# Important Questions Class 12 Maths Chapter 10

## Important Questions for CBSE Class 12 Maths Chapter 10 – Vector Algebra

Vector Algebra is an important chapter not only for the board and competitive examinations, but it also has many real-world applications.

Vector Algebra is particularly useful in situations that involve force and velocity. They are also useful in calculating angles and distances, building network pipes, measuring distances between aircraft, and so on. In civil engineering, vector algebra is widely used.

As a result, students must thoroughly practise this chapter to strengthen its concepts.

Vector Algebra covers a wide range of topics that may be asked on the board exam. The following are the main topics covered in Chapter 10 “Vector Algebra”:

• Types of Vectors
• Multiplication of  a vector by a scalar
• Components of a vector
• Vector joining two points
• Section formula
• Product of two vectors

You can also find CBSE Class 12 Maths Important Questions Chapter-by-Chapter Important Questions here:

 CBSE Class 12 Maths Important Questions Sr No Chapter No Chapter Name 1 Chapter 1 Relations and Functions 2 Chapter 2 Inverse Trigonometric Functions 3 Chapter 3 Matrices 4 Chapter 4 Determinants 5 Chapter 5 Continuity and Differentiability 6 Chapter 6 Application of Derivatives 7 Chapter 7 Integrals 8 Chapter 8 Application of Integrals 9 Chapter 9 Differential Equations 10 Chapter 10 Vector Algebra 11 Chapter 11 Three Dimensional Geometry 12 Chapter 12 Linear Programming 13 Chapter 13 Probability

## Study Important Questions for Class 12 Maths Chapter 10 – Vector Algebra

Vector Algebra is part of the unit Vectors and Three-Dimensional Geometry, which accounts for 14 of the total marks in the exam. Important Questions for “Vector Algebra” include questions with various marking schemes that may appear in board exams.

There are 57 important questions covering the entire chapter of “Vector Algebra” here. This chapter’s practice questions provide you with perfect and thorough practice. These are also beneficial for revision.

In this chapter, Extramarks provides important questions worth 1 and 4 marks. Typically, only such questions are asked in “Vector Algebra” board exams.

This chapter contains four exercises, as well as a miscellaneous exercise, to help students clearly understand the concepts of Vectors and Vector Algebra.

• A vector’s scalar components are its direction ratios, which represent its projections along the respective axes.
• Any vector’s magnitude (r), direction ratios (a, b, c), and direction cosines (l, m, n) are related as l=(a/r), m=(b/r), and n=(c/r).
• The vector sum of a triangle’s three sides taken in order is 0.
• The diagonal of the parallelogram whose adjacent sides are the given vectors gives the vector sum of two initial vectors.
• The multiplication of a given vector by a scalar changes the magnitude of the vector by the multiple || while keeping the direction constant (or reversing it) as the value is positive (or negative).

For a full set of Important Questions for Class 12 Maths Chapter 10 Algebra, click the link provided below.

### Very Short Answer Questions (1 Mark)

Q1. Classify the following measures as scalar and vector quantities:.

(i) 40°

(ii) 50 watt

(iii) 10 gm/cm3

(iv) 20 m/sec towards north

(v) 5 seconds.

A1.

(i) Angle-scalar

(ii) Power-scalar

(iii) Density-scalar

(iv) Velocity-vector

(v) Time-scalar.

Q2. Find the sum of the vectors:

a= i-2j+k, b=-2i+4j+5k and c=i-6j-7k

A2.

Sum of the vectors = a+b+c

=(i-2j+k)+(-2i+4j+5k)+(i-6j-7k

=(i−2i+i) + (-2j+4j−6j)+(k+5k-7k)

=-4j k.

Q3. What are the horizontal and vertical components of a vector a of magnitude 5 making an angle of 150 with the direction of the x-axis?

A3.

The vector a has a magnitude |a|=5 units and makes an angle

θ=150 in the direction of the x-axis. Its horizontal and vertical components can be calculated as follows:

Horizontal component

x =| a |⋅cosθ

⇒x = 5⋅(cos150∘)

⇒x = 5⋅(−cos30∘)

⇒x = 5⋅(−√3/2)

⇒x = −5√3/2

Vertical component

y = | a |⋅sinθ

⇒y = 5⋅(sin150∘)

⇒y = 5⋅(sin30∘)

⇒y = 5⋅(1/2)

⇒y = 5/2

Q1. a , b and c are three mutually perpendicular vectors of equal magnitude. Show that a + b + c makes equal angles with a , b and c with each angle as cos−1(13−−√) .

