The dot product of two vectors questions and solutions are provided here to assist students of Class 12. As we know, dot products (scalar products) of two vectors is one of the essential concepts of Class 12 mathematics. In this article, you will learn how to solve various problems in vector algebra that involve the dot product of two vectors.
What is the dot product of two vectors?
The dot product (or scalar product) of two vectors .i.e non-zero vectors; say a and b is denoted by a.b and is written as:
a.b = |a| |b| cos θ
Here, θ is the angle between two vectors a and b, such that 0 ≤ θ ≤ π.
Also check: Dot product of two vectors
Dot Product of Two Vectors Questions and Answers
1. Suppose a = -2i + 3j + 5k and b = i + 2j + 3k are two vectors, then find the value of the dot product of these two vectors.
Solution:
Given,
a = -2i + 3j + 5k and b = i + 2j + 3k
As we know, the dot product of two vectors a = a1i + a2j + a3k and b = b1i + b2j + b3k is a.b = a1b1 + a2b2 + a3b3.
Thus, a.b = (-2)(1) + (3)(2) + (5)(3)
= -2 + 6 + 15
= 19
2. Find the angle between the vectors i – 2j + 3k and 3i – 2j + k.
Solution:
Let the given vectors be:
a = i – 2j + 3k and b = 3i – 2j + k
|a| = √[12 +(-2)2 + 32] = √(1 + 4 + 9) = √14
And
|b| = √[32 +(-2)2 + 12] = √(9 + 4 + 1) = √14
Now, a.b = (i – 2j + 3k) (3i – 2j + k)
= (1)(3) + (-2)(-2) + (3)(1)
= 3 + 4 + 3
= 10
As we know, a.b = |a| |b| cos θ
cos θ = (a.b)/ |a| |b|
= 10/ (√14 √14)
= 10/14
= 5/7
⇒ θ = cos-1(5/7)
Thus, the angle between the given two vectors is θ = cos-1(5/7).
3. What is the angle between two vectors a and b with magnitudes √3 and 4, respectively, and a.b = 2√3?
Solution:
Let θ be the angle between two vectors a and b.
Given,
|a| = √3, |b| = 4 and a.b = 2√3
We know that,
a.b = |a| |b| cos θ
⇒ cos θ = (a.b)/ |a| |b|
= (2√3)/(√3 × 4)
= ½
⇒ cos θ = cos π/3
⇒ θ = π/3
4. Find |x| if for a unit vector a, (x – a) . (x + a) = 12.
Solution:
Given,
(x – a) . (x + a) = 12
x.x + x.a – a.x – a.a = 12
|x|2 – |a|2 = 12
|x|2 – 12 = 12 {since a is a unit vector such that |a| = 1}
|x|2 = 12 + 1 = 13
⇒ |x| = √13
5. Find the value of λ for which the two vectors 2i – j + 2k and 3i + λj + k are perpendicular.
Solution:
Let the given vectors be:
a = 2i – j + 2k and b = 3i + λj + k
As we know, if two vectors are perpendicular to each other, then their dot or scalar product is equal to 0.
So, a . b = 0
(2i – j + 2k) . (3i + λj + k) = 0
(2)(3) + (-1)(λ) + (2)(1) = 0
6 – λ + 2 = 0
8 – λ = 0
λ = 8
6. If a = 5i – j – 3k and b = i + 3j – 5k, then show that the vectors a + b and a – b are perpendicular.
Solution:
As we know, if two vectors are perpendicular to each other, then their dot or scalar product is equal to 0.
Given,
a = 5i – j – 3k and b = i + 3j – 5k
Now,
a + b = (5i – j – 3k) + (i + 3j – 5k)
= (5 + 1)i (-1 + 3)j (-3 – 5)k
= 6i + 2i – 8k
a – b = (5i – j – 3k) – (i + 3j – 5k)
= (5 – 1)i (-1 – 3)j (-3 + 5)k
= 4i – 4i + 2k
Let’s find the dot product of a + b and a – b.
(a + b) . (a – b) = (6i + 2i – 8k) . (4i – 4i + 2k)
= (6)(4) + (2)(-4) + (-8)(2)
= 24 – 8 – 16
= 0
Therefore, a + b and a – b are perpendicular vectors.
7. If a and b are two vectors, such that |a| = 3, |b| = 2 and a . b = 4, then find the value of |a – b|.
Solution:
Given,
|a| = 3, |b| = 2 and a . b = 4
Consider |a – b|2 = (a – b) . (a – b)
= a.a – a.b – b.a + b.b
= |a|2 – 2(a.b) + |b|2 {since a.b = b.a}
= (3)2 – 2(4) + (2)2
= 9 – 8 + 4
= 5
⇒ |a – b| = 5
8. If a = 2i – j + k, b = i + j – 2k and c = i + 3j – k, find λ such that a is perpendicular to λb + c.
Solution:
Given,
a = 2i – j + k, b = i + j – 2k and c = i + 3j – k
λb + c = λ(i + j – 2k) + (i + 3j – k)
= (λ + 1)i + (λ + 3)j + (-2λ – 1)k
= (λ + 1)i + (λ + 3)j – (2λ + 1)k
If a is perpendicular to λb + c, then we have a . (λb + c) = 0
⇒ (2i – j + k) . [(λ + 1)i + (λ + 3)j – (2λ + 1)k] = 0
⇒ 2(λ + 1) + (-1)(λ + 3) + (1) (-2λ – 1) = 0
⇒ 2λ + 2 – λ – 3 – 2λ – 1 = 0
⇒ -λ – 2 = 0
⇒ λ = -2
9. Find the cosine angle between the vectors i – j and j – k.
Solution:
Let the given vectors be:
a = i – j and b = j = k
Now, |a| = √[12 + (-1)2] = √(1 + 1) = √2
|b| = √[(-1)2 + 12] = √(1 + 1) = √2
a.b = (i – j) . (j – k)
= (1)(0) + (-1)(1) + (0)(-1)
= -1
As we know, a.b = |a| |b| cos θ
cos θ = (a.b)/ |a| |b|
= -1/ (√2 √2)
= -½
⇒ θ = cos-1(-½)
10. Find |a| and |b| if (a + b) . (a – b) = 8 and |a| = 8|b|.
Solution:
Given,
(a + b) . (a – b) = 8
a.a – a.b + b.a – b.b = 8
|a|2 – |b|2 = 8 {since a.b = b.a .i.e dot product of two vectors is commutative}
(8|b|)2 – |b|2 = 8 {given |a| = 8|b|}
64|b|2 – |b|2 = 8
63|b|2 = 8
|b|2 = 8/63
|b| = √(8/63) = 2√2/3√7
|a| = 8|b|
= 8 (2√2/3√7)
= 16√2/3√7
Therefore, |a| = (16√2)/ (3√7) and |b| = (2√2)/(3√7).
Practice Questions on Dot Product of Two Vectors
- Find the angle between two vectors 2i – j + k and 3i + 4j – k.
- If a is a unit vector and (x – a) . (x + a) = 8, then find the value of |x|.
- Find the value of λ such that the vectors a = 2i + λj + k and b = i + 2j + 3k are orthogonal.
- If a, b, c are unit vectors, such that a + b + c = 0, then what will be the value of a.b + b.c + c.a?
- Suppose a, b, c are three vectors, such that |a| = 3, |b| = 4, |c| = 5 and each one of them being perpendicular to the sum of the other two vectors. Find the value of |a + b + c|.
