Question 1

Exercise 4.1
1. Represent graphically the following vectors in the Cartesian plane:
(i) a = 2i + 3j
(ii) b = -i + 4j
(iii) c = 3i - 2j

Accepted Answer

To represent the vectors graphically:
(i) a = 2i + 3j
This vector starts from the origin (0,0) and ends at the point (2,3).
Draw an arrow from (0,0) to (2,3).
(ii) b = -i + 4j
This vector starts from the origin and ends at (-1,4).
Draw an arrow from (0,0) to (-1,4).
(iii) c = 3i - 2j
This vector starts from the origin and ends at (3,-2).
Draw an arrow from (0,0) to (3,-2). — Each vector is written in terms of i and j, which represent the x and y axes respectively. The coefficients give the coordinates of the terminal point when the vector starts at the origin.

Question 2

Exercise 4.1
2. Write each of the following vectors in terms of i, j, k:
(i) The vector from A(1, 2, 3) to B(4, 6, 8)
(ii) The vector from P(-2, 0, 5) to Q(3, -4, 7)

Accepted Answer

(i) The vector from A(1, 2, 3) to B(4, 6, 8) is given by:
AB = (4 - 1)i + (6 - 2)j + (8 - 3)k = 3i + 4j + 5k
(ii) The vector from P(-2, 0, 5) to Q(3, -4, 7) is:
PQ = (3 - (-2))i + (-4 - 0)j + (7 - 5)k = 5i - 4j + 2k — The vector from point A(x1, y1, z1) to B(x2, y2, z2) is (x2 - x1)i + (y2 - y1)j + (z2 - z1)k.

Question 3

Exercise 4.1
3. Find the magnitude of the following vectors:
(i) a = 2i + 3j + 6k
(ii) b = -i + 4j - 2k

Accepted Answer

(i) |a| = sqrt(2^2 + 3^2 + 6^2) = sqrt(4 + 9 + 36) = sqrt(49) = 7
(ii) |b| = sqrt((-1)^2 + 4^2 + (-2)^2) = sqrt(1 + 16 + 4) = sqrt(21) — The magnitude of a vector a = ai + bj + ck is |a| = sqrt(a^2 + b^2 + c^2). Substitute the values and simplify.

Question 4

Exercise 4.1
4. If a = 2i - j + k and b = i + 2j - 3k, find a + b and a - b.

Accepted Answer

a + b = (2i - j + k) + (i + 2j - 3k) = (2+1)i + (-1+2)j + (1-3)k = 3i + j - 2k
a - b = (2i - j + k) - (i + 2j - 3k) = (2-1)i + (-1-2)j + (1-(-3))k = i - 3j + 4k — Add and subtract the corresponding components of the vectors.

Question 5

Exercise 4.1
5. Find a unit vector in the direction of the vector a = 3i - 4j + k.

Accepted Answer

First, find the magnitude of a:
|a| = sqrt(3^2 + (-4)^2 + 1^2) = sqrt(9 + 16 + 1) = sqrt(26)
Unit vector in the direction of a is:
\hat{a} = (1/|a|) * a = (1/sqrt(26)) * (3i - 4j + k) — A unit vector in the direction of a vector a is given by a/|a|.

Question 6

Exercise 4.1
6. If a = 2i + 3j - k and b = -i + 4j + 2k, find the scalar product a · b.

Accepted Answer

a · b = (2i + 3j - k) · (-i + 4j + 2k)
= (2)(-1) + (3)(4) + (-1)(2)
= -2 + 12 - 2 = 8 — The scalar (dot) product is calculated by multiplying corresponding components and adding: a · b = a1b1 + a2b2 + a3b3.

Question 7

Exercise 4.1
7. Find the angle between the vectors a = i + 2j + 2k and b = 2i + j + k.

