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JEE Exam  >  JEE Test  >  Chapter-wise Tests Main & Advanced  >  Test: Introduction To Vector Algebra - JEE MCQ

Introduction To Vector Algebra - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Introduction To Vector Algebra (20 Questions)

You can prepare effectively for JEE Chapter-wise Tests for JEE Main & Advanced with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Introduction To Vector Algebra ". These 20 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 20 minutes
  • - Number of Questions: 20

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Test: Introduction To Vector Algebra - Question 1
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For what values of x and y, the vectorsare equal?

Detailed Solution: Question 1

Two vectors are equal if their corresponding components are equal.
Hence, 2x = 3
or x = 3/2
and 2x = y
or y = 2 * 3/2
y = 3

Test: Introduction To Vector Algebra - Question 2
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Two or more vectors having the same initial point are called​

Detailed Solution: Question 2

Two or more vectors having the same initial point are called coinitial vectors.

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Test: Introduction To Vector Algebra - Question 3
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If  ,the vectors a and b are ______ .​

Detailed Solution: Question 3


Both of these vectors are pointing the same direction but they have different magnitude. Hence, they are collinear vectors.
Collinear Vectors Meaning

Test: Introduction To Vector Algebra - Question 4
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If the magnitude of the position vector is 7, the value of x is:​

Detailed Solution: Question 4


|a| = (x2 + 22 + (2x)2)1/2
7 = (x2 + 22 + (2x)2)1/2
⇒ 49 = x2 + 22 + 4x2
⇒ 49 = 4 + 5x2
⇒ 5x2 = 45
⇒ x2 = 9
x = ±3

Test: Introduction To Vector Algebra - Question 5
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If and  are the position vectors of the points A, B, C and D such that  then ABCD is:

Detailed Solution: Question 5

Given:

If we divide both sides with 2 we get,

⇒ Mid pt. of AC = Mid. pt. of BD
∴ ABCD is a parallelogram.

Test: Introduction To Vector Algebra - Question 6
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If  , then

Detailed Solution: Question 6

a = 2i + 3j - 6k
|a| = √4+9+36 = √49 = 7
b = 6i - 2j + 3k
|b| = √36+4+9 = √49 = 7
|a| = |b|
Hence, option A is correct.

Test: Introduction To Vector Algebra - Question 7
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What is the additive identity of a vector?​

Detailed Solution: Question 7

In the Additive Identity of vectors, the additive identity is zero vector 0.
For any vector V additive identity is defined as,
0 + V = V and V + 0 = V

Test: Introduction To Vector Algebra - Question 8
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The angles α, β, γ made by the vector with the positive directions of X, Y and Z-axes respectively, then the direction cosines of the vector are: 

Detailed Solution: Question 8

Direction Cosines for angle α, β, γ are:
cos α, cos β, cos γ
Direction Cosines and Direction Ratio of a Line | Definition, Examples, Diagrams

Test: Introduction To Vector Algebra - Question 9
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If a and b are the position vectors of two points A and B and C is a point on AB produced such that AC = 3AB, then position vector of C will be​

Detailed Solution: Question 9

Position vector of A = OA = a
Position vector of B = OB = b
Now, AC = OC - OA
Also, AB = OB - OA
Given, AC = 3AB
⇒ OC - OA = 3(OB - OA)
⇒ OC = 3b - 2a

Test: Introduction To Vector Algebra - Question 10
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Vector of magnitude 1 is called.​

Detailed Solution: Question 10

A vector whose magnitude (i.e., length) is equal to 1 is called a unit vector. There are exactly two unit vectors in any given direction and one is the negative of the other.

Test: Introduction To Vector Algebra - Question 11
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If  and , then the value of scalars x and y are:

Detailed Solution: Question 11

Given, a = i + 2j
b = -2i + j
c = 4i +3j
Also, c = xa +yb
Now putting the values in above equation,
4i + 3j  = x(i + 2j) + y(-2i +j)
⇒ xi + 2xj - 2yi + yj
⇒ (x-2y)i + (2x+y)j
We get,
x - 2y = 4
2x + y = 3
After solving,            
x = 2
y = -1

Test: Introduction To Vector Algebra - Question 12
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The direction of zero vector.​

Detailed Solution: Question 12

Zero vector is the unit vector having zero length, hence the direction is undefined.

