RD Sharma Class 12 Ex 24.2 Solutions Chapter 24 Scalar or Dot Product

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	24
Exercise	24.2
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 24.2 Solutions Chapter 24 Scalar or Dot Product

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Question 1. Find $\vec{a}.\vec{b}$ when

(i) $\vec{a} = \hat{i}-2 \hat{j}+ \hat{k}$ and $\vec{b} = 4 \hat{i} -4\hat{j} +7\hat{k}$

Solution:

= $(\hat{i}-2 \hat{j}+ \hat{k})(\hat{i}-2 \hat{j}+ \hat{k})$
= (1)(4) + (-2)(-4) + (1)(7)
= 4 + 8 + 7
= 19

(ii) $\vec{a} = \hat{j}+2 \hat{k}$ and $\vec{b} = 2\hat{i}+\hat{k}$

Solution:

= $(\hat{j}+2 \hat{k} )(2\hat{i}+\hat{k})$
= (0)(2) + (1)(0) + (2)(1)
= 2

(iii) $\vec{a} = \hat{j}-\hat{k}$ and $\vec{b} = 2\hat{i}+3\hat{j}-2 \hat{k}$

Solution:

= $(\hat{j}-\hat{k})(2\hat{i}+3\hat{j}-2 \hat{k} )$
= (0)(2) + (1)(3) + (-1)(-2)
= 0 + 3 + 2
= 5

Question 2. For what value of λ are the vector $\vec{a}$ and $\vec{b}$ perpendicular to each other? where:

(i) $\vec{a} = λ\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b} =4\hat{i}-9\hat{j}+2\hat{k}$

Solution:

$\vec{a}$ and $\vec{b}$ are perpendicular to each other
So $\vec{a} . \vec{b} = 0$
⇒ $(λ\hat{i}+2\hat{j}+\hat{k})(4\hat{i}-9\hat{j}+2\hat{k}) = 0$
⇒ λ(4) + (2)(-9) + (1)(2) = 0
⇒ 4λ – 18 + 2 = 0
⇒ 4λ = 16
⇒ λ = 4

(ii) $\vec{a} = λ\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b} =5\hat{i}-9\hat{j}+2\hat{k}$

Solution:

$\vec{a}$ and $\vec{b}$ are perpendicular to each other
so $\vec{a} . \vec{b}$ = 0
⇒ $(λ\hat{i}+2\hat{j}+\hat{k})(5\hat{i}-9\hat{j}+2\hat{k})$
⇒ λ(5) + (2)(-9) + (1)(2) = 0
⇒ 5λ – 18 + 2 = 0
⇒ 5λ = 16
⇒ λ = 16/5

(iii) $\vec{a} = 2\hat{i}+3\hat{j}+4\hat{k}$ and $\vec{b} =3\hat{i}+2\hat{j}-λ\hat{k}$

Solution:

$\vec{a}$ and $\vec{b}$ are perpendicular to each other
so $\vec{a} . \vec{b}$ = 0
⇒ $(2\hat{i}+3\hat{j}+4\hat{k})(3\hat{i}+2\hat{j}-λ\hat{k})$ =0
⇒ (2)(3) + (3)(2) – (4)λ = 0
⇒ 6 + 6 – 4λ = 0
⇒ 4λ = 12
⇒ λ = 3

(iv) $\vec{a} = λ\hat{i}+3\hat{j}+2\hat{k}$ and $\vec{b} =\hat{i}-\hat{j}+3\hat{k}$

Solution:

$\vec{a}$ and $\vec{b}$ are perpendicular to each other
so $\vec{a} . \vec{b}=0$
⇒ $(λ\hat{i}+3\hat{j}+2\hat{k})(\hat{i}-\hat{j}+3\hat{k})=0$
⇒ λ(1) + (3)(-1) + (2)(3) = 0
⇒ λ – 3 + 6 = 0
⇒ λ = 3

Question 3. If $\vec{a}$ and $\vec{b}$ are two vectors such that | $\vec{a}$ |=4, | $\vec{b}$ | = 3 and $\vec{a}.\vec{b}$ = 6. Find the angle between $\vec{a}$ and $\vec{b}$

Solution:

Let the angle be θ
cos θ = $\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}$
= 6 /(4×3) = 1/2
Therefore, θ = cos^-1(1/2)
= π/3

Question 4. If $\vec{a} = \hat{i}-\hat{j}$ and $\vec{b} =-\hat{j}+2\hat{k}$ , find $(\vec{a}-2\vec{b}). (\vec{a}+\vec{b})$ .

