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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
Addition 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.

Addition of 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

Head to tail vector addition

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’ = (a – b).

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.

Properties of vector addition

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.

Therefore 0 is called the additive identity for the vector addition.

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|

Additive Inverse: –

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 + b3k̂

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

Difference of the vectors is (a –b) =(a1– b1) î + (a2– b2)ĵ + (a3– b3) 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)) =(x2–x1)i+ (y2– y1)j+ (z2–z1) k

The magnitude of P1P2 is given as:-

P1P2= (x2–x1)2+(y2– y1)2+ (z2–z1) 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.

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,

(λa). 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.

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

$Find angle θ between the vectors \vec{a} = \hat{i} + \hat{j} - \hat{k} and \vec{b} = \hat{i} - \hat{j} + \hat{k} .$

Ans

$\begin{array}{l} The angle between the vecotrs \vec{a} = \hat{i} + \hat{j} - \hat{k} and \vec{b} = \hat{i} - \hat{j} + \hat{k} is \\ cosθ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} \\ = \frac{(\hat{i} + \hat{j} - \hat{k}) (\hat{i} - \hat{j} + \hat{k})}{\sqrt{1^{2} + 1^{2} + 1^{2}} \sqrt{1^{2} + 1^{2} + 1^{2}}} \\ = \frac{1 - 1 - 1}{\sqrt{3} \sqrt{3}} = - \frac{1}{3} \\ cosθ = - \frac{1}{3} \Rightarrow θ = \cos^{- 1} (- \frac{1}{3}) \end{array}$

Q.2

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

Ans

$\begin{array}{l} Let \vec{a} = \vec{p} + \vec{d} where \vec{p} is parallel to \vec{b} and \vec{d} is perpendicular \vec{b} . \\ \vec{p} = λ \vec{b} [\vec{p} | | \vec{b}] \\ \Rightarrow \vec{p} = λ (3 \hat{i} + \hat{k}) = 3 λ \hat{i} + λ \hat{k} \\ \vec{d} = \vec{a} - \vec{p} \\ = 5 \hat{i} - 2 \hat{j} + 5 \hat{k} - 3 λ \hat{i} + λ \hat{k} \\ = (5 - 3 λ) \hat{i} - 2 \hat{j} + (5 - λ) \hat{k} \\ \vec{d} . \vec{b} = 0 [\vec{d} ⊥ \vec{b}] \\ \Rightarrow [(5 - 3 λ) \hat{i} - 2 \hat{j} + (5 - λ) \hat{k}] . (3 \hat{i} + \hat{k}) = 0 \\ \Rightarrow (5 - 3 λ) 3 + (5 - λ) . 1 = 0 \\ \Rightarrow 15 - 9 λ + 5 - λ = 0 \\ \Rightarrow 20 - 10 λ = 0 \\ \Rightarrow λ = 2 \\ Hence, \\ \vec{p} = 3 λ \hat{i} + λ \hat{k} = 6 \hat{i} + 2 \hat{k} \\ \vec{d} = (5 - 3 λ) \hat{i} - 2 \hat{j} + (5 - λ) \hat{k} \\ = (5 - 6) \hat{i} - 2 \hat{j} + (5 - 2) \hat{k} \\ = \hat{i} - 2 \hat{j} + 3 \hat{k} \end{array}$

Q.3

$Find unit vector in the direction of vector \vec{a} = \hat{i} + 2 \hat{j} + \hat{k} .$

Ans

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

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

Ans

$\begin{array}{l} The given points are A (1, 2, 7), B (2, 6, 3) and C (3, 10, - 1) . \\ ∴ \vec{AB} = (2 - 1) \hat{i} + (6 - 2) \hat{j} + (3 - 7) \hat{k} = \hat{i} + 4 \hat{j} - 4 \hat{k} \\ \vec{BC} = (3 - 2) \hat{i} + (10 - 6) \hat{j} + (- 1 - 3) \hat{k} = \hat{i} + 4 \hat{j} - 4 \hat{k} \\ \vec{AC} = (3 - 1) \hat{i} + (10 - 2) \hat{j} + (- 1 - 7) \hat{k} = 2 \hat{i} + 8 \hat{j} - 8 \hat{k} \\ |\vec{AB}| = \sqrt{1^{2} + 4^{2} + {(- 4)}^{2}} = \sqrt{1 + 16 + 16} = \sqrt{33} \\ |\vec{BC}| = \sqrt{1^{2} + 4^{2} + {(- 4)}^{2}} = \sqrt{1 + 16 + 16} = \sqrt{33} \\ |\vec{AC}| = \sqrt{2^{2} + 8^{2} + {(- 8)}^{2}} = \sqrt{4 + 64 + 64} = \sqrt{132} = 2 \sqrt{33} \\ ∴ |\vec{AC}| = |\vec{AB}| + |\vec{BC}| \\ Hence, the given points A, B and C are collinear . \end{array}$