Test: Introduction To Vector Algebra - Question 13
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The unit vector in the direction of , where A and B are the points (2, – 3, 7) and (1, 3, – 4) is:

Detailed Solution: Question 13

Given, Point A (2,-3,7)

Point B (1,3,-4)

Let vector in the direction of AB be C.

∴ C = B - A

⇒ (1,3,-4) - (2,-3,7)

⇒ ( 1-2 , 3+3 , -4-7 )

⇒ (-1,6,-11)

⇒ -1i + 6j -11k
Magnitude of vector C
|C| = √(-1)2 + 62 + (-11)2
⇒ √1+36+121
⇒ √158

Unit vector = (Vector)/(Magnitude of vector)
Unit vector C = (C vector)/(Magnitude of C vector)  = (-1i + 6j -11k)/√158

Test: Introduction To Vector Algebra - Question 14
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If a be magnitude of vector  then​

Detailed Solution: Question 14

Since a is the magnitude of the vector, it is always positive and it can be 0 in case of zero vectors.        
So, a ≥ 0

Test: Introduction To Vector Algebra - Question 15
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A vector of magnitude 14 units, which is parallel to the vector

Detailed Solution: Question 15

Given vector = i + 2j - 3k
Magnitude = √12 + 22 + (-3)2 = √14
Unit vector in direction of resultant = (i + 2j - 3k) / √14
Vector of magnitude 14​ unit in direction of resultant,
⇒ 14[ (i + 2j - 3k) / √14 ]
⇒ √14(i + 2j - 3k)

Test: Introduction To Vector Algebra - Question 16
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For any two vectors a and b​, we always have

Detailed Solution: Question 16

|a + b|2 = |a|2 + |b|2 + 2|a||b|.cosθ
|a|2 + |b|2 = |a|2 + |b|2 + 2|a| + |b|  ∵ −1 ⩽ cosθ ⩽ 1
⇒ 2|a||b|.cosθ ⩽ 2|a||b|
So, |a + b|2 ⩽ (|a| + |b|)2
⇒ |a + b| ≤ |a| + |b|
This is also known as Triangle Inequality of vectors.

Test: Introduction To Vector Algebra - Question 17
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If l, m, n are the direction cosines of a position vector  then which of the following is true?

Detailed Solution: Question 17

Consider   is the position vector of a point M(x,y,z) and α, β, γ are the angles, made by the vector   with the positive directions of x, y and z respectively. The cosines of the angles, cos⁡α, cos⁡β, cos⁡γ are the direction cosines of the vector   denoted by l, m, n, then
cos2⁡α + cos2⁡β+ cos2⁡γ =1  i.e.l2 + m2 + n2 = 1.

Test: Introduction To Vector Algebra - Question 18
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A vector whose initial and terminal points coincide, is called​

Detailed Solution: Question 18

A vector whose initial and terminal points coincide has no particular direction and 0 magnitude. Therefore, it is called zero vector.

Test: Introduction To Vector Algebra - Question 19
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A point from a vector starts is called______and where it ends is called its______.​

Detailed Solution: Question 19

A vector is a specific quantity drawn as a line segment with an arrowhead at one end. It has an initial point, where it begins, and a terminal point, where it ends. A vector is defined by its magnitude, or the length of the line, and its direction, indicated by an arrowhead at the terminal point.

Test: Introduction To Vector Algebra - Question 20
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If  are position vectors of the points (- 1, 1) and (m, – 2). then for what value of m, the vectors  are collinear. 

Detailed Solution: Question 20

To determine the value of m for which the vectors a and b are collinear, we use the condition that two vectors are collinear if one is a scalar multiple of the other.

The position vector of the point (-1, 1) is:

a = (-1, 1)

The position vector of the point (m, -2) is:

b = (m, -2)

For the vectors to be collinear, the direction ratios must be proportional. That is:

m / -1 = -2 / 1

Solving this:

m = 2

Thus, the vectors are collinear when m = 2.

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This test contains 20 MCQs on Introduction To Vector Algebra with detailed solutions for each question. It is part of the Chapter-wise Tests for JEE Main & Advanced course on EduRev, designed to align with the current JEE syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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