\to }{\mathrm{AB}}\text{}=\text{}\stackrel{^}{\mathrm{j}}\text{}+2\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{AC}}\text{}=\text{}\stackrel{^}{\mathrm{i}}\text{}+2\stackrel{^}{\mathrm{j}}.\\ \mathrm{The}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{given}\text{}\mathrm{triangle}\text{}\mathrm{is}\text{}\frac{1}{2}\text{}\left|\stackrel{\to }{\mathrm{AB}}\text{}×\text{}\stackrel{\to }{\mathrm{AC}}\right|\\ \stackrel{\to }{\mathrm{AB}}\text{}×\text{}\stackrel{\to }{\mathrm{AC}}\text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ 0\text{\hspace{0.17em}}1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\\ 1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0\end{array}\right|\\ =-4\stackrel{^}{\mathrm{i}}\text{}-\left(0-2\right)\stackrel{^}{\mathrm{j}}\text{}+\text{}\left(0-1\right)\stackrel{^}{\mathrm{k}}\\ =-4\stackrel{^}{\mathrm{i}}\text{}+2\stackrel{^}{\mathrm{j}}\text{}-\text{}\stackrel{^}{\mathrm{k}}\\ \left|\stackrel{\to }{\mathrm{AB}}\text{}×\text{}\stackrel{\to }{\mathrm{AC}}\right|\text{}=\text{}\sqrt{{4}^{2}\text{}+\text{}{2}^{2\text{}}+\text{}{1}^{2}}\text{}=\text{}\sqrt{21}\\ \therefore \text{}\frac{1}{2}\text{}\left|\stackrel{\to }{\mathrm{AB}}\text{}×\text{}\stackrel{\to }{\mathrm{AC}}\right|\text{}=\text{}\frac{\sqrt{21}}{2}\end{array}$

Q.18

$\mathrm{Prove}\text{}\mathrm{that}\text{}\left[\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}},\text{}\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}},\text{}\stackrel{\to }{\mathrm{c}}+\stackrel{\to }{\mathrm{a}}\right]\text{}=\text{}2\text{}\left[\stackrel{\to }{\mathrm{a}},\text{}\stackrel{\to }{\mathrm{b}},\text{}\stackrel{\to }{\mathrm{c}}\right].$

Ans

$\begin{array}{l}\mathrm{LHS}\text{}=\text{}\left[\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}},\text{}\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}},\text{}\stackrel{\to }{\mathrm{c}}+\stackrel{\to }{\mathrm{a}}\right]\\ =\text{}\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right).\text{}\left[\left(\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right)\text{}×\left(\stackrel{\to }{\mathrm{c}}+\stackrel{\to }{\mathrm{a}}\right)\right]\\ =\text{}\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right).\text{}\left[\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}×\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}×\stackrel{\to }{\mathrm{a}}\right]\\ =\text{}\left(\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}\right).\text{}\left[\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}×\stackrel{\to }{\mathrm{a}}\right]\text{}\left[\text{}\stackrel{\to }{\mathrm{c}}×\stackrel{\to }{\mathrm{c}}\text{}=0\right]\\ =\text{}\stackrel{\to }{\mathrm{a}}.\left(\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}×\stackrel{\to }{\mathrm{a}}\right)\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\left(\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}×\stackrel{\to }{\mathrm{a}}\right)\\ =\text{}\stackrel{\to }{\mathrm{a}}.\left(\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{c}}\right)\text{}+\text{}\stackrel{\to }{\mathrm{a}}.\left(\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{a}}\right)\text{}+\text{}\stackrel{\to }{\mathrm{a}}.\left(\stackrel{\to }{\mathrm{c}}×\stackrel{\to }{\mathrm{a}}\right)\text{}+\stackrel{\to }{\mathrm{b}}.\left(\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{c}}\right)\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\left(\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{a}}\right)\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\left(\stackrel{\to }{\mathrm{c}}×\stackrel{\to }{\mathrm{a}}\right)\\ =\text{}\stackrel{\to }{\mathrm{a}}.\left(\stackrel{\to }{\mathrm{b}}×\stackrel{\to }{\mathrm{c}}\right)\text{}+\text{}0\text{}+\text{}0\text{}+0\text{}+\text{}0\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\left(\stackrel{\to }{\mathrm{c}}×\stackrel{\to }{\mathrm{a}}\right)\\ =\text{}2\left[\stackrel{\to }{\mathrm{a}},\text{}\stackrel{\to }{\mathrm{b}},\text{}\stackrel{\to }{\mathrm{c}}\right]\text{}=\text{}\mathrm{RHS}\end{array}$

Q.19

$\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{unit}\text{}\mathrm{vector}\text{}\mathrm{in}\text{}\mathrm{the}\text{}\mathrm{direction}\text{}\mathrm{of}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}.$

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{unit}\text{}\mathrm{vector}\text{}\mathrm{in}\text{}\mathrm{the}\text{}\mathrm{direction}\text{}\mathrm{of}\text{}\mathrm{vector},\\ \stackrel{^}{\mathrm{a}}=\frac{\stackrel{\to }{\mathrm{a}}}{\left|\stackrel{\to }{\mathrm{a}}\right|}\\ =\frac{2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}}{\sqrt{4+9+3}}\\ =\frac{2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}}{\sqrt{16}}\\ =\frac{1}{4}\left(2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}\right)\\ =\frac{1}{2}\stackrel{^}{\mathrm{i}}\text{}+\text{}\frac{3}{4}\stackrel{^}{\mathrm{j}}\text{}+\text{}\frac{\sqrt{3}}{4}\stackrel{^}{\mathrm{k}}\end{array}$

Q.20

$\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{direction}\text{}\mathrm{cosine}\text{}\mathrm{of}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}.$

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{unit}\text{}\mathrm{vector}\text{}\mathrm{in}\text{}\mathrm{the}\text{}\mathrm{direction}\text{}\mathrm{of}\text{}\mathrm{vector},\\ \stackrel{^}{\mathrm{a}}=\frac{\stackrel{\to }{\mathrm{a}}}{\left|\stackrel{\to }{\mathrm{a}}\right|}\\ =\frac{2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}}{\sqrt{4+9+3}}\\ =\frac{2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}}{\sqrt{16}}\\ =\frac{1}{4}\left(2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}\right)\\ =\frac{1}{2}\stackrel{^}{\mathrm{i}}\text{}+\text{}\frac{3}{4}\stackrel{^}{\mathrm{j}}\text{}+\text{}\frac{\sqrt{3}}{4}\stackrel{^}{\mathrm{k}}\\ \mathrm{Therefore}\text{}\mathrm{the}\text{}\mathrm{direction}\text{}\mathrm{cosines}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{given}\text{}\mathrm{vectors}\text{}\mathrm{are}\\ \frac{1}{2},\text{}\frac{3}{4},\text{}\frac{\sqrt{3}}{4}\end{array}$

Q.21

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{scalar}\text{}\mathrm{product}\text{}\mathrm{of}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}8\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\\ \stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}\text{}-\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\\ \stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}8\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\\ \stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}\text{}-\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{scalar}\text{}\mathrm{prodct}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\\ \stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left(2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}8\stackrel{^}{\mathrm{k}}\right).\left(3\stackrel{^}{\mathrm{i}}\text{}-\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}\right)\\ =\text{}6-6+16\\ =\text{}16\end{array}$