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

$\begin{array}{l} \vec{AB} = Position vector of B - Position vector of A \\ \Rightarrow \vec{AB} = (0 \hat{i} + \hat{j} + 0 \hat{k}) - (2 \hat{i} + 0 \hat{j} + 0 \hat{k}) = - 2 \hat{i} + \hat{j} + 0 \hat{k} \\ \Rightarrow AB = |\vec{AB}| = \sqrt{{(- 2)}^{2} + 1^{2} + 0^{2}} = \sqrt{5} \\ \vec{BC} = Position vector of C - Position vector of B \\ \Rightarrow \vec{BC} = (0 \hat{i} + 0 \hat{j} + 2 \hat{k}) - (0 \hat{i} + \hat{j} + 0 \hat{k}) = 0 \hat{i} - \hat{j} + 2 \hat{k} \\ \Rightarrow BC = |\vec{BC}| = \sqrt{{(0)}^{2} + {(- 1)}^{2} + 2^{2}} = \sqrt{5} \\ So, AB = BC \\ Hence, ΔABC is an isosceles triangle . \end{array}$

Q.6

$\begin{array}{l} Find the scalar projection of the vector \vec{a} = \hat{i} - 2 \hat{j} + \hat{k} on the \\ vector \vec{b} = 4 \hat{i} - 4 \hat{j} + 7 \hat{k} . \end{array}$

Ans

$\begin{array}{l} Scalar projection of \vec{a} on \vec{b} is given by \\ \frac{\vec{a} . \vec{b}}{|\vec{b}|} = \frac{(\hat{i} - 2 \hat{j} + \hat{k}) . (4 \hat{i} - 4 \hat{j} + 7 \hat{k})}{\sqrt{4^{2} + 4^{2} + 7^{2}}} \\ = \frac{4 + 8 + 7}{\sqrt{16 + 16 + 49}} \\ = \frac{19}{\sqrt{81}} = \frac{19}{9} \end{array}$

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.

$\begin{array}{l} Now, \\ \vec{CA} . \vec{CB} = (\vec{CO} + \vec{OA}) . (\vec{CO} + \vec{OB}) \\ = (- \vec{c} + \vec{a}) . (- \vec{c} + \vec{b}) \\ = (- \vec{c} + \vec{a}) . (- \vec{c} - \vec{a}) [\vec{b} = - \vec{a}] \\ = {|- \vec{c}|}^{2} - {|\vec{a}|}^{2} \\ = c^{2} - a^{2} \\ = r^{2} - r^{2} [Radius of same circle] \\ = 0 \\ ∴ \vec{CA} . \vec{CB} = 0 \\ \Rightarrow \vec{CA} ⊥ \vec{CB} \end{array}$

Q.8

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

Ans

$\begin{array}{l} \vec{a} + \vec{b} + \vec{c} = \vec{0} \\ or \\ (\vec{a} + \vec{b} + \vec{c}) . (\vec{a} + \vec{b} + \vec{c}) = 0 \\ \Rightarrow \vec{a} . \vec{a} + \vec{b} . \vec{b} + \vec{c} . \vec{c} + 2 \vec{a} . \vec{b} + 2 \vec{b} . \vec{c} + 2 \vec{c} . \vec{a} = 0 \\ \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = 0 \\ \Rightarrow 1 + 1 + 1 + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = 0 \\ \Rightarrow 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = - 3 \\ \Rightarrow \vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a} = - \frac{3}{2} \end{array}$

Q.9

$Find |\vec{a} \times \vec{b}|, if \vec{a} = 2 \hat{i} + \hat{j} + 3 \hat{k} and \vec{b} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k} .$

Ans

$\begin{array}{l} \vec{a} \times \vec{b} = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 2 1 3 \\ 3 5 - 2 \end{array}| \\ = (- 2 - 15) \hat{i} - (- 4 - 9) \hat{j} + (10 - 3) \hat{k} \\ = - 17 \hat{i} + 13 \hat{j} + 7 \hat{k} \\ |\vec{a} \times \vec{b}| = \sqrt{17^{2} + 13^{2} + 7^{2}} = \sqrt{507} \end{array}$

Q.10

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

Ans

$\begin{array}{l} Area of parallelogram \\ \vec{a} \times \vec{b} = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 3 1 4 \\ 1 - 1 1 \end{array}| \\ = (1 + 4) \hat{i} - (3 - 4) \hat{j} + (- 3 - 1) \hat{k} \\ = 5 \hat{i} + \hat{j} - 4 \hat{k} \\ |\vec{a} \times \vec{b}| = \sqrt{5^{2} + 1^{2} + 4^{2}} = \sqrt{42} \end{array}$

Q.11

$If \hat{a} is a unit vector and (\vec{x} - \hat{a}) . (\vec{x} + \hat{a}) = 8, find |\vec{x}| .$

Ans

$\begin{array}{l} Since \hat{a} is a unit vector so |\hat{a}| = 1 \\ (\vec{x} - \hat{a}) . (\vec{x} + \hat{a}) = 8 \\ \vec{x} . \vec{x} + \vec{x} . \hat{a} - \hat{a} . \vec{x} - \hat{a} . \hat{a} = 8 \\ {|\vec{x}|}^{2} - 1 = 8 \\ \Rightarrow |\vec{x}| = 3 \end{array}$

