Class 12 RD Sharma Solutions – Chapter 24 Scalar or Dot Product – Exercise 24.1 | Set 2

Last Updated : 15 Feb, 2022

Question 17. If $\vec{α}=3\hat{i}+4\hat{j}+5\hat{k}$ and $\vec{β}=2\hat{i}+\hat{j}-4\hat{k}$ , then express $\vec{β}$ in the form of $\vec{β}=\vec{β_1}+\vec{β_2}$ where $\vec{β_1}$ is parallel to $\vec{α}$ and $\vec{β_2}$ is perpendicular to $\vec{α}$ .

Solution:

Given, $\vec{α}=3\hat{i}+4\hat{j}+5\hat{k}$
$\vec{β}=2\hat{i}+\hat{j}-4\hat{k}$
According to question
$\vec{β_1} = λ\vec{α}$ also $\vec{β_2}.\vec{α}$ = 0
Now,
$\vec{β_1} = λ(3\hat{i}+4\hat{j}+5\hat{k})$
⇒ $\vec{β_1} = 3λ\hat{i}+4λ\hat{j}+5λ\hat{k}$
$\vec{β_2}=\vec{β}-\vec{β_1}$
⇒ $\vec{β_2}=2\hat{i}+\hat{j}-4\hat{k}-(3λ\hat{i}+4λ\hat{j}+5λ\hat{k})$
⇒ $\vec{β_2}=(2-3λ)\hat{i}+(1-4λ)\hat{j}-(4+5λ)\hat{k}$
Now,
$\vec{β_2}.\vec{α}=0$
⇒ $[(2-3λ)\hat{i}+(1-4λ)\hat{j}-(4+5λ)\hat{k}](3\hat{i}+4\hat{j}+5\hat{k})=0$
⇒ 3(2-3λ)+4(1-4λ)-5(4+5λ) = 0
⇒ 6-9λ+4-16λ-20-25λ = 0
⇒ -10 -50λ = 0
⇒ λ = -1/5
$\vec{β_1} = -3/5\hat{i}-4/5\hat{j}-1\hat{k}$
$\vec{β_2} = 13/5\hat{i}+9/5\hat{j}-3\hat{k}$

Question 18. If either $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}$ , then $\vec{a}.\vec{b}=0$ . But, The converse need not be true. Justify your answer with an example.

Solution:

Given,
$\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}$ then $\vec{a}.\vec{b}=0$
Suppose $\vec{a}=2\hat{i}+\hat{j}+\hat{k}$
$\vec{b}=\hat{i}-\hat{j}-\hat{k}$
$\vec{a}.\vec{b}=0$
But,
$|\vec{a}|$ = √(2)²+(1)²+(1)²
= √4+1+1
= √6 ≠ 0
$|\vec{b}|$ = √(1)²+(1)²+(1)²
= √3 ≠ 0
Hence Proved

Question 19. Show that the vectors $\vec{a}=3\hat{i}-2\hat{j}+\hat{k}, \vec{b}=\hat{i}-3\hat{j}+5\hat{k},\vec{c}=2\hat{i}+\hat{j}-4\hat{k}$ form a right-angled triangle.

Solution:

Given, $\vec{a}=3\hat{i}-2\hat{j}+\hat{k}$
$\vec{b}=\hat{i}-3\hat{j}+5\hat{k}$
$\vec{c}=2\hat{i}+\hat{j}-4\hat{k}$
To prove given vectors form a right angle triangle
$|\vec{a}|$ = √(3²+(-2)²+1²) = √14
$|\vec{b}|$ = √(1²+(-3)²+5²) = √35
$|\vec{c}|$ = √(2²+1²+(-4)²) = √21
$|\vec{a}|^2+|\vec{c}|^2 = (\sqrt14)^2+(\sqrt21)^2$
= 14 + 21 = 35
Since, $|\vec{b}|^2=|\vec{a}|^2+|\vec{c}|^2$ (Pythagoras Theorem)
Hence, $\vec{a},\vec{b}$ and $\vec{c}$ form a right angled triangle.

Question 20. If $\vec{a}=2\hat{i}+2\hat{j}+3\hat{k}$ , $\vec{b}=-\hat{i}+2\hat{j}+\hat{k}$ and $\vec{c}=3\hat{i}+\hat{j}$ are such that $\vec{a}+λ\vec{b}$ is perpendicular to $\vec{c}$ , then find the value of λ.