Q.22

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{value}\text{}\mathrm{of}\text{}\mathrm{\lambda },\text{}\mathrm{so}\text{}\mathrm{that}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}\text{}+\text{}\mathrm{\lambda }\stackrel{^}{\mathrm{j}}\text{}+\text{}\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\\ \stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{}-\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\text{}\mathrm{are}\text{}\mathrm{perpendicular}\text{}\mathrm{to}\text{}\mathrm{each}\text{}\mathrm{other}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Since}\text{}\mathrm{the}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{are}\text{}\mathrm{perpendicular}\text{}\mathrm{to}\text{}\mathrm{each}\text{}\mathrm{other},\\ \therefore \text{}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}=\text{}0\\ ⇒\left(2\stackrel{^}{\mathrm{i}}\text{}+\text{}\mathrm{\lambda }\stackrel{^}{\mathrm{j}}\text{}+\text{}\stackrel{^}{\mathrm{k}}\right).\left(\stackrel{^}{\mathrm{i}}\text{}-\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\right)=0\\ ⇒\text{}2\text{}-\text{}2\mathrm{\lambda }\text{}+\text{}3=0\\ ⇒\text{}5\text{}-\text{}2\mathrm{\lambda }\text{}=0\\ ⇒\text{}2\mathrm{\lambda }\text{}=5\\ ⇒\text{}\mathrm{\lambda }\text{}=\frac{5}{2}\end{array}$

Q.23

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{vector}\text{}\mathrm{product}\text{}\mathrm{of}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\\ \stackrel{\to }{\mathrm{b}}=4\stackrel{^}{\mathrm{i}}\text{}+\text{}6\stackrel{^}{\mathrm{j}}\text{}+\text{}8\stackrel{^}{\mathrm{k}}.\text{}\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\\ \stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\\ \stackrel{\to }{\mathrm{b}}=4\stackrel{^}{\mathrm{i}}\text{}+\text{}6\stackrel{^}{\mathrm{j}}\text{}+\text{}8\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{vector}\text{}\mathrm{product}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\\ \stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ 2\text{\hspace{0.17em}}3\text{}4\\ 4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6\text{}8\end{array}\right|\\ =\stackrel{^}{\mathrm{i}}\left(24-24\right)\text{}-\stackrel{^}{\mathrm{j}}\left(16-16\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(12-12\right)\\ =\stackrel{^}{\mathrm{i}}\left(0\right)\text{}-\stackrel{^}{\mathrm{j}}\left(0\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(0\right)\\ =\text{}\stackrel{\to }{0}\end{array}$

Q.24

$\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\mathrm{the}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}=4\stackrel{^}{\mathrm{i}}\text{}+\text{}\sqrt{5}\stackrel{^}{\mathrm{j}}\text{}-\text{}2\stackrel{^}{\mathrm{k}}.\text{}$

Ans

$\begin{array}{l}\mathrm{Let}\text{}\mathrm{\theta }\text{}\mathrm{be}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}},\text{}\mathrm{then}\\ \mathrm{cos\theta }\text{}=\text{}\frac{\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}}{\left|\stackrel{\to }{\mathrm{a}}\right|\left|\stackrel{\to }{\mathrm{b}}\right|}\text{}\\ =\frac{\left(3\stackrel{^}{\mathrm{i}}\text{}+\text{}0\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}\right).\left(4\stackrel{^}{\mathrm{i}}\text{}+\text{}\sqrt{5}\stackrel{^}{\mathrm{j}}\text{}-\text{}2\stackrel{^}{\mathrm{k}}\right)}{\sqrt{9+0+16}\sqrt{16+5+4}}\\ =\frac{12+0-8}{\sqrt{25}\sqrt{25}}\\ =\frac{4}{5×5}\\ ⇒\text{}\mathrm{cos\theta }\text{}=\text{}\frac{4}{25}\\ ⇒\text{}\mathrm{\theta }\text{}=\text{}{\mathrm{cos}}^{-1}\text{}\left(\frac{4}{25}\right)\end{array}$

Q.25

$\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\mathrm{the}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{if}\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}.$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\\ \left|\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\\ ⇒\text{}\left|\stackrel{\to }{\mathrm{a}}\right|\left|\stackrel{\to }{\mathrm{b}}\right|\text{}\mathrm{sin\theta }=\text{}\left|\stackrel{\to }{\mathrm{a}}\right|\left|\stackrel{\to }{\mathrm{b}}\right|\text{}\mathrm{cos\theta }\\ \mathrm{where}\text{}\mathrm{\theta }\text{}\mathrm{is}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}.\\ ⇒\text{}\mathrm{sin\theta }=\text{}\mathrm{cos\theta }\\ ⇒\text{}\mathrm{tan\theta }\text{}=\text{}1\\ ⇒\text{}\mathrm{tan\theta }\text{}=\text{}\mathrm{tan}{45}^{\mathrm{o}}\\ ⇒\text{}\mathrm{tan\theta }\text{}=\text{}\mathrm{tan}\frac{\mathrm{\pi }}{4}\\ ⇒\text{}\mathrm{\theta }\text{}=\text{}\frac{\mathrm{\pi }}{4}\end{array}$

Q.26

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{a}\text{}\mathrm{vector}\text{}\mathrm{in}\text{}\mathrm{the}\text{}\mathrm{direction}\text{}\mathrm{of}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{a}}\text{}=2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}},\\ \mathrm{which}\text{}\mathrm{has}\text{}\mathrm{a}\text{}\mathrm{magnitude}\text{}8\text{}\mathrm{units}.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{unit}\text{}\mathrm{vector}\text{}\mathrm{in}\text{}\mathrm{the}\text{}\mathrm{direction}\text{}\mathrm{of}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{is}\\ \stackrel{^}{\mathrm{a}}=\frac{\stackrel{\to }{\mathrm{a}}}{\left|\stackrel{\to }{\mathrm{a}}\right|}\\ =\frac{2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}}{\sqrt{4+9+3}}\\ =\frac{2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}}{\sqrt{16}}\\ =\frac{1}{4}\left(2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}\right)\\ \mathrm{The}\text{}\mathrm{required}\text{}\mathrm{vector}\text{}=\text{}8\text{}×\frac{1}{4}\text{}\left(2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}\right)\\ =2\text{}\left(2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{3}\stackrel{^}{\mathrm{k}}\right)\\ =\text{}\left(4\stackrel{^}{\mathrm{i}}\text{}+\text{}6\stackrel{^}{\mathrm{j}}\text{}+\text{}2\sqrt{3}\stackrel{^}{\mathrm{k}}\right)\end{array}$

Q.27

$\begin{array}{l}\mathrm{Prove}\text{}\mathrm{that}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}\text{}=7\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}+\text{}8\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}=12\stackrel{^}{\mathrm{i}}\text{}-\text{}25\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}\\ \mathrm{are}\text{}\mathrm{perpendicular}\text{}\mathrm{to}\text{}\mathrm{each}\text{}\mathrm{other}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\\ \stackrel{\to }{\mathrm{a}}=7\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}+\text{}8\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\\ \stackrel{\to }{\mathrm{b}}=12\stackrel{^}{\mathrm{i}}\text{}-25\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{scalar}\text{}\mathrm{product}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\\ \stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}=\left(7\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}+\text{}8\stackrel{^}{\mathrm{k}}\right).\left(12\stackrel{^}{\mathrm{i}}\text{}-25\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}\right)\\ =84-100+16\\ =\text{}100-100\\ =0\\ \text{}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}=\text{}0\\ \therefore \text{}\mathrm{The}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{are}\text{}\mathrm{perpendicular}\text{}\mathrm{to}\text{}\mathrm{each}\text{}\mathrm{other}.\end{array}$