Q.12

$Find |\vec{a}| and |\vec{b}|, if (\vec{a} - \vec{b}) . (\vec{a} + \vec{b}) = 27 and |\vec{a}| = 2 |\vec{b}| .$

Ans

$\begin{array}{l} (\vec{a} - \vec{b}) . (\vec{a} + \vec{b}) = 27 \\ \Rightarrow {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = 27 \\ \Rightarrow 4 {|\vec{b}|}^{2} - {|\vec{b}|}^{2} = 27 \\ \Rightarrow 3 {|\vec{b}|}^{2} = 27 \\ \Rightarrow {|\vec{b}|}^{2} = 9 \\ \Rightarrow |\vec{b}| = 3 \\ Now, |\vec{a}| = 2 |\vec{b}| \\ \Rightarrow |\vec{a}| = 6 \\ Thus, |\vec{a}| = 6 and |\vec{b}| = 3 \end{array}$

Q.13

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

Ans

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

Q.14

$If \vec{a} and \vec{b} are any vectors, then prove that |\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}| .$

Ans

$\begin{array}{l} {|\vec{a} + \vec{b}|}^{2} = (\vec{a} + \vec{b}) . (\vec{a} + \vec{b}) = \vec{a} . \vec{a} + \vec{a} . \vec{b} + \vec{b} . \vec{a} + \vec{b} . \vec{b} \\ = {|\vec{a}|}^{2} + 2 |\vec{a}| |\vec{b}| cosθ + {|\vec{b}|}^{2} \\ \leq {|\vec{a}|}^{2} + 2 |\vec{a}| |\vec{b}| + {|\vec{b}|}^{2} = {(|\vec{a}| + |\vec{b}|)}^{2} \\ \Rightarrow |\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}| \end{array}$

Q.15

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

Ans

$\begin{array}{l} Let \vec{a} = \hat{i} + 3 \hat{j} - 8 \hat{k} and \vec{b} = - 3 \hat{i} - 5 \hat{j} + 4 \hat{k} \\ \vec{a} . \vec{b} = (\hat{i} + 3 \hat{j} - 8 \hat{k}) . (- 3 \hat{i} - 5 \hat{j} + 4 \hat{k}) \\ = - 3 - 15 - 32 = - 50 \end{array}$

Q.16

$\begin{array}{l} Find the value of λ such that the vectors \vec{a} and \vec{b} are \\ penpendicular where \vec{a} = λ \hat{i} + 2 \hat{j} + \hat{k}, \vec{b} = 4 \hat{i} - 9 \hat{j} + 2 \hat{k} . \end{array}$

Ans

$\begin{array}{l} \vec{a} ⊥ \vec{b} \Rightarrow \vec{a} . \vec{b} = 0 \\ \Rightarrow (λ \hat{i} + 2 \hat{j} + \hat{k}) . (4 \hat{i} - 9 \hat{j} + 2 \hat{k}) = 0 \\ \Rightarrow 4 λ - 18 + 2 = 0 \\ \Rightarrow 4 λ - 16 = 0 \\ \Rightarrow 4 λ = 16 \\ \Rightarrow λ = 4 \end{array}$

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

$\begin{array}{l} Here \vec{AB} = \hat{j} + 2 \hat{k} and \vec{AC} = \hat{i} + 2 \hat{j} . \\ The area of given triangle is \frac{1}{2} |\vec{AB} \times \vec{AC}| \\ \vec{AB} \times \vec{AC} = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 0 1 2 \\ 1 2 0 \end{array}| \\ = - 4 \hat{i} - (0 - 2) \hat{j} + (0 - 1) \hat{k} \\ = - 4 \hat{i} + 2 \hat{j} - \hat{k} \\ |\vec{AB} \times \vec{AC}| = \sqrt{4^{2} + 2^{2} + 1^{2}} = \sqrt{21} \\ ∴ \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{\sqrt{21}}{2} \end{array}$

Q.18

$Prove that [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] = 2 [\vec{a}, \vec{b}, \vec{c}] .$

Ans

$\begin{array}{l} LHS = [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] \\ = (\vec{a} + \vec{b}) . [(\vec{b} + \vec{c}) \times (\vec{c} + \vec{a})] \\ = (\vec{a} + \vec{b}) . [\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a}] \\ = (\vec{a} + \vec{b}) . [\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a}] [\vec{c} \times \vec{c} = 0] \\ = \vec{a} . (\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a}) + \vec{b} . (\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a}) \\ = \vec{a} . (\vec{b} \times \vec{c}) + \vec{a} . (\vec{b} \times \vec{a}) + \vec{a} . (\vec{c} \times \vec{a}) + \vec{b} . (\vec{b} \times \vec{c}) + \vec{b} . (\vec{b} \times \vec{a}) + \vec{b} . (\vec{c} \times \vec{a}) \\ = \vec{a} . (\vec{b} \times \vec{c}) + 0 + 0 + 0 + 0 + \vec{b} . (\vec{c} \times \vec{a}) \\ = 2 [\vec{a}, \vec{b}, \vec{c}] = RHS \end{array}$

Q.19

$Find the unit vector in the direction of vector \vec{a} = 2 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k} .$

