NCERT Solutions for Chapter 10 Vector Algebra Class 12 Maths Exercise 10.4

Class 12 Maths NCERT Solutions for Chapter 10 Vector Algebra Exercise 10.4

Vector Algebra Exercise 10.4 Solutions

1. Find |a⃗×b⃗|, if a⃗ = i ^-7j ^+7k^ and b⃗=3i ^-2j ^+2k^ .

Solution

We have,
a⃗ = i ^-7j ^+7k^  and b⃗ = 3i ^-2j ^+2k^ . 

2. Find a unit vector perpendicular to each of the vector a⃗+b⃗ and a⃗-b⃗, where a⃗ = 3i ^+2j ^+2k^ and b⃗ = i ^-2j ^-2k^ . 

Solution

We have,  
 a⃗ = 3i ^+2j ^+2k^ and b⃗ = i ^-2j ^-2k^ . 
∴  a⃗+ b⃗ = 4i ^+4j ^,  a⃗ - b⃗ = 2i ^ + 4k^ . 

Hence, the unit vector perpendicular to each of the vector a⃗+ b⃗ and a⃗ - b⃗, is given by the relation, 

3. If a unit vector a⃗ makes an angles π/3 with i ^, π/4 with j ^ and an acute angle θ with k^ , then find θ and hence, the compounds of  a⃗ .

Solution

Let unit vector  a⃗  have (a₁, a₂, a₃) components. 

Since a⃗ is a unit vector, |a⃗| = 1.
Also, it is given that a⃗ makes angles π/3 with i ^, π/4 with j ^ and an acute angle θ with k^ .
Then, we have :  

Hence, θ = π/3 and the components of a⃗ are (1/2, 1/√2, 1/2).

4. Show that (a⃗ -  b⃗) × (a⃗+ b⃗) = 2(a⃗ × b⃗)  

Solution

5. Find λ and μ if (2i ^-6j ^-27k^) × (i ^+ λj ^- μk^) = 0⃗ .

Solution

On comparing the corresponding components, we have : 

6μ - 27λ = 0 
2μ - 27 = 0 
2λ - 6 = 0 
Now, 

2λ - 6 = 0

⇒ λ = 3
2μ - 27 = 0

⇒ μ = 27/2
Hence, λ = 3 and μ = 27/2. 

6. Given that a⃗ . b⃗ = 0 and a⃗× b⃗ = 0. What can you conclude about the vectors a⃗ and b⃗ .

Solution

a⃗ . b⃗ = 0
Then,

7. Let the vectors a⃗ , b⃗ , c⃗ = given as . Then show that = a⃗×(b⃗ + c⃗) = a⃗×b⃗ + a⃗×c⃗ .

Solution

We have, 

On adding (2) and (3), we get : 

Now, from (1) and (4), we have :
 a⃗×(b⃗ + c⃗) = a⃗ × b⃗ +  a⃗× c⃗

Hence, the given result is proved.

8. If either a⃗ = 0⃗ or b⃗ = 0⃗ then a⃗ × b⃗ = 0⃗. Is the converse true? Justify your answer with an example. 

Solution

Take any parallel non - zero vectors so that a⃗ × b⃗ = 0⃗  

It can now be observed that :  

Hence, the converse of the given statement need not be true. 

9. Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

Solution

The vertices of triangle ABC are given as  A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

The adjacent sides AB⃗ ×BC⃗ of ΔABC are given as : 

Hence, the area of ΔABC is √61/2 square units.

10. Find the area of the parallelogram whose adjacent sides are determined by the vector a⃗ = i ^- j ^+3k^ and b⃗ = 2i ^-7j ^+k^ .

Solution

The area of the parallelogram whose adjacent sides are a⃗ and b⃗ is | a⃗× b⃗|.

Adjacent sides are given as : 
  a⃗ = i ^ - j ^ +3k^  and b⃗ = 2i ^ - 7j ^ + k^  .  

Hence, the area of the given parallelogram is 15√2 square units.

11. Let the vectors a⃗ and b⃗ be such that |a⃗| = 3 and |b⃗| = √2/3, then a⃗ × b⃗ is a unit vector, if the angle between a⃗ and b⃗ is:
(A) π/6 
(B) π/4
(C) π/3
(D) π/2 

Solution 

It is given that |a⃗| = 3 and |b⃗| = √2/3. 
We know that a⃗×b⃗ = |a⃗||b⃗| sinθ n ^, where n ^ is a unit vector perpendicular to both a⃗ and b⃗ θ is the angle between a⃗ and b⃗ . 
Now, a⃗ × b⃗ is a unit vector if |a⃗ × b⃗| = 1. 

Hence, a⃗ × b⃗ is a unit vector if the angle between a⃗ and b⃗ is π/4. 
The correct answer is B.

12. Area of a rectangle having vertices, A, B, C and D with position vectors -i ^+ (1/2)j ^+ 4k^ ,  i ^+(1/2)j ^ +4k^ , and -i ^-(1/2)j ^+4k^ and respectively is 
(A) 1/2 
(B) 1 
(C) 2 
(D) 4 

Solution

The position vectors of vertices A, B, C,, and D of rectangle ABCD are given as : 

Now, it is known that the area of a parallelogram whose adjacent sides are a⃗ and b⃗ is |a⃗×b⃗|.
Hence, the area of the given rectangle is |AB⃗×BC⃗| = 2 square units.
The correct answer is C.