Q.28

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\mathrm{the}\text{}\mathrm{two}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{with}\text{}\mathrm{magnitudes}\\ 1\text{}\mathrm{and}\text{}\frac{2}{\sqrt{3}}\text{}\mathrm{respectively},\text{}\mathrm{whose}\text{}\mathrm{dot}\text{}\mathrm{product}\text{}\mathrm{is}\text{}\mathrm{unity}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\text{}\left|\stackrel{\to }{\mathrm{a}}\right|\text{}=1,\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}\frac{2}{\sqrt{3}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}=\text{}1\\ \mathrm{Let}\text{}\mathrm{\theta }\text{}\mathrm{is}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}},\text{}\mathrm{then}\\ \mathrm{cos\theta }\text{}=\frac{\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}}{\left|\stackrel{\to }{\mathrm{a}}\right|\left|\stackrel{\to }{\mathrm{b}}\right|}\\ ⇒\mathrm{cos\theta }\text{}=\frac{1}{1\text{}×\frac{2}{\sqrt{3}}}\\ ⇒\mathrm{cos\theta }\text{}=\frac{\sqrt{3}}{2}\\ ⇒\mathrm{cos\theta }\text{}=\mathrm{cos}{30}^{\mathrm{o}}\\ ⇒\mathrm{cos\theta }\text{}=\mathrm{cos}\frac{\mathrm{\pi }}{6}\\ ⇒\mathrm{\theta }\text{}=\frac{\mathrm{\pi }}{6}\\ \therefore \text{}\mathrm{The}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\text{}\frac{\mathrm{\pi }}{6}.\end{array}$

Q.29

$\begin{array}{l}\mathrm{Prove}\text{}\mathrm{that}\text{}\mathrm{the}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}=\text{}2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\\ \stackrel{\to }{\mathrm{b}}\text{}=\text{}4\stackrel{^}{\mathrm{i}}\text{}+\text{}6\stackrel{^}{\mathrm{j}}\text{}+\text{}8\stackrel{^}{\mathrm{k}}\text{}\mathrm{are}\text{}\mathrm{parallel}\text{}\mathrm{to}\text{}\mathrm{each}\text{}\mathrm{other}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\\ \stackrel{\to }{\mathrm{a}}=\text{}2\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\\ \stackrel{\to }{\mathrm{b}}\text{}=\text{}4\stackrel{^}{\mathrm{i}}\text{}+\text{}6\stackrel{^}{\mathrm{j}}\text{}+\text{}8\stackrel{^}{\mathrm{k}}\text{}\\ \mathrm{The}\text{}\mathrm{vectors}\text{}\mathrm{product}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\\ \stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{}\stackrel{^}{\mathrm{j}}\text{}\stackrel{^}{\mathrm{k}}\\ 2\text{}3\text{}4\\ 4\text{}6\text{}8\end{array}\right|\\ =\text{}\stackrel{^}{\mathrm{i}}\left(24-24\right)\text{}-\text{}\stackrel{^}{\mathrm{j}}\left(16-16\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(12-12\right)\\ =\text{}\stackrel{^}{\mathrm{i}}\left(0\right)\text{}-\text{}\stackrel{^}{\mathrm{j}}\left(0\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(0\right)\\ =\stackrel{\to }{0}\\ \text{}\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\stackrel{\to }{0}\\ \therefore \text{}\mathrm{The}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{are}\text{}\mathrm{parallel}\text{}\mathrm{to}\text{}\mathrm{each}\text{}\mathrm{other}.\end{array}$

Q.30

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\mathrm{two}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{such}\text{}\mathrm{that}\text{}\mathrm{the}\text{}\\ \mathrm{magnitude}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{is}\text{}\mathrm{unity}\text{}\mathrm{and}\text{}\mathrm{their}\text{}\mathrm{dot}\text{}\mathrm{product}\text{}\mathrm{is}\text{}\mathrm{also}\text{}\mathrm{unity}.\\ \mathrm{The}\text{}\mathrm{magnitude}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\text{}\sqrt{2}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\text{}\left|\stackrel{\to }{\mathrm{a}}\right|=\text{}1,\text{}\left|\stackrel{\to }{\mathrm{b}}\right|=\text{}\sqrt{2}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}=1\\ \mathrm{Let}\text{}\mathrm{\theta }\text{}\mathrm{is}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}},\text{}\mathrm{then}\\ \mathrm{cos\theta }\text{}=\text{}\frac{\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}}{\left|\stackrel{\to }{\mathrm{a}}\right|\left|\stackrel{\to }{\mathrm{b}}\right|}\\ ⇒\text{}\mathrm{cos\theta }\text{}=\text{}\frac{1}{1\text{}×\text{}\sqrt{2}}\\ ⇒\text{}\mathrm{cos\theta }\text{}=\text{}\frac{1}{\sqrt{2}}\\ ⇒\text{}\mathrm{cos\theta }\text{}=\text{}\mathrm{cos}{45}^{\mathrm{o}}\\ ⇒\text{}\mathrm{cos\theta }\text{}=\text{}\mathrm{cos}\frac{\mathrm{\pi }}{4}\\ ⇒\text{}\mathrm{\theta }\text{}=\text{}\frac{\mathrm{\pi }}{4}\\ \therefore \text{}\mathrm{The}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\text{}\frac{\mathrm{\pi }}{4}.\end{array}$

Q.31

$\mathrm{Evaluate}\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}×\stackrel{\to }{\mathrm{b}}\right|,\text{}\mathrm{where}\text{}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}\text{}-\text{}3\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}=-\stackrel{^}{\mathrm{j}}\text{}+\text{}\stackrel{^}{\mathrm{k}}.$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}\text{}-\text{}3\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\\ \stackrel{\to }{\mathrm{b}}=-\stackrel{^}{\mathrm{j}}\text{}+\text{}\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{vector}\text{}\mathrm{product}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\\ \stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{}\stackrel{^}{\mathrm{j}}\text{}\stackrel{^}{\mathrm{k}}\\ 1\text{}2\text{}-3\\ 0\text{}-1\text{}1\end{array}\right|\\ =\text{}\stackrel{^}{\mathrm{i}}\left(2-3\right)-\text{}\stackrel{^}{\mathrm{j}}\left(1+0\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(-1\right)\\ =\text{}\stackrel{^}{\mathrm{i}}\left(-1\right)-\text{}\stackrel{^}{\mathrm{j}}\left(1\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(-1\right)\\ =\text{}-\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\\ ⇒\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}-\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\\ ⇒\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}\sqrt{1+1+1}\\ =\text{}\sqrt{3}\end{array}$

Q.32

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{projection}\text{}\mathrm{of}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{a}}=4\stackrel{^}{\mathrm{i}}\text{}-\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{2}\stackrel{^}{\mathrm{k}}\\ \mathrm{on}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{the}\text{}\mathrm{projection}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{on}\text{}\stackrel{\to }{\mathrm{b}}=\text{}\frac{\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}}{\left|\stackrel{\to }{\mathrm{b}}\right|}\\ \stackrel{\to }{\mathrm{a}}=4\stackrel{^}{\mathrm{i}}\text{}-\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{2}\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}\\ \text{}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left(4\stackrel{^}{\mathrm{i}}\text{}-\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}\sqrt{2}\stackrel{^}{\mathrm{k}}\right)\text{}.\text{}\left(3\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}\right)\\ =\text{}12-12+5\sqrt{2}\\ \mathrm{and}\text{}\left|\stackrel{\to }{\mathrm{b}}\right|=\sqrt{9+16+25}\\ =\text{}\sqrt{50}\text{}=\text{}5\sqrt{2}\\ ⇒\text{}\mathrm{The}\text{}\mathrm{projection}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{on}\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\frac{\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}}{\left|\stackrel{\to }{\mathrm{b}}\right|}\\ =\text{}\frac{5\sqrt{2}}{5\sqrt{2}}\text{}=\text{}1\end{array}$