Accepted Answer

Let θ be the angle between a and b.
First, compute a · b:
a · b = (1)(2) + (2)(1) + (2)(1) = 2 + 2 + 2 = 6
|a| = sqrt(1^2 + 2^2 + 2^2) = sqrt(1 + 4 + 4) = sqrt(9) = 3
|b| = sqrt(2^2 + 1^2 + 1^2) = sqrt(4 + 1 + 1) = sqrt(6)
cosθ = (a · b)/(|a||b|) = 6/(3*sqrt(6)) = 2/sqrt(6) = sqrt(6)/3
θ = cos^{-1}(sqrt(6)/3) — The angle θ between two vectors is given by cosθ = (a · b)/(|a||b|). Calculate dot product and magnitudes, then use inverse cosine.

Question 8

Exercise 4.1
8. Find the vector product of the vectors a = i + j + k and b = 2i - j + k.

Accepted Answer

a × b = |i   j   k|
         |1   1   1|
         |2  -1   1|
= i[(1)(1) - (1)(-1)] - j[(1)(1) - (1)(2)] + k[(1)(-1) - (1)(2)]
= i[1 + 1] - j[1 - 2] + k[-1 - 2]
= 2i + j - 3k — The cross product is calculated using the determinant of a 3x3 matrix with unit vectors and the components of the two vectors.

Question 9

Exercise 4.1
9. Find the scalar triple product of the vectors a = i + 2j + 3k, b = 2i + j + k, c = i + j + k.

Accepted Answer

The scalar triple product is a · (b × c).
First, compute b × c:
|i   j   k|
|2   1   1|
|1   1   1|
= i[(1)(1) - (1)(1)] - j[(2)(1) - (1)(1)] + k[(2)(1) - (1)(1)]
= i[1 - 1] - j[2 - 1] + k[2 - 1]
= 0i - 1j + 1k
So, b × c = -j + k
Now, a · (b × c) = (i + 2j + 3k) · (-j + k) = (1)(0) + (2)(-1) + (3)(1) = 0 - 2 + 3 = 1 — First, compute the cross product b × c, then take the dot product with a.

Question 10

Which of the following quantities is a vector?

Accepted Answer

Displacement — Displacement is a vector quantity as it has both magnitude and direction. Mass and temperature are scalars, and speed is also a scalar as it does not have direction.

Question 11

A vector is represented graphically by a _____ with an arrow indicating direction.

Accepted Answer

directed line segment — A vector is shown as a directed line segment where the length represents magnitude and the arrow indicates direction.

Question 12

Which of the following best describes a free vector?

Accepted Answer

A vector that can be moved anywhere in space without changing its magnitude or direction — A free vector can be translated parallel to itself in space without changing its effect, as only its magnitude and direction matter.

Question 13

If vector $\vec{a}$ has components (2, 3, -1), what is its magnitude?

Accepted Answer

\sqrt{14} — Given: $\vec{a} = (2, 3, -1)$
Find: Magnitude $|\vec{a}|$
Formula: $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$
Solution:
Step 1: Substitute values: $|\vec{a}| = \sqrt{2^2 + 3^2 + (-1)^2}$
Step 2: $= \sqrt{4 + 9 + 1}$
Step 3: $= \sqrt{14}$
Answer: $\sqrt{14}$
Note: Students may forget to square the negative component.

Question 14

Assertion (A): The zero vector has no specific direction.
Reason (R): The magnitude of the zero vector is zero.

Accepted Answer

A) Both A and R are true and R is the correct explanation of A. The zero vector has zero magnitude, and hence, its direction is not defined. The reason correctly explains the assertion. — Both statements are correct. The zero vector's magnitude is zero, and because direction is undefined for a vector with zero length, the reason explains the assertion.

Question 15

Which of the following pairs of vectors are equal?

Accepted Answer

Two vectors with the same magnitude and direction, regardless of initial points — Equal vectors have both the same magnitude and direction, even if their initial points differ.

Vector Algebra

Vector Algebra — Study Notes

Introduction

Types of Vectors

Algebra of Vectors

Practice Questions — Vector Algebra

All 7 Chapters in Mathematics Part-II