Solution:

Given:
$\vec{a}=2\hat{i}+2\hat{j}+3\hat{k}$
$\vec{b}=-\hat{i}+2\hat{j}+\hat{k}$
$\vec{c}=3\hat{i}+\hat{j}$
Now, $(\vec{a}+λ\vec{b}).\vec{c}=0$
⇒ $[2\hat{i}+2\hat{j}+3\hat{k}+λ(-\hat{i}+2\hat{j}+\hat{k})](3\hat{i}+\hat{j})=0$
⇒ $[(2-λ)\hat{i}+(2+2λ)\hat{j}+(3+λ)\hat{k}](3\hat{i}+\hat{j})=0$
⇒ (2 – λ)3 + (2 + 2λ) + 0 = 0
⇒ 6 – 3λ + 2 + 2λ =0
⇒ λ = 8

Question 21. Find the angles of a triangle whose vertices are A (0, -1. -2), B (3, 1, 4), and C(5, 7, 1).

Solution:

Given that angle of a triangle whose vertices are A (0, -1. -2), B (3, 1, 4) and C(5, 7, 1).
$\vec{AB}=(\vec{B}-\vec{A}) =3\hat{i}+2\hat{j}+6\hat{k}$
$\vec{BC}=(\vec{C}-\vec{B}) =2\hat{i}+6\hat{j}-3\hat{k}$
$\vec{AC}=(\vec{C}-\vec{A}) =5\hat{i}+8\hat{j}+3\hat{k}$
$|\vec{AB}|=\sqrt{3^2+2^2+6^2}=7$
$|\vec{BC}|=\sqrt{2^2+6^2+(-3)^2}=7$
$|\vec{AC}|=\sqrt{5^2+8^2+3^2}$
= √98 = 7√2
Now,
$\vec{AB}.\vec{BC}$ = (3 × 2 + 2 × 6 – 6 × 3) = 0
Thus, we can say AB is perpendicular to BC.
Hence, AB = BC = 7, ∠A =∠C and ∠B = 90°
∠A + ∠B + ∠C = 180°
2∠A = 180° – 90°
∠A = 45°
∠C = 45°
∠B = 90°

Question 22. Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$ , having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.

Solution:

We know $\vec{a}.\vec{b}=|\vec{a}||\vec{b}|cos θ$
⇒ 1/ 2 = $|\vec{a}||\vec{a}|cos θ$
⇒ 1/2 = $|\vec{a}|^2$ (1/2)
⇒ $|\vec{a}| = 1$
or
⇒ $|\vec{a}| = |\vec{b}| = 1$

Question 23. Show that the points whose position vector are $\vec{a} =4\hat{i}-3\hat{j}+\hat{k},\vec{b} =2\hat{i}-4\hat{j}+5\hat{k}, \vec{c} =\hat{i}-\hat{j}$ form a right triangle.

Solution:

Given that positions vectors
$\vec{a} =4\hat{i}-3\hat{j}+\hat{k}$
$\vec{b} =2\hat{i}-4\hat{j}+5\hat{k}$
$\vec{c} =\hat{i}-\hat{j}$
Now,
$\vec{AB} = (\vec{b}-\vec{a})$
⇒ $\vec{AB} =-2\hat{i}-\hat{j}+4\hat{k}$
$\vec{BC} = (\vec{c}-\vec{b})$
⇒ $\vec{BC} =-\hat{i}+3\hat{j}-5\hat{k}$
$\vec{CA} = (\vec{a}-\vec{c})$
⇒ $\vec{CA} =3\hat{i}-2\hat{j}+\hat{k}$
Now, $\vec{AB}.\vec{BC} =(-2\hat{i}-\hat{j}+4\hat{k})(-\hat{i}+3\hat{j}-5\hat{k})$
= 2 – 3 – 20 = -21
$\vec{BC}.\vec{CA} =(-\hat{i}+3\hat{j}-5\hat{k})(3\hat{i}-2\hat{j}+\hat{k})$
= -3 – 6 – 5 = -14
$\vec{AB}.\vec{CA} =(-2\hat{i}-\hat{j}+4\hat{k})(3\hat{i}-2\hat{j}+\hat{k})$
= -6 + 2 + 4 = 0
So, AB is perpendicular to CA or the given position vectors form a right-angled triangle.

Question 24. If the vertices A, B, C of △ABC have position vectors (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively, what is the magnitude of ∠ABC?