Q.33

$\begin{array}{l}\mathrm{Let}\text{}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}},\text{}\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}\text{}-\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}7\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{c}}=2\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}.\\ \mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{d}}\text{}\mathrm{which}\text{}\mathrm{is}\text{}\mathrm{perpendicular}\text{}\mathrm{to}\text{}\mathrm{both}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\\ \mathrm{and}\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{d}}\text{}=\text{}18.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{cross}\text{}\mathrm{product}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\\ \stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{}\stackrel{^}{\text{}\mathrm{j}}\text{}\stackrel{^}{\text{}\mathrm{k}}\\ 1\text{}4\text{}2\\ 3\text{}-2\text{}7\end{array}\right|\\ =\stackrel{^}{\mathrm{i}}\left(28+4\right)\text{}-\stackrel{^}{\mathrm{j}}\left(7-6\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(-2-12\right)\\ =\stackrel{^}{\mathrm{i}}\left(32\right)\text{}-\stackrel{^}{\mathrm{j}}\left(1\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(-14\right)\\ =32\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\text{}-\text{}14\stackrel{^}{\mathrm{k}}\\ \mathrm{Let}\text{}\stackrel{\to }{\mathrm{d}}\text{}=\text{}\mathrm{\lambda }\text{}\left(32\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\text{}-\text{}14\stackrel{^}{\mathrm{k}}\right)\text{}...\left(1\right)\\ \left[\text{}\stackrel{\to }{\mathrm{d}}\text{}\mathrm{is}\text{}\mathrm{also}\text{}\mathrm{perpendicular}\text{}\mathrm{to}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\right]\\ \mathrm{Now},\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{d}}\text{}=\text{}18\\ ⇒\text{}\left(2\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}\right).\mathrm{\lambda }\text{}\left(32\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\text{}-\text{}14\stackrel{^}{\mathrm{k}}\right)=18\\ ⇒\text{}\mathrm{\lambda }\text{}\left(64+1-56\right)=18\\ ⇒9\mathrm{\lambda }\text{}=18\\ ⇒\mathrm{\lambda }\text{}=\text{}2\\ \mathrm{Putting}\text{}\mathrm{the}\text{}\mathrm{value}\text{}\mathrm{in}\text{}\left(1\right),\text{}\mathrm{we}\text{}\mathrm{get}\\ \stackrel{\to }{\mathrm{d}}=2\text{}\left(32\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\text{}-\text{}14\stackrel{^}{\mathrm{k}}\right)\\ ⇒\text{}\stackrel{\to }{\mathrm{d}}=\text{}\left(64\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}\text{}-\text{}28\stackrel{^}{\mathrm{k}}\right)\end{array}$

Q.34

$\begin{array}{l}\mathrm{Let}\text{}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}},\text{}\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}\text{}-\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}7\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{c}}=2\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}.\\ \mathrm{Find}\text{}\mathrm{a}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{d}}\text{}\mathrm{which}\text{}\mathrm{is}\text{}\mathrm{perpendicular}\text{}\mathrm{to}\text{}\mathrm{both}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\\ \mathrm{and}\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{d}}\text{}=\text{}15.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{cross}\text{}\mathrm{product}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\\ \stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ 1\text{}4\text{}2\\ 3\text{}-2\text{}7\end{array}\right|\\ =\stackrel{^}{\mathrm{i}}\left(28+4\right)\text{}-\stackrel{^}{\mathrm{j}}\left(7-6\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(-2-12\right)\\ =\stackrel{^}{\mathrm{i}}\left(32\right)\text{}-\stackrel{^}{\mathrm{j}}\left(1\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(-14\right)\\ =32\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\text{}-\text{}14\stackrel{^}{\mathrm{k}}\\ \mathrm{Let}\text{}\stackrel{\to }{\mathrm{d}}\text{}=\text{}\mathrm{\lambda }\text{}\left(32\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\text{}-\text{}14\stackrel{^}{\mathrm{k}}\right)\text{}...\left(1\right)\\ \left[\text{}\stackrel{\to }{\mathrm{d}}\text{}\mathrm{is}\text{}\mathrm{also}\text{}\mathrm{perpendicular}\text{}\mathrm{to}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\right]\\ \mathrm{Now},\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{d}}\text{}=\text{}15\\ ⇒\text{}\left(2\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}\right).\mathrm{\lambda }\text{}\left(32\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\text{}-\text{}14\stackrel{^}{\mathrm{k}}\right)=15\\ ⇒\text{}\mathrm{\lambda }\text{}\left(64+1-56\right)=15\\ ⇒9\mathrm{\lambda }\text{}=15\\ ⇒\mathrm{\lambda }\text{}=\text{}\frac{5}{3}\\ \mathrm{Putting}\text{}\mathrm{the}\text{}\mathrm{value}\text{}\mathrm{in}\text{}\left(1\right),\text{}\mathrm{we}\text{}\mathrm{get}\\ \stackrel{\to }{\mathrm{d}}=\frac{5}{3}\text{}\left(32\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\text{}-\text{}14\stackrel{^}{\mathrm{k}}\right)\end{array}$

Q.35

$\begin{array}{l}\mathrm{Let}\text{}\stackrel{\to }{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}.\text{}\mathrm{Find}\text{}\mathrm{a}\text{}\mathrm{vector}\text{}\mathrm{which}\text{}\mathrm{is}\text{}\\ \mathrm{perpendicular}\text{}\mathrm{to}\text{}\mathrm{both}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{and}\text{}\mathrm{having}\text{}\mathrm{a}\text{}\mathrm{magnitude}\text{}\mathrm{of}\text{}\sqrt{780}.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{vector}\text{}\mathrm{which}\text{}\mathrm{is}\text{}\mathrm{perpendicular}\text{}\mathrm{to}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\\ \stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ 3\text{\hspace{0.17em}}1\text{}2\\ 1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\text{}5\end{array}\right|\\ =\stackrel{^}{\mathrm{i}}\left(5-4\right)\text{}-\stackrel{^}{\mathrm{j}}\left(15-2\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(6-1\right)\\ =\stackrel{^}{\mathrm{i}}\text{}-13\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{unit}\text{}\mathrm{vector}\text{}\mathrm{in}\text{}\mathrm{the}\text{}\mathrm{direction}\text{}\mathrm{of}\text{}\left(\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right)\text{}=\text{}\frac{\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}}{\left|\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|}\\ =\text{}\frac{\stackrel{^}{\mathrm{i}}\text{}-13\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}}{\sqrt{1+169+25}}\\ =\text{}\frac{\stackrel{^}{\mathrm{i}}\text{}-13\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}}{\sqrt{195}}\\ \mathrm{The}\text{}\mathrm{required}\text{\hspace{0.17em}}\mathrm{vector}\text{}=\text{}\sqrt{780}×\text{}\frac{\left(\stackrel{^}{\mathrm{i}}\text{}-13\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}\right)}{\sqrt{195}}\\ =2\sqrt{195}\text{}×\text{}\frac{\left(\stackrel{^}{\mathrm{i}}\text{}-13\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}\right)}{\sqrt{195}}\\ =\text{}2\left(\stackrel{^}{\mathrm{i}}\text{}-13\stackrel{^}{\mathrm{j}}\text{}+\text{}5\stackrel{^}{\mathrm{k}}\right)\\ =\text{}2\stackrel{^}{\mathrm{i}}\text{}-26\stackrel{^}{\mathrm{j}}\text{}+\text{}10\stackrel{^}{\mathrm{k}}\end{array}$