A1. Let |a→|=|b→|=|c→|=λ and since they are mutually perpendicular,

a→⋅b→=b→⋅c→=a→⋅c→=0 .

Consider |a→+b→+c→|2 :

⇒|a→+b→+c→|2=|a→|2+|b→|2+|c→|2+2(a→⋅b→+b→⋅c→+a→⋅c→)

⇒|a→+b→+c→|2=λ2+λ2+λ2+2(0+0+0)

⇒|a→+b→+c→|2=3λ2

⇒|a→+b→+c→|=3–√λ

Suppose a→,b→,c→ make angles θ1,θ2,θ3 with a→+b→+c→ respectively.

Then, cosθ1=a→⋅(a→+b→+c→)|a→||a→+b+c→|

⇒cosθ1=a→⋅a→+a→⋅b→+a→⋅c→|a→|⋅3–√λ

⇒cosθ1=|a→|2|a→|⋅3–√λ

⇒cosθ1=λ3–√λ

⇒cosθ1=13–√

Similarly, cosθ2=cosθ3=13–√ .

Therefore, θ1=θ2=θ3 .

Q2. If →α =3 ˆi − ˆj and →β =2 ˆi + ˆj +3 ˆk then express →β in the form of →β = →β1 + →β2, where →β1 is parallel to →α and →β2 is perpendicular to →α .

A2. It is given that →α =3 ˆi − ˆj and →β =2 ˆi + ˆj +3 ˆk .

Since →α is parallel to →β1 , that implies →β1 =λ →α .

⇒ →β1 =λ(3 ˆi − ˆj )

Further, →β = →β1 + →β2⇒2 ˆi + ˆj +3 ˆk =[λ(3 ˆi − ˆj )]+ →β2

⇒ →β2 =(2−3λ) ˆi +(1+λ) ˆj +3 ˆk

Now, →β2 is perpendicular to →α , hence their dot product will be equal to zero.

→α ⋅ →β2 =0

⇒(3 ˆi − ˆj )⋅[(2−3λ) ˆi +(1+λ) ˆj +3 ˆk ]=0

⇒3(2−3λ)+(−1−λ)=0

⇒6−9λ−1−λ=0

⇒5−10λ=0

⇒λ=1/2

We calculate the value of

→β1 and →β2 :

→β1 =λ(3 ˆi − ˆj )

⇒ →β1=1/2(3 ˆi − ˆj )

⇒ →β1 =3/2ˆi −1/2ˆj

Also, →β2 =(2−3λ) ˆi+(1+λ) ˆj +3 ˆk

⇒ →β2=[2−3(1/2)] ˆi +(1+1/2) ˆj +3 ˆk

⇒ →β2 =1/2ˆi +3/2ˆj +3 ˆk

Hence, we can say β =(3/2ˆi−1/2ˆj )+(1/2ˆi +3/2ˆj +3 ˆk ).

Q3. Let a^,b^,c^ are unit vectors such that a^⋅b^=a^⋅c^=0 and the angle between b^ and c^ is π6 , then prove that a^=±2(b^×c^) .

A3.

Since a^⋅b^=a^⋅c^=0 , it means that a^

is perpendicular to both b^ and c^ .

This further implies that a^ is perpendicular to the plane in which b^ and c^ lie.

Also, b^×c^=|b^||c^|sinθn^  , where θ is the angle between b^ and c^ , and n^ is a unit vector perpendicular to the plane in which b^ and c^ lie.

⇒b^×c^=(1)(1)sinπ6a^

(Since b^ and c^ are unit vectors, the angle between them is given to be π6 , and a^ satisfies the conditions for n^ )

⇒b^×c^=sinπ6a^

⇒b^×c^=12a^

⇒a^=±2(b^×c^)

Hence proved.

Q4. If a→×b→=c→×d→ and a→×c→=b→×d→ , then prove that a→−d→ is parallel to b→−c→ provided a→≠d→ and b→≠c→ .