Ans

$\begin{array}{l} The unit vector in the direction of vector, \\ \hat{a} = \frac{\vec{a}}{|\vec{a}|} \\ = \frac{2 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}}{\sqrt{4 + 9 + 3}} \\ = \frac{2 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}}{\sqrt{16}} \\ = \frac{1}{4} (2 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}) \\ = \frac{1}{2} \hat{i} + \frac{3}{4} \hat{j} + \frac{\sqrt{3}}{4} \hat{k} \end{array}$

Q.20

$Find the direction cosine of vector \vec{a} = 2 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k} .$

Ans

Q.21

$\begin{array}{l} Find the scalar product of vectors \vec{a} = 2 \hat{i} + 3 \hat{j} + 8 \hat{k} and \\ \vec{b} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} . \end{array}$

Ans

$\begin{array}{l} We have, \\ \vec{a} = 2 \hat{i} + 3 \hat{j} + 8 \hat{k} and \\ \vec{b} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} \\ The scalar prodct of \vec{a} and \vec{b} \\ \vec{a} . \vec{b} = (2 \hat{i} + 3 \hat{j} + 8 \hat{k}) . (3 \hat{i} - 2 \hat{j} + 2 \hat{k}) \\ = 6 - 6 + 16 \\ = 16 \end{array}$

Q.22

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

Ans

$\begin{array}{l} Since the vectors \vec{a} and \vec{b} are perpendicular to each other, \\ ∴ \vec{a} . \vec{b} = 0 \\ \Rightarrow (2 \hat{i} + λ \hat{j} + \hat{k}) . (\hat{i} - 2 \hat{j} + 3 \hat{k}) = 0 \\ \Rightarrow 2 - 2 λ + 3 = 0 \\ \Rightarrow 5 - 2 λ = 0 \\ \Rightarrow 2 λ = 5 \\ \Rightarrow λ = \frac{5}{2} \end{array}$

Q.23

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

Ans

$\begin{array}{l} We have, \\ \vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} and \\ \vec{b} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} \\ The vector product of \vec{a} and \vec{b} is \\ \vec{a} \times \vec{b} = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 2 3 4 \\ 4 6 8 \end{array}| \\ = \hat{i} (24 - 24) - \hat{j} (16 - 16) + \hat{k} (12 - 12) \\ = \hat{i} (0) - \hat{j} (0) + \hat{k} (0) \\ = \vec{0} \end{array}$

Q.24

$Find the angle between the vectors \vec{a} = 3 \hat{i} + 4 \hat{k} and \vec{b} = 4 \hat{i} + \sqrt{5} \hat{j} - 2 \hat{k} .$

Ans

$\begin{array}{l} Let θ be the angle between \vec{a} and \vec{b}, then \\ cosθ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} \\ = \frac{(3 \hat{i} + 0 \hat{j} + 4 \hat{k}) . (4 \hat{i} + \sqrt{5} \hat{j} - 2 \hat{k})}{\sqrt{9 + 0 + 16} \sqrt{16 + 5 + 4}} \\ = \frac{12 + 0 - 8}{\sqrt{25} \sqrt{25}} \\ = \frac{4}{5 \times 5} \\ \Rightarrow cosθ = \frac{4}{25} \\ \Rightarrow θ = \cos^{- 1} (\frac{4}{25}) \end{array}$

Q.25

$Find the angle between the vectors \vec{a} and \vec{b} if |\vec{a} \times \vec{b}| = \vec{a} . \vec{b} .$

Ans

$\begin{array}{l} We have, \\ |\vec{a} \times \vec{b}| = \vec{a} . \vec{b} \\ \Rightarrow |\vec{a}| |\vec{b}| sinθ = |\vec{a}| |\vec{b}| cosθ \\ where θ is the angle between \vec{a} and \vec{b} . \\ \Rightarrow sinθ = cosθ \\ \Rightarrow tanθ = 1 \\ \Rightarrow tanθ = \tan 45^{o} \\ \Rightarrow tanθ = \tan \frac{π}{4} \\ \Rightarrow θ = \frac{π}{4} \end{array}$

Q.26

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

Ans

$\begin{array}{l} The unit vector in the direction of vector \vec{a} is \\ \hat{a} = \frac{\vec{a}}{|\vec{a}|} \\ = \frac{2 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}}{\sqrt{4 + 9 + 3}} \\ = \frac{2 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}}{\sqrt{16}} \\ = \frac{1}{4} (2 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}) \\ The required vector = 8 \times \frac{1}{4} (2 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}) \\ = 2 (2 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}) \\ = (4 \hat{i} + 6 \hat{j} + 2 \sqrt{3} \hat{k}) \end{array}$

Q.27

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

Ans

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

Q.28

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

Ans

$\begin{array}{l} We have, |\vec{a}| = 1, |\vec{b}| = \frac{2}{\sqrt{3}} and \vec{a} . \vec{b} = 1 \\ Let θ is the angle between \vec{a} and \vec{b}, then \\ cosθ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} \\ \Rightarrow cosθ = \frac{1}{1 \times \frac{2}{\sqrt{3}}} \\ \Rightarrow cosθ = \frac{\sqrt{3}}{2} \\ \Rightarrow cosθ = \cos 30^{o} \\ \Rightarrow cosθ = \cos \frac{π}{6} \\ \Rightarrow θ = \frac{π}{6} \\ ∴ The angle between \vec{a} and \vec{b} is \frac{π}{6} . \end{array}$