Q.36

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{parallelogram}\text{}\mathrm{whose}\text{}\mathrm{adjacent}\text{}\mathrm{sides}\text{}\mathrm{are}\text{}\\ \mathrm{given}\text{}\mathrm{by}\text{}\mathrm{the}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{}\stackrel{^}{\mathrm{k}}.\text{}\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\\ \stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+\text{}\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{vector}\text{}\mathrm{which}\text{}\mathrm{is}\text{}\mathrm{perpendicular}\text{}\mathrm{to}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\\ \stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ 1\text{\hspace{0.17em}}2\text{}3\\ 1\text{\hspace{0.17em}\hspace{0.17em}}-1\text{}1\end{array}\right|\\ =\stackrel{^}{\mathrm{i}}\left(2+3\right)\text{}-\stackrel{^}{\mathrm{j}}\left(1-3\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(-1-2\right)\\ =\stackrel{^}{\mathrm{i}}\left(5\right)\text{}-\stackrel{^}{\mathrm{j}}\left(-2\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(-3\right)\\ =5\stackrel{^}{\mathrm{i}}\text{}+2\stackrel{^}{\mathrm{j}}\text{}-\text{}3\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{parallelogram}\text{}\mathrm{whose}\text{}\mathrm{adjacent}\text{}\mathrm{sides}\text{}\mathrm{are}\\ \text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}=\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\\ =\sqrt{25+4+9}\\ =\text{\hspace{0.17em}}\sqrt{38}\text{}\mathrm{square}\text{}\mathrm{unit}.\end{array}$

Q.37

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{triangle}\text{}\mathrm{whose}\text{}\mathrm{adjacent}\text{}\mathrm{sides}\text{}\mathrm{are}\text{}\\ \mathrm{given}\text{}\mathrm{by}\text{}\mathrm{the}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{}-\text{}5\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}.\text{}\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\\ \stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{}-\text{}5\stackrel{^}{\mathrm{j}}\text{}+\text{}2\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{vector}\text{}\mathrm{product}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\\ \stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ 1\text{\hspace{0.17em}}2\text{}3\\ 1\text{\hspace{0.17em}\hspace{0.17em}}-5\text{}2\end{array}\right|\\ =\stackrel{^}{\mathrm{i}}\left(4+15\right)\text{}-\stackrel{^}{\mathrm{j}}\left(2-3\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(-5-2\right)\\ =\stackrel{^}{\mathrm{i}}\left(19\right)\text{}-\stackrel{^}{\mathrm{j}}\left(-1\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(-7\right)\\ =19\stackrel{^}{\mathrm{i}}\text{}+\stackrel{^}{\mathrm{j}}\text{}-\text{}7\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{triangle}\text{}\mathrm{whose}\text{}\mathrm{adjacent}\text{}\mathrm{sides}\text{}\mathrm{are}\\ \text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}=\frac{1}{2}\left|\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|\\ =\frac{1}{2}\sqrt{{\left(19\right)}^{2}\text{}+\text{}{\left(1\right)}^{2}\text{}+\text{}{\left(-7\right)}^{2}}\\ =\frac{1}{2}\sqrt{361+1+49}\\ =\text{\hspace{0.17em}}\frac{1}{2}\sqrt{411}\text{}\mathrm{square}\text{}\mathrm{unit}.\end{array}$

Q.38

$\begin{array}{l}\mathrm{Let}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{are}\text{}\mathrm{two}\text{}\mathrm{unit}\text{}\mathrm{vectors}\text{}\mathrm{and}\text{}\mathrm{\theta }\text{}\mathrm{is}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\\ \mathrm{them}.\text{}\mathrm{Then},\text{}\mathrm{prove}\text{}\mathrm{that}\text{}\mathrm{sin}\text{}\left(\frac{\mathrm{\theta }}{2}\right)\text{}=\text{}\frac{1}{2}\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}-\text{}\stackrel{\to }{\mathrm{b}}\right|.\end{array}$

Ans

$\begin{array}{l}\mathrm{Since}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{are}\text{}\mathrm{two}\text{}\mathrm{unit}\text{}\mathrm{vectors},\\ \therefore \text{}\left|\stackrel{\to }{\mathrm{a}}\right|\text{}=1,\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}1\\ \mathrm{We}\text{}\mathrm{have},\\ {\left|\stackrel{\to }{\mathrm{a}}\text{}-\text{}\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}=\text{}{\left|\stackrel{\to }{\mathrm{a}}\right|}^{2}\text{}+\text{}{\left|\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}-\text{}2\left|\stackrel{\to }{\mathrm{a}}\right|\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\mathrm{cos\theta }\\ \mathrm{where}\text{}\mathrm{\theta }\text{}\mathrm{is}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\\ ⇒\text{}{\left|\stackrel{\to }{\mathrm{a}}\text{}-\text{}\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}=\text{}{1}^{2}\text{}+\text{}{1}^{2}\text{}-\text{}2×1×1\mathrm{cos\theta }\\ ⇒\text{}{\left|\stackrel{\to }{\mathrm{a}}\text{}-\text{}\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}=\text{}2\text{}-\text{}2\mathrm{cos\theta }\\ ⇒\text{}{\left|\stackrel{\to }{\mathrm{a}}\text{}-\text{}\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}=\text{}2\text{}\left(1-\text{}\mathrm{cos\theta }\right)\\ ⇒\text{}{\left|\stackrel{\to }{\mathrm{a}}\text{}-\text{}\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}=\text{}2\text{}×\text{}2{\mathrm{sin}}^{2}\frac{\mathrm{\theta }}{2}\text{}\left[\text{}1-\mathrm{cos\theta }=\text{}2{\mathrm{sin}}^{2}\frac{\mathrm{\theta }}{2}\right]\\ \mathrm{Taking}\text{}\mathrm{square}\text{}\mathrm{root}\text{}\mathrm{of}\text{}\mathrm{both}\text{}\mathrm{sides},\\ ⇒\text{}\sqrt{{\left|\stackrel{\to }{\mathrm{a}}\text{}-\text{}\stackrel{\to }{\mathrm{b}}\right|}^{2}}=\text{}\sqrt{2\text{}×\text{}2{\mathrm{sin}}^{2}\frac{\mathrm{\theta }}{2}}\\ ⇒\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}-\text{}\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}2\mathrm{sin}\frac{\mathrm{\theta }}{2}\\ ⇒\text{}\mathrm{sin}\left(\frac{\mathrm{\theta }}{2}\right)\text{}=\text{}\frac{1}{2}\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}-\text{}\stackrel{\to }{\mathrm{b}}\right|\\ \mathrm{Hence}\text{}\mathrm{Proved}.\end{array}$

Q.39

$\begin{array}{l}\mathrm{Let}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{are}\text{}\mathrm{two}\text{}\mathrm{unit}\text{}\mathrm{vectors}\text{}\mathrm{and}\text{}\mathrm{\theta }\text{}\mathrm{is}\text{}\mathrm{the}\text{}\mathrm{angle}\\ \mathrm{between}\text{}\mathrm{them}.\text{}\mathrm{Then},\text{}\mathrm{prove}\text{}\mathrm{that}\text{}\mathrm{cos}\left(\frac{\mathrm{\theta }}{2}\right)\text{}=\text{}\frac{1}{2}\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|.\end{array}$

Ans

$\begin{array}{l}\mathrm{Since}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{are}\text{}\mathrm{two}\text{}\mathrm{unit}\text{}\mathrm{vectors},\\ \therefore \text{}\left|\stackrel{\to }{\mathrm{a}}\right|=1,\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}1\\ \left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|=\text{}{\left|\stackrel{\to }{\mathrm{a}}\right|}^{2}+\text{}{\left|\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}+\text{}2\left|\stackrel{\to }{\mathrm{a}}\right|\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\text{}\mathrm{cos\theta }\\ \mathrm{where}\text{}\mathrm{\theta }\text{}\mathrm{is}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\\ ⇒{\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|}^{2}=\text{}{1}^{2}+\text{}{1}^{2}\text{}+\text{}2×1×1\text{}\mathrm{cos\theta }\\ ⇒{\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|}^{2}=\text{}2\text{}+\text{}2\mathrm{cos\theta }\\ ⇒{\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|}^{2}=\text{}2\text{}\left(1+\text{}\mathrm{cos\theta }\right)\\ ⇒{\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|}^{2}=\text{}2\text{}×\text{}2{\mathrm{cos}}^{2}\frac{\mathrm{\theta }}{2}\text{}\left[\text{}1+\text{}\mathrm{cos\theta }\text{}=\text{}2{\mathrm{cos}}^{2}\frac{\mathrm{\theta }}{2}\right]\\ \mathrm{Taking}\text{}\mathrm{square}\text{}\mathrm{root}\text{}\mathrm{of}\text{}\mathrm{both}\text{}\mathrm{sides},\\ ⇒\sqrt{{\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|}^{2}}=\text{}\sqrt{2\text{}×\text{}2{\mathrm{cos}}^{2}\frac{\mathrm{\theta }}{2}}\\ ⇒\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|=\text{}2\mathrm{cos}\frac{\mathrm{\theta }}{2}\\ ⇒\text{}\mathrm{cos}\left(\frac{\mathrm{\theta }}{2}\right)\text{}=\text{}\frac{1}{2}\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\right|\\ \mathrm{Hence}\text{}\mathrm{Proved}.\end{array}$