Solution:

Given the vertices of △ABC are A(1, 2, 3), B(-1, 0, 0), C(0, 1, 2)
Now, $\vec{AB} = (-1-1)\hat{i}+(0-2)\hat{j}+(0-3)\hat{k}$
= $-2\hat{i}-2\hat{j}-3\hat{k}$
Or, $\vec{BA}=2\hat{i}+2\hat{j}+3\hat{k}$
$\vec{BC}=\hat{i}+\hat{j}+2\hat{k}$
We know that $\vec{BA}.\vec{BC} = |\vec{BA}||\vec{BC}|$
$\vec{BA}.\vec{BC} =$ (2 × 1) + (2 × 1) + (3 × 2)
= 2 + 2 + 6 = 10
Now, $|\vec{BA}|=\sqrt{4+4+9}$ = √17
$|\vec{BC}|=\sqrt{1+1+4}$ = √6
Therefore,
cos θ = $= \frac{\vec{BA}.\vec{BC}}{|\vec{BA}||\vec{BC}|}$
cos θ = 10/ √(17×6)
θ = cos^-1(10/√102)

Question 25. If A, B, C have position vectors (0, 1, 1), (3, 1, 5), (0, 3, 3) respectively, show that △ABC is right-angled at C.

Solution:

Given, position vectors A(0, 1, 1), B(3, 1, 5), C(0, 3, 3)
Now, $\vec{AC} = (0-0)\hat{i}+(3-1)\hat{j}+(3-1)\hat{k}$
$=2\hat{j}+2\hat{k}$
$\vec{BC}=(0-3)\hat{i}+(3-1)\hat{j}+(3-5)\hat{k}$
= $-3\hat{i}+2\hat{j}-2\hat{k}$
$\vec{AC}.\vec{BC}=(2\hat{j}+2\hat{k}).(-3\hat{i}+2\hat{j}-2\hat{k})$
= 2 × 2 – 2 × 2 = 0
Thus, $\vec{AC}$ and $\vec{BC}$ are perpendicular hence △ABC is right-angled at C

Question 26. Find the projection of $\vec{b}+\vec{c}$ on $\vec{a}$ , where $\vec{a}=2\hat{i}-2\hat{j}+\hat{k}, \vec{b}=\hat{i}+2\hat{j}-2\hat{k}$ and $\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$ .

Solution:

Given:
$\vec{a}=2\hat{i}-2\hat{j}+\hat{k}$
$|\vec{a}|=\sqrt{4+4+1} = 3$
$\vec{b}=\hat{i}+2\hat{j}-2\hat{k}$
$\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$
To find the projection of $\vec{b}+\vec{c}$ on $\vec{a}$
$\vec{b}+\vec{c} = 3\hat{i}+\hat{j}+2\hat{k}$
Now, Projection of $\vec{b}+\vec{c}$ = $[\frac{(\vec{b}+\vec{c})\vec{a}}{|\vec{a}|^2}]\vec{a}$
= $[\frac{6-2+2}{3^2}](2\hat{i}-2\hat{j}+\hat{k})$
= 6/9 × 3
= 2

Question 27. If $\vec{a}=5\hat{i}-\hat{j}-3\hat{k}$ and $\vec{b}=\hat{i}+3\hat{j}-5\hat{k}$ , then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are orthogonal.

Solution:

Given: $\vec{a}=5\hat{i}-\hat{j}-3\hat{k}$
$\vec{b}=\hat{i}+3\hat{j}-5\hat{k}$
To prove
$(\vec{a}+\vec{b})(\vec{a}-\vec{b})=0$
Taking LHS
= $|\vec{a}|^2-|\vec{b}|^2$
= $\sqrt{5^2+1+3^2}-\sqrt{1+3^2+5^2}$
= √35 – √35
= 0
Thus, the given vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are orthogonal.

Question 28. A unit vector $\vec{a}$ makes angle π/2 and π/3 with $\hat{i}$ and $\hat{j}$ respectively and an acute angle θ with $\hat{k}$ . Find the angle θ and components of $\vec{a}$ .