Q.29

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

Ans

$\begin{array}{l} We have, \\ \vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} and \\ \vec{b} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} \\ The vectors product of \vec{a} and \vec{b} is \\ \vec{a} \times \vec{b} = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 2 3 4 \\ 4 6 8 \end{array}| \\ = \hat{i} (24 - 24) - \hat{j} (16 - 16) + \hat{k} (12 - 12) \\ = \hat{i} (0) - \hat{j} (0) + \hat{k} (0) \\ = \vec{0} \\ \vec{a} \times \vec{b} = \vec{0} \\ ∴ The vectors \vec{a} and \vec{b} are parallel to each other . \end{array}$

Q.30

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

Ans

$\begin{array}{l} We have, |\vec{a}| = 1, |\vec{b}| = \sqrt{2} and \vec{a} . \vec{b} = 1 \\ Let θ is the angle between \vec{a} and \vec{b}, then \\ cosθ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} \\ \Rightarrow cosθ = \frac{1}{1 \times \sqrt{2}} \\ \Rightarrow cosθ = \frac{1}{\sqrt{2}} \\ \Rightarrow cosθ = \cos 45^{o} \\ \Rightarrow cosθ = \cos \frac{π}{4} \\ \Rightarrow θ = \frac{π}{4} \\ ∴ The angle between \vec{a} and \vec{b} is \frac{π}{4} . \end{array}$

Q.31

$Evaluate |\vec{a} \times \vec{b}|, where \vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k} and \vec{b} = - \hat{j} + \hat{k} .$

Ans

$\begin{array}{l} We have, \\ \vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k} and \\ \vec{b} = - \hat{j} + \hat{k} \\ The vector product of \vec{a} and \vec{b} is \\ \vec{a} \times \vec{b} = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 1 2 - 3 \\ 0 - 1 1 \end{array}| \\ = \hat{i} (2 - 3) - \hat{j} (1 + 0) + \hat{k} (- 1) \\ = \hat{i} (- 1) - \hat{j} (1) + \hat{k} (- 1) \\ = - \hat{i} - \hat{j} - \hat{k} \\ \Rightarrow |\vec{a} \times \vec{b}| = - \hat{i} - \hat{j} - \hat{k} \\ \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{1 + 1 + 1} \\ = \sqrt{3} \end{array}$

Q.32

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

Ans

$\begin{array}{l} We have, \\ the projection of \vec{a} on \vec{b} = \frac{\vec{a} . \vec{b}}{|\vec{b}|} \\ \vec{a} = 4 \hat{i} - 3 \hat{j} + \sqrt{2} \hat{k} and \vec{b} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \\ \vec{a} . \vec{b} = (4 \hat{i} - 3 \hat{j} + \sqrt{2} \hat{k}) . (3 \hat{i} + 4 \hat{j} + 5 \hat{k}) \\ = 12 - 12 + 5 \sqrt{2} \\ and |\vec{b}| = \sqrt{9 + 16 + 25} \\ = \sqrt{50} = 5 \sqrt{2} \\ \Rightarrow The projection of \vec{a} on \vec{b} = \frac{\vec{a} . \vec{b}}{|\vec{b}|} \\ = \frac{5 \sqrt{2}}{5 \sqrt{2}} = 1 \end{array}$

Q.33

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

Ans

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

Q.34

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

Ans

$\begin{array}{l} The cross product of \vec{a} and \vec{b} is \\ \vec{a} \times \vec{b} = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 1 4 2 \\ 3 - 2 7 \end{array}| \\ = \hat{i} (28 + 4) - \hat{j} (7 - 6) + \hat{k} (- 2 - 12) \\ = \hat{i} (32) - \hat{j} (1) + \hat{k} (- 14) \\ = 32 \hat{i} - \hat{j} - 14 \hat{k} \\ Let \vec{d} = λ (32 \hat{i} - \hat{j} - 14 \hat{k}) . .. (1) \\ [\vec{d} is also perpendicular to \vec{a} and \vec{b}] \\ Now, \vec{c} . \vec{d} = 15 \\ \Rightarrow (2 \hat{i} - \hat{j} + 4 \hat{k}) . λ (32 \hat{i} - \hat{j} - 14 \hat{k}) = 15 \\ \Rightarrow λ (64 + 1 - 56) = 15 \\ \Rightarrow 9 λ = 15 \\ \Rightarrow λ = \frac{5}{3} \\ Putting the value in (1), we get \\ \vec{d} = \frac{5}{3} (32 \hat{i} - \hat{j} - 14 \hat{k}) \end{array}$