Q.40

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{magnitude}\text{}\mathrm{of}\text{}\mathrm{moment}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{force}\text{}\left(4\stackrel{^}{\mathrm{i}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\right)\text{}\mathrm{about}\text{}\mathrm{a}\\ \mathrm{point}\text{}\mathrm{whose}\text{}\mathrm{position}\text{}\mathrm{vector}\text{}\mathrm{is}\text{}\left(\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\\ \mathrm{the}\text{}\mathrm{position}\text{}\mathrm{vector}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{point}\text{}\mathrm{P}\text{}=\text{}\stackrel{\to }{\mathrm{OP}}\text{}=\text{}\stackrel{\to }{\mathrm{r}}=\text{}\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\\ \text{}\mathrm{and}\text{}\mathrm{force},\text{}\stackrel{\to }{\mathrm{F}}\text{}=\text{}4\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\text{}\\ =\text{}0\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\\ \text{}\mathrm{The}\text{}\mathrm{moment}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{force}\text{}=\text{}\mathrm{r}\text{}×\text{}\mathrm{F}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ 1\text{\hspace{0.17em}\hspace{0.17em}}2\text{\hspace{0.17em}\hspace{0.17em}}3\\ 0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\end{array}\right|\\ =\text{}\stackrel{^}{\mathrm{i}}\text{}\left(6-12\right)-\text{}\stackrel{^}{\mathrm{j}}\left(3-0\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(4-0\right)\\ =\text{}\stackrel{^}{\mathrm{i}}\text{}\left(-6\right)-\text{}\stackrel{^}{\mathrm{j}}\left(3\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(4\right)\\ =\text{}-6\stackrel{^}{\mathrm{i}}\text{}-\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{magnitude}\text{}\mathrm{of}\text{}\mathrm{moment}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{force}\text{}=\text{}\left|\stackrel{\to }{\mathrm{r}}\text{}×\text{}\stackrel{\to }{\mathrm{F}}\right|\\ =\text{}\sqrt{{\left(-6\right)}^{2}\text{}+\text{}{\left(-3\right)}^{2}\text{}+\text{}{\left(4\right)}^{2}}\\ =\text{}\sqrt{36\text{}+\text{}9\text{}+\text{}16}\\ =\text{}\sqrt{61}\text{}\mathrm{Nm}.\end{array}$

Q.41 The one end of a wrench acts at point O (origin), the other end pulled by a force is represented in magnitude and direction by the line joining the point A(1,-2,4) to point B(5,2,3). Find the magnitude of torque acting on the wrench.

Ans

$\begin{array}{l}\mathrm{Since}\text{}\mathrm{the}\text{}\mathrm{wrench}\text{}\mathrm{is}\text{}\mathrm{along}\text{}\mathrm{line}\text{}\mathrm{OA},\\ \therefore \text{}\mathrm{the}\text{}\mathrm{position}\text{}\mathrm{vector}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{point}\text{}\mathrm{A}\text{}=\text{}\stackrel{\to }{\mathrm{OA}}\text{}=\text{}\stackrel{\to }{\mathrm{r}}=\text{}\stackrel{^}{\mathrm{i}}\text{}-\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}\\ \text{}\mathrm{and}\text{}\mathrm{the}\text{}\mathrm{force}\text{}\mathrm{acts}\text{}\mathrm{along}\text{}\mathrm{line}\text{}\mathrm{AB},\\ \therefore \text{}\stackrel{\to }{\mathrm{AB}}\text{}=\text{}\stackrel{\to }{\mathrm{OB}}\text{}-\text{}\stackrel{\to }{\mathrm{OA}}\text{}=\text{}4\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}-\text{}\stackrel{^}{\mathrm{k}}\text{}\\ ⇒\stackrel{\to }{\mathrm{F}}\text{}=4\stackrel{^}{\mathrm{i}}\text{}+\text{}4\stackrel{^}{\mathrm{j}}\text{}-\text{}\stackrel{^}{\mathrm{k}}\text{}\\ \mathrm{We}\text{}\mathrm{have},\\ \mathrm{The}\text{}\mathrm{torque}\text{}=\text{}\stackrel{\to }{\mathrm{r}}\text{}×\text{}\stackrel{\to }{\mathrm{F}}\\ \text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ 1\text{}-2\text{\hspace{0.17em}}4\\ 4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\text{\hspace{0.17em}\hspace{0.17em}}-1\end{array}\right|\\ =\text{}\stackrel{^}{\mathrm{i}}\text{}\left(2-16\right)-\text{}\stackrel{^}{\mathrm{j}}\left(-1-16\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(4+8\right)\\ =\text{}\stackrel{^}{\mathrm{i}}\text{}\left(-14\right)-\text{}\stackrel{^}{\mathrm{j}}\left(-17\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(12\right)\\ =\text{}-14\stackrel{^}{\mathrm{i}}\text{}+\text{}17\stackrel{^}{\mathrm{j}}\text{}+\text{}12\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{magnitude}\text{}\mathrm{of}\text{}\mathrm{torque}\text{}=\text{}\left|\stackrel{\to }{\mathrm{r}}\text{}×\text{}\stackrel{\to }{\mathrm{F}}\right|\\ =\text{}\sqrt{{\left(-14\right)}^{2}\text{}+\text{}{\left(17\right)}^{2}\text{}+\text{}{\left(12\right)}^{2}}\\ =\text{}\sqrt{196\text{}+\text{}289\text{}+\text{}144}\\ =\text{}\sqrt{629}\text{}\mathrm{Nm}.\end{array}$