Solution:

Let us assume $\vec{a}=a_1i + a_2j+a_3k$
We know that
a₁²+ a₂²+ a₃² = 1 ….(1)
So,
$\vec{a}i=a_1$
$|\vec{a}||i|cos\frac{π}{4}=a_1$
(1)(1)(1/√2) = a₁
a₁= 1/√2
Again we take
$\vec{a}j=a_2$
$|\vec{a}||j|cos\frac{π}{3}=a_2$
(1)(1)(1/2) = a₂
a₂= 1/2
Put all these values in eq(1) to find the value of a₃
(1/√2)²+ (1/2)²+ a₃² = 1 ….(1)
a₃² = 1/4
a₃ = 1/2
Now we find the value of θ
$\vec{a}k=a_3$
$|\vec{a}||k|cos\theta=a_3$
(1)(1)cosθ = 1/2
cosθ = 1/2
cosθ = π/3
and components of $\vec{a}$
$\vec{a}=\frac{1}{\sqrt{2}}i + \frac{1}{2}j+\frac{1}{2}k$

Question 29. If two vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}|$ = 2, $|\vec{b}|$ = 1, and $\vec{a}.\vec{b}$ =1. Find the value of $(3\vec{a}-5\vec{b}).(2\vec{a}+7\vec{b}).$

Solution:

Given, $(3\vec{a}-5\vec{b}).(2\vec{a}+7\vec{b})$
= $6|\vec{a}|^2+21\vec{a}.\vec{b}-10\vec{a}.\vec{b}-35|\vec{b}|^2$
= $6|\vec{a}|^2+11\vec{a}.\vec{b}-35|\vec{b}|^2$
= 6(2)² + 11(1) – 35(1)²
= 24 + 11 – 35
= 35 – 35 = 0

Question 30. If $\vec{a}$ is a unit vector, then find $|\vec{x}|$ in each of the following:

(i) $(\vec{x}-\vec{a})(\vec{x}+\vec{a})=8$

Solution:

Given, $(\vec{x}-\vec{a})(\vec{x}+\vec{a})=8$
⇒ $|\vec{x}|^2-\vec{x}.\vec{a}+\vec{a}\vec{x}+|\vec{a}|^2=8$
⇒ $|\vec{x}|^2-|\vec{a}|^2=8$
⇒ $|\vec{x}|^2-1=8$
⇒ $|\vec{x}|^2=9$
⇒ $|\vec{x}|=3$

(ii) $(\vec{x}-\vec{a})(\vec{x}+\vec{a})=12$

Solution:

Given, $(\vec{x}-\vec{a})(\vec{x}+\vec{a})=12$
⇒ $|\vec{x}|^2-\vec{x}.\vec{a}+\vec{a}\vec{x}+|\vec{a}|^2=12$
⇒ $|\vec{x}|^2-|\vec{a}|^2=12$
⇒ $|\vec{x}|^2-1=12$
⇒ $|\vec{x}|^2=13$
⇒ $|\vec{x}|$ =√13

Question 31. Find $|\vec{a}|$ and $|\vec{b}|$ , if

(i) $(\vec{a}+\vec{b})(\vec{a}-\vec{b})$ = 12 and $|\vec{a}| = 2 |\vec{b}|$

Solution:

Given, $(\vec{a}+\vec{b})(\vec{a}-\vec{b})$ = 12
⇒ $|\vec{a}|^2-\vec{a}.\vec{b}+\vec{a}\vec{b}-|\vec{b}|^2=12$
⇒ $|\vec{a}|^2-|\vec{b}|^2$ = 12
⇒ $4|\vec{b}|^2-|\vec{b}|^2$ = 12
⇒ $3|\vec{b}|^2$ = 12
⇒ $|\vec{b}|$ = 2
So,
$|\vec{a}|$ = 4

(ii) $(\vec{a}+\vec{b})(\vec{a}-\vec{b})$ = 8 and $|\vec{a}|$ = 8 $|\vec{b}|$

Solution:

Given, $(\vec{a}+\vec{b})(\vec{a}-\vec{b})$ = 8
⇒ $|\vec{a}|^2-\vec{a}.\vec{b}+\vec{a}\vec{b}-|\vec{b}|^2=8$
⇒ $|\vec{a}|^2-|\vec{b}|^2=8$
⇒ $64|\vec{b}|^2-|\vec{b}|^2=8$
⇒ $63|\vec{b}|^2=8$
⇒ $|\vec{b}|$ = √(8/63)
So,
$|\vec{a}|$ = 8√(8/63)

(iii) $(\vec{a}+\vec{b})(\vec{a}-\vec{b})$ = 3 and $|\vec{a}|$ = 2 $|\vec{b}|$

Solution:

Given, $(\vec{a}+\vec{b})(\vec{a}-\vec{b})= 3$
⇒ $|\vec{a}|^2-\vec{a}.\vec{b}+\vec{a}\vec{b}-|\vec{b}|^2=3$
⇒ $|\vec{a}|^2-|\vec{b}|^2 = 3$
⇒ $4|\vec{b}|^2-|\vec{b}|^2=3$
⇒ 3 $|\vec{b}|^2$ = 3
⇒ $|\vec{b}|$ = 1
So,
$|\vec{a}|$ = 2