Q.35

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

Ans

$\begin{array}{l} The vector which is perpendicular to \vec{a} and \vec{b} is \\ \vec{a} \times \vec{b} = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 3 1 2 \\ 1 2 5 \end{array}| \\ = \hat{i} (5 - 4) - \hat{j} (15 - 2) + \hat{k} (6 - 1) \\ = \hat{i} - 13 \hat{j} + 5 \hat{k} \\ The unit vector in the direction of (\vec{a} \times \vec{b}) = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \\ = \frac{\hat{i} - 13 \hat{j} + 5 \hat{k}}{\sqrt{1 + 169 + 25}} \\ = \frac{\hat{i} - 13 \hat{j} + 5 \hat{k}}{\sqrt{195}} \\ The required vector = \sqrt{780} \times \frac{(\hat{i} - 13 \hat{j} + 5 \hat{k})}{\sqrt{195}} \\ = 2 \sqrt{195} \times \frac{(\hat{i} - 13 \hat{j} + 5 \hat{k})}{\sqrt{195}} \\ = 2 (\hat{i} - 13 \hat{j} + 5 \hat{k}) \\ = 2 \hat{i} - 26 \hat{j} + 10 \hat{k} \end{array}$

Q.36

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

Ans

$\begin{array}{l} We have, \\ \vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} and \\ \vec{b} = \hat{i} - \hat{j} + \hat{k} \\ The vector which is perpendicular to \vec{a} and \vec{b} is \\ \vec{a} \times \vec{b} = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 1 2 3 \\ 1 - 1 1 \end{array}| \\ = \hat{i} (2 + 3) - \hat{j} (1 - 3) + \hat{k} (- 1 - 2) \\ = \hat{i} (5) - \hat{j} (- 2) + \hat{k} (- 3) \\ = 5 \hat{i} + 2 \hat{j} - 3 \hat{k} \\ The area of parallelogram whose adjacent sides are \\ \vec{a} and \vec{b} = \vec{a} \times \vec{b} \\ = \sqrt{25 + 4 + 9} \\ = \sqrt{38} square unit . \end{array}$

Q.37

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

Ans

$\begin{array}{l} We have, \\ \vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} and \\ \vec{b} = \hat{i} - 5 \hat{j} + 2 \hat{k} \\ The vector product of \vec{a} and \vec{b} is \\ \vec{a} \times \vec{b} = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 1 2 3 \\ 1 - 5 2 \end{array}| \\ = \hat{i} (4 + 15) - \hat{j} (2 - 3) + \hat{k} (- 5 - 2) \\ = \hat{i} (19) - \hat{j} (- 1) + \hat{k} (- 7) \\ = 19 \hat{i} + \hat{j} - 7 \hat{k} \\ The area of triangle whose adjacent sides are \\ \vec{a} and \vec{b} = \frac{1}{2} |\vec{a} \times \vec{b}| \\ = \frac{1}{2} \sqrt{{(19)}^{2} + {(1)}^{2} + {(- 7)}^{2}} \\ = \frac{1}{2} \sqrt{361 + 1 + 49} \\ = \frac{1}{2} \sqrt{411} square unit . \end{array}$

Q.38

$\begin{array}{l} Let \vec{a} and \vec{b} are two unit vectors and θ is the angle between \\ them . Then, prove that \sin (\frac{θ}{2}) = \frac{1}{2} |\vec{a} - \vec{b}| . \end{array}$

Ans

$\begin{array}{l} Since \vec{a} and \vec{b} are two unit vectors, \\ ∴ |\vec{a}| = 1, |\vec{b}| = 1 \\ We have, \\ {|\vec{a} - \vec{b}|}^{2} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 |\vec{a}| |\vec{b}| cosθ \\ where θ is the angle between \vec{a} and \vec{b} \\ \Rightarrow {|\vec{a} - \vec{b}|}^{2} = 1^{2} + 1^{2} - 2 \times 1 \times 1 cosθ \\ \Rightarrow {|\vec{a} - \vec{b}|}^{2} = 2 - 2 cosθ \\ \Rightarrow {|\vec{a} - \vec{b}|}^{2} = 2 (1 - cosθ) \\ \Rightarrow {|\vec{a} - \vec{b}|}^{2} = 2 \times 2 \sin^{2} \frac{θ}{2} [1 - cosθ = 2 \sin^{2} \frac{θ}{2}] \\ Taking square root of both sides, \\ \Rightarrow \sqrt{{|\vec{a} - \vec{b}|}^{2}} = \sqrt{2 \times 2 \sin^{2} \frac{θ}{2}} \\ \Rightarrow |\vec{a} - \vec{b}| = 2 \sin \frac{θ}{2} \\ \Rightarrow \sin (\frac{θ}{2}) = \frac{1}{2} |\vec{a} - \vec{b}| \\ Hence Proved . \end{array}$

Q.39

$\begin{array}{l} Let \vec{a} and \vec{b} are two unit vectors and θ is the angle \\ between them . Then, prove that \cos (\frac{θ}{2}) = \frac{1}{2} |\vec{a} + \vec{b}| . \end{array}$