Q.42

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{parallelogram}\text{}\mathrm{whose}\text{}\mathrm{diagonals}\text{}\mathrm{are}\text{}\mathrm{givne}\text{}\mathrm{by}\\ \mathrm{the}\text{}\mathrm{vecotrs}\text{}\stackrel{\to }{\mathrm{a}}\text{}=\text{}3\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{j}}\text{}-\text{}2\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\stackrel{^}{\mathrm{i}}\text{}-\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\\ \text{}\stackrel{\to }{\mathrm{a}}\text{}=\text{}3\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{j}}\text{}-\text{}2\stackrel{^}{\mathrm{k}}\text{}\mathrm{and}\text{}\\ \stackrel{\to }{\mathrm{b}}\text{}=\text{}\stackrel{^}{\mathrm{i}}\text{}-\text{}3\stackrel{^}{\mathrm{j}}\text{}+\text{}4\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{vector}\text{}\mathrm{product}\text{}\mathrm{of}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\\ \stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left|\begin{array}{l}\stackrel{^}{\mathrm{i}}\text{}\stackrel{^}{\mathrm{j}}\text{}\stackrel{^}{\mathrm{k}}\\ 3\text{}1\text{}-2\\ 1\text{}-3\text{}4\end{array}\right|\\ =\text{}\stackrel{^}{\mathrm{i}}\text{}\left(4-6\right)-\text{}\stackrel{^}{\mathrm{j}}\left(12+2\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(-9-1\right)\\ =\text{}\stackrel{^}{\mathrm{i}}\text{}\left(-2\right)-\text{}\stackrel{^}{\mathrm{j}}\left(14\right)\text{}+\text{}\stackrel{^}{\mathrm{k}}\left(-10\right)\\ =\text{}-2\stackrel{^}{\mathrm{i}}\text{}-\text{}14\stackrel{^}{\mathrm{j}}\text{}+\text{}10\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{parallelogram}\text{}\mathrm{whose}\text{}\mathrm{diagonals}\\ \mathrm{are}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\frac{1}{2}\left|\text{}\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|\\ =\text{}\frac{1}{2}\sqrt{{\left(-2\right)}^{2}\text{}+\text{}{\left(-14\right)}^{2}\text{}+\text{}{\left(-10\right)}^{2}}\\ =\text{}\frac{1}{2}\sqrt{4\text{}+\text{}196\text{}+\text{}100}\\ =\text{}\frac{1}{2}\sqrt{300}\\ =\text{}\frac{1}{2}×10\sqrt{3}\\ =\text{}5\sqrt{3}\text{}\mathrm{square}\text{}\mathrm{unit}.\end{array}$

Q.43

$\begin{array}{l}\mathrm{Let}\text{}\stackrel{\to }{\mathrm{a}},\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{c}}\text{}\mathrm{are}\text{}\mathrm{three}\text{}\mathrm{vectors}\text{}\mathrm{that}\text{}\mathrm{satisfy}\text{}\mathrm{the}\text{}\mathrm{condition}\text{}\\ \stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}\text{}=\text{}\stackrel{\to }{0}.\text{}\mathrm{Evaluate}\text{}\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}\text{}\mathrm{if}\text{}\left|\stackrel{\to }{\mathrm{a}}\right|=\text{}3\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}2\text{}\mathrm{and}\text{}\left|\stackrel{\to }{\mathrm{c}}\right|\text{}=\text{}\sqrt{2}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Since}\text{}\stackrel{\to }{\mathrm{a}},\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{c}}\text{}\mathrm{are}\text{}\mathrm{vectors}\text{}\mathrm{such}\text{}\mathrm{that}\\ \therefore \text{}\left|\stackrel{\to }{\mathrm{a}}\right|=\text{}3\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}2\text{}\mathrm{and}\text{}\left|\stackrel{\to }{\mathrm{c}}\right|\text{}=\text{}\sqrt{2}\\ \mathrm{We}\text{}\mathrm{have},\\ \stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}\text{}=\text{}\stackrel{\to }{0}\text{}\\ ⇒\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}+\text{}\stackrel{\to }{\mathrm{b}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}\right|\text{}=\text{}0\\ \mathrm{Squaring}\text{}\mathrm{both}\text{}\mathrm{sides},\\ ⇒{\left|\stackrel{\to }{\mathrm{a}}\right|}^{2}\text{}+\text{}{\left|\stackrel{\to }{\mathrm{b}}\right|}^{2}\text{}+\text{}{\left|\stackrel{\to }{\mathrm{c}}\right|}^{2}\text{}+\text{}2\text{}\left(\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}\right)\text{}=\text{}0\\ ⇒{3}^{2}\text{}+\text{}{2}^{2}\text{}+\text{}{\left(\sqrt{2}\right)}^{2}\text{}+\text{}2\text{}\left(\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}\right)\text{}=\text{}0\\ ⇒9\text{}+\text{}4\text{}+\text{}2\text{}+\text{}2\text{}\left(\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}\right)\text{}=\text{}0\\ ⇒15\text{}+\text{}2\text{}\left(\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}\right)\text{}=\text{}0\\ ⇒\text{}2\text{}\left(\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}\right)\text{}=\text{}-15\\ ⇒\text{}\left(\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\text{}+\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{\mathrm{c}}\text{}+\text{}\stackrel{\to }{\mathrm{c}}.\stackrel{\to }{\mathrm{a}}\right)\text{}=\text{}-\frac{15}{2}\end{array}$

Q.44

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\mathrm{the}\text{}\mathrm{two}\text{}\mathrm{vectors}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{with}\text{}\mathrm{magnitudes}\text{}\\ 3\text{}\mathrm{and}\text{}\frac{\sqrt{2}}{3}\text{}\mathrm{respectively},\text{}\mathrm{whose}\text{}\mathrm{cross}\text{}\mathrm{product}\text{}\mathrm{is}\text{}\mathrm{unit}\text{}\mathrm{vector}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\text{}\left|\stackrel{\to }{\mathrm{a}}\right|\text{}=\text{}3,\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}\frac{\sqrt{2}}{3}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\text{}\mathrm{a}\text{}\mathrm{unit}\text{}\mathrm{vector}.\\ ⇒\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|\text{}=1\\ \mathrm{Let}\text{}\mathrm{\theta }\text{}\mathrm{is}\text{}\mathrm{the}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}},\text{}\mathrm{then}\\ \stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\left|\stackrel{\to }{\mathrm{a}}\right|\text{}\left|\stackrel{\to }{\mathrm{b}}\right|\text{}\mathrm{sin\theta }\text{}\stackrel{^}{\mathrm{n}},\text{}\mathrm{where}\text{}\mathrm{n}\text{}\mathrm{is}\text{}\mathrm{a}\text{}\mathrm{unit}\text{}\mathrm{vector}.\\ ⇒\text{}\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}3\text{}×\text{}\frac{\sqrt{2}}{3}\text{}\mathrm{sin\theta }\text{}\stackrel{^}{\mathrm{n}}\\ ⇒\text{}\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\text{}=\text{}\sqrt{2}\text{}\mathrm{sin\theta }\text{}\stackrel{^}{\mathrm{n}}\\ \mathrm{Using}\text{}\mathrm{modulus}\text{}\mathrm{on}\text{}\mathrm{both}\text{}\mathrm{sides},\\ \text{}\left|\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}\sqrt{2}\text{}\mathrm{sin\theta }\text{}\left|\stackrel{^}{\mathrm{n}}\right|\\ ⇒\text{}1\text{}=\text{}\sqrt{2}\text{}\mathrm{sin\theta }\text{}×\text{}1\text{}\left[\text{}\left|\stackrel{\to }{\mathrm{a}}\text{}×\text{}\stackrel{\to }{\mathrm{b}}\right|\text{}=\text{}1,\text{}\left|\stackrel{^}{\mathrm{n}}\right|\text{}=1\right]\\ ⇒\text{}\mathrm{sin\theta }\text{}=\text{}\frac{1}{\sqrt{2}}\\ ⇒\text{}\mathrm{sin\theta }\text{}=\text{}\mathrm{sin}{45}^{\mathrm{o}}\\ ⇒\text{}\mathrm{sin\theta }\text{}=\text{}\mathrm{sin}\frac{\mathrm{\pi }}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\theta }\text{}=\text{}\frac{\mathrm{\pi }}{4}\\ \therefore \text{}\mathrm{The}\text{}\mathrm{angle}\text{}\mathrm{between}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\text{}\frac{\mathrm{\pi }}{4}.\end{array}$