Question 32. Find $|\vec{a}-\vec{b}|$ , if

(i) $|\vec{a}| = 2, |\vec{b}| = 5$ and $\vec{a}.\vec{b} = 8$

Solution:

We have, $|\vec{a}-\vec{b}|$
⇒ $|\vec{a}-\vec{b}|^2 =|\vec{a}|^2-2\vec{a}.\vec{b}+|\vec{b}|^2$
⇒ $|\vec{a}-\vec{b}|^2$ = 2²– 2 × 8 + 5²
⇒ $|\vec{a}-\vec{b}|^2$ = 4 – 16 + 25
⇒ $|\vec{a}-\vec{b}|^2$ = 13
⇒ $|\vec{a}-\vec{b}|$ = √13

(ii) $|\vec{a}|$ = 3, $|\vec{b}|$ = 4 and $\vec{a}.\vec{b}$ = 1

Solution:

We have, $|\vec{a}-\vec{b}|$
⇒ $|\vec{a}-\vec{b}|^2 =|\vec{a}|^2-2\vec{a}.\vec{b}+|\vec{b}|^2$
⇒ $|\vec{a}-\vec{b}|^2$ = 3²– 2 × 1 + 4²
⇒ $|\vec{a}-\vec{b}|^2$ = 9 – 2 + 16
⇒ $|\vec{a}-\vec{b}|^2$ = 23
⇒ $|\vec{a}-\vec{b}|$ = √23

(iii) $|\vec{a}|=2, |\vec{b}| = 3$ and $\vec{a}.\vec{b}$ = 4

Solution:

We have, $|\vec{a}-\vec{b}|$
⇒ $|\vec{a}-\vec{b}|^2 =|\vec{a}|^2-2\vec{a}.\vec{b}+|\vec{b}|^2$
⇒ $|\vec{a}-\vec{b}|^2$ = 2²– 2 × 4 + 3²
⇒ $|\vec{a}-\vec{b}|^2$ = 4 – 8 + 9
⇒ $|\vec{a}-\vec{b}|^2$ = 5
⇒ $|\vec{a}-\vec{b}|$ = √5

My Personal Notes

1. Class 12 RD Sharma Solutions - Chapter 24 Scalar or Dot Product - Exercise 24.1 | Set 1

2. Class 12 RD Sharma Solutions - Chapter 24 Scalar or Dot Product - Exercise 24.1 | Set 3

3. Class 12 RD Sharma Solutions - Chapter 24 Scalar or Dot Product - Exercise 24.2

4. Class 12 RD Sharma Solutions - Chapter 26 Scalar Triple Product - Exercise 26.1

5. Class 12 RD Sharma Solutions - Chapter 25 Vector or Cross Product - Exercise 25.1 | Set 3

6. Class 12 RD Sharma Solutions - Chapter 25 Vector or Cross Product - Exercise 25.1 | Set 2

7. Class 12 RD Sharma Solutions - Chapter 25 Vector or Cross Product - Exercise 25.1 | Set 1

8. Class 12 RD Sharma Solutions - Chapter 9 Continuity - Exercise 9.1 | Set 2

9. Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 1

10. Class 12 RD Sharma Solutions - Chapter 33 Binomial Distribution - Exercise 33.1 | Set 1

Class 12 RD Sharma Solutions - Chapter 24 Scalar or Dot Product - Exercise 24.1 | Set 1

Class 12 RD Sharma Solutions - Chapter 24 Scalar or Dot Product - Exercise 24.1 | Set 3

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Class 12 RD Sharma Solutions – Chapter 24 Scalar or Dot Product – Exercise 24.1 | Set 2

Question 17. If and , then express in the form of where is parallel to and is perpendicular to .

Question 18. If either or , then . But, The converse need not be true. Justify your answer with an example.

Question 19. Show that the vectors form a right-angled triangle.

Question 20. If , and are such that is perpendicular to , then find the value of λ.

Question 21. Find the angles of a triangle whose vertices are A (0, -1. -2), B (3, 1, 4), and C(5, 7, 1).

Question 22. Find the magnitude of two vectors and , having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.

Question 23. Show that the points whose position vector are form a right triangle.

Question 24. If the vertices A, B, C of △ABC have position vectors (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively, what is the magnitude of ∠ABC?