Ans

$\begin{array}{l} Since \vec{a} and \vec{b} are two unit vectors, \\ ∴ |\vec{a}| = 1, |\vec{b}| = 1 \\ |\vec{a} + \vec{b}| = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 |\vec{a}| |\vec{b}| cosθ \\ where θ is the angle between \vec{a} and \vec{b} \\ \Rightarrow {|\vec{a} + \vec{b}|}^{2} = 1^{2} + 1^{2} + 2 \times 1 \times 1 cosθ \\ \Rightarrow {|\vec{a} + \vec{b}|}^{2} = 2 + 2 cosθ \\ \Rightarrow {|\vec{a} + \vec{b}|}^{2} = 2 (1 + cosθ) \\ \Rightarrow {|\vec{a} + \vec{b}|}^{2} = 2 \times 2 \cos^{2} \frac{θ}{2} [1 + cosθ = 2 \cos^{2} \frac{θ}{2}] \\ Taking square root of both sides, \\ \Rightarrow \sqrt{{|\vec{a} + \vec{b}|}^{2}} = \sqrt{2 \times 2 \cos^{2} \frac{θ}{2}} \\ \Rightarrow |\vec{a} + \vec{b}| = 2 \cos \frac{θ}{2} \\ \Rightarrow \cos (\frac{θ}{2}) = \frac{1}{2} |\vec{a} + \vec{b}| \\ Hence Proved . \end{array}$

Q.40

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

Ans

$\begin{array}{l} We have, \\ the position vector of a point P = \vec{OP} = \vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} \\ and force, \vec{F} = 4 \hat{j} + 3 \hat{k} \\ = 0 \hat{i} + 4 \hat{j} + 3 \hat{k} \\ The moment of a force = r \times F = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 1 2 3 \\ 0 4 3 \end{array}| \\ = \hat{i} (6 - 12) - \hat{j} (3 - 0) + \hat{k} (4 - 0) \\ = \hat{i} (- 6) - \hat{j} (3) + \hat{k} (4) \\ = - 6 \hat{i} - 3 \hat{j} + 4 \hat{k} \\ The magnitude of moment of a force = |\vec{r} \times \vec{F}| \\ = \sqrt{{(- 6)}^{2} + {(- 3)}^{2} + {(4)}^{2}} \\ = \sqrt{36 + 9 + 16} \\ = \sqrt{61} Nm . \end{array}$

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

$\begin{array}{l} Since the wrench is along line OA, \\ ∴ the position vector of a point A = \vec{OA} = \vec{r} = \hat{i} - 2 \hat{j} + 4 \hat{k} \\ and the force acts along line AB, \\ ∴ \vec{AB} = \vec{OB} - \vec{OA} = 4 \hat{i} + 4 \hat{j} - \hat{k} \\ \Rightarrow \vec{F} = 4 \hat{i} + 4 \hat{j} - \hat{k} \\ We have, \\ The torque = \vec{r} \times \vec{F} \\ = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 1 - 2 4 \\ 4 4 - 1 \end{array}| \\ = \hat{i} (2 - 16) - \hat{j} (- 1 - 16) + \hat{k} (4 + 8) \\ = \hat{i} (- 14) - \hat{j} (- 17) + \hat{k} (12) \\ = - 14 \hat{i} + 17 \hat{j} + 12 \hat{k} \\ The magnitude of torque = |\vec{r} \times \vec{F}| \\ = \sqrt{{(- 14)}^{2} + {(17)}^{2} + {(12)}^{2}} \\ = \sqrt{196 + 289 + 144} \\ = \sqrt{629} Nm . \end{array}$

Q.42

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

Ans

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

Q.43

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

Ans

$\begin{array}{l} Since \vec{a}, \vec{b} and \vec{c} are vectors such that \\ ∴ |\vec{a}| = 3 |\vec{b}| = 2 and |\vec{c}| = \sqrt{2} \\ We have, \\ \vec{a} + \vec{b} + \vec{c} = \vec{0} \\ \Rightarrow |\vec{a} + \vec{b} + \vec{c}| = 0 \\ Squaring both sides, \\ \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = 0 \\ \Rightarrow 3^{2} + 2^{2} + {(\sqrt{2})}^{2} + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = 0 \\ \Rightarrow 9 + 4 + 2 + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = 0 \\ \Rightarrow 15 + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = 0 \\ \Rightarrow 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = - 15 \\ \Rightarrow (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = - \frac{15}{2} \end{array}$

Q.44

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

Ans

$\begin{array}{l} We have, |\vec{a}| = 3, |\vec{b}| = \frac{\sqrt{2}}{3} and \vec{a} \times \vec{b} is a unit vector . \\ \Rightarrow |\vec{a} \times \vec{b}| = 1 \\ Let θ is the angle between \vec{a} and \vec{b}, then \\ \vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| sinθ \hat{n}, where n is a unit vector . \\ \Rightarrow \vec{a} \times \vec{b} = 3 \times \frac{\sqrt{2}}{3} sinθ \hat{n} \\ \Rightarrow \vec{a} \times \vec{b} = \sqrt{2} sinθ \hat{n} \\ Using modulus on both sides, \\ |\vec{a} \times \vec{b}| = \sqrt{2} sinθ |\hat{n}| \\ \Rightarrow 1 = \sqrt{2} sinθ \times 1 [|\vec{a} \times \vec{b}| = 1, |\hat{n}| = 1] \\ \Rightarrow sinθ = \frac{1}{\sqrt{2}} \\ \Rightarrow sinθ = \sin 45^{o} \\ \Rightarrow sinθ = \sin \frac{π}{4} \\ \Rightarrow θ = \frac{π}{4} \\ ∴ The angle between \vec{a} and \vec{b} is \frac{π}{4} . \end{array}$