Q.45

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{a}\text{}\mathrm{vector}\text{}\mathrm{in}\text{}\mathrm{the}\text{}\mathrm{direction}\text{}\mathrm{of}\text{}\mathrm{vector}\text{}2\text{}\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+3\text{}\stackrel{^}{\mathrm{k}}\\ \mathrm{which}\text{}\mathrm{has}\text{}\mathrm{magnitude}\text{}7\text{}\mathrm{units}.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{unit}\text{}\mathrm{vector}\text{}\mathrm{of}\text{}2\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+3\stackrel{^}{\mathrm{k}}\text{}=\text{\hspace{0.17em}}\frac{2\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+3\stackrel{^}{\mathrm{k}}}{\left|2\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+3\stackrel{^}{\mathrm{k}}\right|}\\ \text{}=\text{\hspace{0.17em}}\frac{2\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+3\stackrel{^}{\mathrm{k}}}{\sqrt{{2}^{2}\text{}-\text{}{\left(-1\right)}^{2}\text{}+{3}^{2}}}\\ \text{}=\text{\hspace{0.17em}}\frac{2\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+3\stackrel{^}{\mathrm{k}}}{\sqrt{4\text{}+\text{}1\text{}+9}}\\ \text{}=\text{\hspace{0.17em}}\frac{2\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+3\stackrel{^}{\mathrm{k}}}{\sqrt{14}}\\ \mathrm{Thus},\text{}\mathrm{the}\text{}\mathrm{vector}\text{}\mathrm{in}\text{}\mathrm{the}\text{}\mathrm{direction}\text{}\mathrm{of}\text{}\left(2\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+3\stackrel{^}{\mathrm{k}}\right)\text{}\mathrm{which}\text{}\mathrm{has}\\ \mathrm{magnitude}\text{}7\text{}\mathrm{units}\text{}\mathrm{is}\\ \text{}=\text{\hspace{0.17em}}7\text{}×\text{}\frac{2\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\text{}+3\stackrel{^}{\mathrm{k}}}{\sqrt{14}}\\ \text{}=\text{\hspace{0.17em}}\frac{14\stackrel{^}{\mathrm{i}}\text{}-\text{}7\stackrel{^}{\mathrm{j}}\text{}+\text{}21\stackrel{^}{\mathrm{k}}}{\sqrt{14}}\\ \text{}=\text{\hspace{0.17em}}\frac{14}{\sqrt{14}}\stackrel{^}{\mathrm{i}}\text{}-\text{}\frac{7}{\sqrt{14}}\stackrel{^}{\mathrm{j}}\text{}+\text{}\frac{21}{\sqrt{14}}\stackrel{^}{\mathrm{k}}\end{array}$

Q.46

$\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{projection}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{vector}\text{}\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{j}}\text{}\mathrm{on}\text{}\mathrm{the}\text{}\mathrm{vector}\text{}\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}.$

Ans

$\begin{array}{l}\mathrm{Let}\text{}\stackrel{\to }{\mathrm{a}}\text{}\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{j}}\text{}\mathrm{and}\text{}\stackrel{\to }{\mathrm{b}}\text{}\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\\ \mathrm{Now},\text{}\mathrm{projection}\text{}\mathrm{of}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{on}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\text{}\mathrm{given}\text{}\mathrm{by},\\ \frac{1}{\left|\stackrel{\to }{\mathrm{b}}\right|}\text{}\left(\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{b}}\right)\text{}=\text{}\frac{1}{\left|\text{}\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\right|}\text{}\left(\text{}\stackrel{^}{\mathrm{i}}\text{}+\text{}\stackrel{^}{\mathrm{j}}\right).\left(\text{}\stackrel{^}{\mathrm{i}}\text{}-\text{}\stackrel{^}{\mathrm{j}}\right)\\ \text{}=\text{}\frac{1}{\sqrt{{1}^{2}\text{}+\text{}{1}^{2}}}\text{}\left(1-1\right)\text{}=\text{}0\\ \mathrm{Hence},\text{}\mathrm{the}\text{}\mathrm{projection}\text{}\mathrm{of}\text{}\mathrm{vector}\text{}\stackrel{\to }{\mathrm{a}}\text{}\mathrm{on}\text{}\stackrel{\to }{\mathrm{b}}\text{}\mathrm{is}\text{}0.\end{array}$

Q.47

$If a ^ is a unit vector and ( x → − a ^ )( x → + a ^ )=15, then the value of | x → | is ____. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@7F17@$

Ans

$\begin{array}{l}\text{Since}\stackrel{^}{\text{a}}\text{is a unit vector, so}|\stackrel{^}{\text{a}}|=1.\\ \therefore \left(\stackrel{\to }{x}-\stackrel{^}{\text{a}}\right)\left(\stackrel{\to }{x}+\stackrel{^}{\text{a}}\right)=15\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\to }{x}.\stackrel{\to }{x}+\stackrel{\to }{x}.\stackrel{^}{\text{\hspace{0.17em}}\text{a}}-\stackrel{^}{\text{a}}.\stackrel{\to }{x}-\stackrel{^}{\text{a}}.\text{\hspace{0.17em}}\stackrel{^}{\text{a}}=15\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{|\stackrel{\to }{x}|}^{2}-1=15\\ \left[\stackrel{\to }{x}.\stackrel{^}{\text{a}}=\stackrel{^}{\text{a}}.\stackrel{\to }{x}\text{and}\stackrel{^}{\text{a}}.\stackrel{^}{\text{a}}=1\right]\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{|\stackrel{\to }{x}|}^{2}=16\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\stackrel{\to }{x}|=4\\ \therefore \text{If}\stackrel{^}{\text{a}}\text{is a unit vector and}\left(\stackrel{\to }{x}-\stackrel{^}{a}\right)\left(\stackrel{\to }{x}+\stackrel{^}{a}\right)=15,\text{then the}\\ \text{value of}|\stackrel{\to }{\text{x}}|\text{is}\underset{¯}{\text{4}}\text{.}\end{array}$

Q.48

$\begin{array}{l}\text{The scalar projection of the vector}\stackrel{\to }{a}=\text{4}\text{\hspace{0.17em}}\stackrel{^}{\text{i}}-\text{3}\stackrel{^}{\text{j}}+\text{2}\stackrel{^}{\text{k}}\text{on the}\\ \text{vector}\stackrel{\to }{b}=\text{4}\text{\hspace{0.17em}}\stackrel{^}{\text{i}}-\text{4}\stackrel{^}{\text{j}}+\text{7}\stackrel{^}{\text{k}}\text{is _______}\text{.}\end{array}$

Ans

$\begin{array}{l}\text{Scalar projection of}\stackrel{\to }{\text{a}}\text{on}\stackrel{\to }{\text{b}}\text{is given by}\\ \frac{\stackrel{\to }{\text{a}}\text{.}\stackrel{\to }{\text{b}}}{|\stackrel{\to }{\text{b}}|}=\frac{\left(\text{4}\text{\hspace{0.17em}}\stackrel{^}{\text{i}}-\text{3}\stackrel{^}{\text{j}}+\text{2}\stackrel{^}{\text{k}}\right).\left(\text{4}\text{\hspace{0.17em}}\stackrel{^}{\text{i}}-\text{4}\stackrel{^}{\text{j}}+\text{7}\stackrel{^}{\text{k}}\right)}{|\text{4}\text{\hspace{0.17em}}\stackrel{^}{\text{i}}-\text{4}\stackrel{^}{\text{j}}+\text{7}\stackrel{^}{\text{k}}|}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{16+12+14}{\sqrt{{4}^{2}+{4}^{2}+{7}^{2}}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{42}{\sqrt{81}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{42}{9}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{14}{3}\\ \therefore \text{The scalar projection of the vector}\stackrel{\to }{a}=\text{4}\text{\hspace{0.17em}}\stackrel{^}{\text{i}}-\text{3}\stackrel{^}{\text{j}}+\text{2}\stackrel{^}{\text{k}}\text{on}\\ \text{the vector}\stackrel{\to }{b}=\text{4}\text{\hspace{0.17em}}\stackrel{^}{\text{i}}-\text{4}\stackrel{^}{\text{j}}+\text{7}\stackrel{^}{\text{k}}\text{is}\underset{¯}{\frac{14}{3}}\text{.}\end{array}$

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