Question 25. If A, B, C have position vectors (0, 1, 1), (3, 1, 5), (0, 3, 3) respectively, show that △ABC is right-angled at C.

Question 26. Find the projection of on , where and.

Question 27. If and , then show that the vectors and are orthogonal.

Question 28. A unit vector makes angle π/2 and π/3 with and respectively and an acute angle θ with . Find the angle θ and components of .

Question 29. If two vectors and are such that = 2, = 1, and =1. Find the value of

Question 30. If is a unit vector, then find in each of the following:

(i)

(ii)

Question 31. Find and , if

(i) = 12 and

(ii) = 8 and = 8

(iii)= 3 and = 2

Question 32. Find , if

(i) and

(ii) = 3, = 4 and = 1

(iii) and = 4

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Question 17. If $\vec{α}=3\hat{i}+4\hat{j}+5\hat{k}$ and $\vec{β}=2\hat{i}+\hat{j}-4\hat{k}$ , then express $\vec{β}$ in the form of $\vec{β}=\vec{β_1}+\vec{β_2}$ where $\vec{β_1}$ is parallel to $\vec{α}$ and $\vec{β_2}$ is perpendicular to $\vec{α}$ .

Question 18. If either $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}$ , then $\vec{a}.\vec{b}=0$ . But, The converse need not be true. Justify your answer with an example.

Question 19. Show that the vectors $\vec{a}=3\hat{i}-2\hat{j}+\hat{k}, \vec{b}=\hat{i}-3\hat{j}+5\hat{k},\vec{c}=2\hat{i}+\hat{j}-4\hat{k}$ form a right-angled triangle.

Question 20. If $\vec{a}=2\hat{i}+2\hat{j}+3\hat{k}$ , $\vec{b}=-\hat{i}+2\hat{j}+\hat{k}$ and $\vec{c}=3\hat{i}+\hat{j}$ are such that $\vec{a}+λ\vec{b}$ is perpendicular to $\vec{c}$ , then find the value of λ.

Question 22. Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$ , having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.

Question 23. Show that the points whose position vector are $\vec{a} =4\hat{i}-3\hat{j}+\hat{k},\vec{b} =2\hat{i}-4\hat{j}+5\hat{k}, \vec{c} =\hat{i}-\hat{j}$ form a right triangle.

Question 26. Find the projection of $\vec{b}+\vec{c}$ on $\vec{a}$ , where $\vec{a}=2\hat{i}-2\hat{j}+\hat{k}, \vec{b}=\hat{i}+2\hat{j}-2\hat{k}$ and $\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$ .

Question 27. If $\vec{a}=5\hat{i}-\hat{j}-3\hat{k}$ and $\vec{b}=\hat{i}+3\hat{j}-5\hat{k}$ , then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are orthogonal.

Question 28. A unit vector $\vec{a}$ makes angle π/2 and π/3 with $\hat{i}$ and $\hat{j}$ respectively and an acute angle θ with $\hat{k}$ . Find the angle θ and components of $\vec{a}$ .

Question 29. If two vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}|$ = 2, $|\vec{b}|$ = 1, and $\vec{a}.\vec{b}$ =1. Find the value of $(3\vec{a}-5\vec{b}).(2\vec{a}+7\vec{b}).$

Question 30. If $\vec{a}$ is a unit vector, then find $|\vec{x}|$ in each of the following:

(i) $(\vec{x}-\vec{a})(\vec{x}+\vec{a})=8$

(ii) $(\vec{x}-\vec{a})(\vec{x}+\vec{a})=12$

Question 31. Find $|\vec{a}|$ and $|\vec{b}|$ , if

(i) $(\vec{a}+\vec{b})(\vec{a}-\vec{b})$ = 12 and $|\vec{a}| = 2 |\vec{b}|$

(ii) $(\vec{a}+\vec{b})(\vec{a}-\vec{b})$ = 8 and $|\vec{a}|$ = 8 $|\vec{b}|$

(iii) $(\vec{a}+\vec{b})(\vec{a}-\vec{b})$ = 3 and $|\vec{a}|$ = 2 $|\vec{b}|$

Question 32. Find $|\vec{a}-\vec{b}|$ , if

(i) $|\vec{a}| = 2, |\vec{b}| = 5$ and $\vec{a}.\vec{b} = 8$

(ii) $|\vec{a}|$ = 3, $|\vec{b}|$ = 4 and $\vec{a}.\vec{b}$ = 1

(iii) $|\vec{a}|=2, |\vec{b}| = 3$ and $\vec{a}.\vec{b}$ = 4