Q.45

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

Ans

$\begin{array}{l} The unit vector of 2 \hat{i} - \hat{j} + 3 \hat{k} = \frac{2 \hat{i} - \hat{j} + 3 \hat{k}}{|2 \hat{i} - \hat{j} + 3 \hat{k}|} \\ = \frac{2 \hat{i} - \hat{j} + 3 \hat{k}}{\sqrt{2^{2} - {(- 1)}^{2} + 3^{2}}} \\ = \frac{2 \hat{i} - \hat{j} + 3 \hat{k}}{\sqrt{4 + 1 + 9}} \\ = \frac{2 \hat{i} - \hat{j} + 3 \hat{k}}{\sqrt{14}} \\ Thus, the vector in the direction of (2 \hat{i} - \hat{j} + 3 \hat{k}) which has \\ magnitude 7 units is \\ = 7 \times \frac{2 \hat{i} - \hat{j} + 3 \hat{k}}{\sqrt{14}} \\ = \frac{14 \hat{i} - 7 \hat{j} + 21 \hat{k}}{\sqrt{14}} \\ = \frac{14}{\sqrt{14}} \hat{i} - \frac{7}{\sqrt{14}} \hat{j} + \frac{21}{\sqrt{14}} \hat{k} \end{array}$

Q.46

$Find the projection of the vector \hat{i} + \hat{j} on the vector \hat{i} - \hat{j} .$

Ans

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

Q.47

$\begin{array}{l} If \hat{a} is a unit vector and (\vec{x} - \hat{a}) (\vec{x} + \hat{a}) = 15, then the value of \\ | \vec{x} | is ____. \end{array}$

Ans

$\begin{array}{l} Since \hat{a} is a unit vector, so | \hat{a} | = 1. \\ ∴ (\vec{x} - \hat{a}) (\vec{x} + \hat{a}) = 15 \\ \vec{x} . \vec{x} + \vec{x} . \hat{a} - \hat{a} . \vec{x} - \hat{a} . \hat{a} = 15 \\ \Rightarrow {| \vec{x} |}^{2} - 1 = 15 \\ [\vec{x} . \hat{a} = \hat{a} . \vec{x} and \hat{a} . \hat{a} = 1] \\ \Rightarrow {| \vec{x} |}^{2} = 16 \\ \Rightarrow | \vec{x} | = 4 \\ ∴ If \hat{a} is a unit vector and (\vec{x} - \hat{a}) (\vec{x} + \hat{a}) = 15, then the \\ value of | \vec{x} | is \underline{4} . \end{array}$

Q.48

$\begin{array}{l} The scalar projection of the vector \vec{a} = 4 \hat{i} - 3 \hat{j} + 2 \hat{k} on the \\ vector \vec{b} = 4 \hat{i} - 4 \hat{j} + 7 \hat{k} is _______. \end{array}$

Ans

$\begin{array}{l} Scalar projection of \vec{a} on \vec{b} is given by \\ \frac{\vec{a} . \vec{b}}{| \vec{b} |} = \frac{(4 \hat{i} - 3 \hat{j} + 2 \hat{k}) . (4 \hat{i} - 4 \hat{j} + 7 \hat{k})}{| 4 \hat{i} - 4 \hat{j} + 7 \hat{k} |} \\ = \frac{16 + 12 + 14}{\sqrt{4^{2} + 4^{2} + 7^{2}}} \\ = \frac{42}{\sqrt{81}} \\ = \frac{42}{9} \\ = \frac{14}{3} \\ ∴ The scalar projection of the vector \vec{a} = 4 \hat{i} - 3 \hat{j} + 2 \hat{k} on \\ the vector \vec{b} = 4 \hat{i} - 4 \hat{j} + 7 \hat{k} is \underline{\frac{14}{3}} . \end{array}$

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CBSE Class 12 Maths Revision Notes Chapter 10

Class 12 Mathematics Chapter 10 notes

NCERT Class 12 Mathematics Chapter 10: Key Topics

Introduction

Vectors

The right-handed rectangular coordinate system

Position Vector

Direction Angle & Direction Cosines

Zero Vector:

Unit Vector:

Coinitial Vectors:

Collinear Vectors:

Equal Vectors:

Negative of a Vector:

Free Vectors:

Addition of Vectors:

Head to tail vector addition

Parallelogram law of vector addition:

Properties of vector addition

Commutative property:

Associative property:

Multiplication of any Vector by any Scalar:-

Additive Inverse: –

Unit Vector in a direction

Components of a vector:

Unit Vector:

Position vector:

Vectors Operations

Distributive law

Collinear Vectors

Section Formula

Product of Two Vectors

Properties of scalar product:

Projection of a vector on a line

Right-handed rectangular coordinate systems

Area of triangle

Class 12 Mathematics Chapter 10 notes: Exercise & Solutions

Class 12 Mathematics Chapter 10 Notes : Key Features

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