A4.

For vectors a→−d→

and b→−c→

to be parallel, their cross-product must be equal to zero.

⇒(a→−d→)×(b→−c→)

⇒(a→−d→)×b→−(a→−d→)×c→

⇒a→×b→−d→×b→−a→×c→+d→×c→

⇒a→×b→+b→×d→−a→×c→−c→×d→

(Since −d→×b→=b→×d→

and d→×c→=−c→×d→ )

⇒0

(Since a→×b→=c→×d→ and a→×c→=b→×d→ )

Hence proved.

Q5. Dot product of a vector with vectors i^+j^−3k^ , i^+3j^−2k^ and 2i^+j^+4k^ is 0 , 5 and 8 respectively. Find the vector.

A5. Let the required vector be h→=xi^+yj^+zk^ . We have given a→=i^+j^−3k^ , b→=i^+3j^−2k^ , c→=2i^+j^+4k^ and a→⋅h→=0 ,b→⋅h→=5 , c→⋅h→=8 .

Since a→⋅h→=0 :

(i^+j^−3k^)⋅(xi^+yj^+zk^)=0

⇒x+y−3z=0  ……(1)

Since b→⋅h→=5 :

(i^+3j^−2k^)⋅(xi^+yj^+zk^)=5

⇒x+3y−2z=5 ……(2)

Since c→⋅h→=8 :

(2i^+j^+4k^)⋅(xi^+yj^+zk^)=8

⇒2x+y+4z=8 ……(3)

Subtract equation (1) from (2) :

(x+3y−2z)−(x+y−3z)=5

⇒2y+z=5

⇒y=5−z2

Subtract equation (1) from (3) :

(2x+y+4z)−(x+y−3z)=8

⇒x+7z=8

⇒x=8−7z

Substituting in (1) :

x+y−3z=0

⇒8−7z+5−z2−3z=0

⇒8−10z+5−z2=0

⇒16−20z+5−z2=0

⇒21−21z2=0

⇒21−21z=0

⇒z=1

Now, x=8−7(1)

⇒x=1

Also, y=5−1/2

⇒y=2

Hence, we have h→=i^+2j^+k^.

Q1.

Opt.

Ans.

Q2.

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

Opt.

Ans.

## CBSE Class 12 Maths Important Questions

### 1. Explain the different types of vectors.

The different types of vectors include the following:

• Zero Vector: A Zero Vector (or null vector) is a vector whose initial and terminal points coincide and are denoted as 0. Because it has no magnitude, a zero vector cannot be assigned a definite direction. Alternatively, it could be regarded as having no direction. The zero vector is represented by the vectors AA and BB
• Unit Vector: A Unit Vector is a vector with the magnitude of one (i.e., one unit). a denotes the unit vector in the direction of a given vector.
• Coinitial Vectors: Coinitial vectors are two or more vectors that have the same initial point.
• Collinear Vectors: Two or more vectors are said to be collinear if their magnitudes and directions are parallel to the same line.
• Equal Vectors: Two vectors are said to be equal if they have the same magnitude and direction regardless of their initial point positions and are written as.
• Negative of a Vector: A vector with the same magnitude as a given vector (say,) but the opposite direction is referred to as the negative of the given vector. Vector, for example, is the inverse of vector and is denoted as = -.

### 2. What are the observations of the scalar product of two vectors?

The scalar product of two vectors observations are-

• A real number is the dot product of two vectors.
• If a and b are non-zero vectors, their dot product is zero only if they are perpendicular to each other.
• If the angle between a and b vectors is zero, the dot product is equal to the magnitude of the vectors’ individual products.
• If the angle between a and b vectors is 180 degrees, the dot product is equal to the magnitude of the vectors’ individual products but in the opposite direction.

### 3. What are the observations of the cross product of two vectors?

The observations of the cross product of two vectors are as follows-

• A vector quantity is always the cross product of two vectors.
• If a and b vectors are not equal to zero vectors, their cross product equals the product of the magnitudes of the individual vectors, the sin angle between the two vectors, and the normal vector.
• If the angle formed by two vectors, a and b, is 90 degrees, sin 90 equals 1. As a result, the vector product in this case equals the product of the magnitudes of the individual vectors and the normal vector.