Solution:

$(\vec{a}-2\vec{b})$ = $(\hat{i}-\hat{j})-2(-\hat{j}+2\hat{k})$
= $\hat{i}-\hat{j}+2\hat{j}-4\hat{k}$
= $\hat{i}+\hat{j}-4\hat{k}$
$(\vec{a}+\vec{b})$ = $(\hat{i}-\hat{j})+(-\hat{j}+2\hat{k})$
= $\hat{i}-\hat{j}-\hat{j}+2\hat{k}$
= $\hat{i}-2\hat{j}+2\hat{k}$
Now, $(\vec{a}-2\vec{b}).(\vec{a}+\vec{b})$
= $(\hat{i}+\hat{j}-4\hat{k})(\hat{i}-2\hat{j}+2\hat{k})$
= (1)(1) + (1)(-2) + (-4)(2)
= 1 – 2 – 8
= -9
Therefore, $(\vec{a}-2\vec{b}).(\vec{a}+\vec{b})$ = -9

Question 5. Find the angle between the vectors $\vec{a}$ and $\vec{b}$ where :

(i) $\vec{a} = \hat{i}-\hat{j}$ and $\vec{b} = \hat{j}+\hat{k}$

Solution:

Let the angle be θ between $\vec{a}$ and $\vec{b}$
cos θ = $\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}$
Now, $\vec{a} . \vec{b}$
= $(\hat{i}-\hat{j})(\hat{j}+\hat{k})$
= (1)(0) + (-1)(1) + (0)(1)
= 0 – 1 + 0 = -1
| $\vec{a}$ |= | $\hat{i}-\hat{j}$ |
= $\sqrt{(1)^2+(-1)^2}$
= √2
$|\vec{b}|$ = | $\hat{j}+\hat{k}$ |
= $\sqrt{(1)^2+(1)^2}$
= √2
Now, cos θ = -1/(√2×√2)
= -1/2
θ = cos^-1(-1/2)
= 2π/3

(ii) $\vec{a} =3\hat{i}-2\hat{j}-6\hat{k}$ and $\vec{b} =4\hat{i}-\hat{j}+8\hat{k}$

Solution:

Let the angle be θ between $\vec{a}$ and $\vec{b}$

Now, $\vec{a} . \vec{b}$

= $(3\hat{i}-2\hat{j}-6\hat{k})(4\hat{i}-\hat{j}+8\hat{k})$

=(3)(4) + (-2)(-1) + (-6)(8)

= 12 + 2 – 48

= -34

| $\vec{a}$ | = | $3\hat{i}-2\hat{j}-6\hat{k}$ |

= $\sqrt{(3)^2+(-2)^2+(-6)^2}$

= √49 = 7

$|\vec{b}| = |4\hat{i}-\hat{j}+8\hat{k}|$

= $\sqrt{(4)^2+(-1)^2+(8)^2}$

= √81 = 9

cos θ = $\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}$

Now, cos θ = -34/(7×9)

= -34/63

θ = cos^-1(-34/63)

(iii) $\vec{a} =2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b} =4\hat{i}+4\hat{j}-2\hat{k}$

Solution:

Let the angle be θ between $\vec{a}$ and $\vec{b}$
Now, $\vec{a} . \vec{b}$
= $(2\hat{i}-\hat{j}+2\hat{k})(4\hat{i}+4\hat{j}-2\hat{k})$
= (2)(4) + (-1)(4) + (2)(-2)
= 8 – 4 – 4 = 0
| $\vec{a}$ | = | $2\hat{i}-\hat{j}+2\hat{k}$ |
= $\sqrt{(2)^2+(-1)^2+(2)^2}$
= √9 = 3
| $\vec{b}$ | = | $4\hat{i}+4\hat{j}-2\hat{k}$ |
= $\sqrt{(4)^2+(4)^2+(-2)^2}$
= √36 = 6
Now, cos θ = $\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}$
cos θ = 0/(3×6) = 0
θ = cos^-1(0)
θ = π/2

(iv) $\vec{a} =2\hat{i}-3\hat{j}+\hat{k}$ and $\vec{b} =\hat{i}+\hat{j}-2\hat{k}$

Solution:

Let the angle be θ between $\vec{a}$ and $\vec{b}$
Now, $\vec{a} . \vec{b}$
= $(2\hat{i}-3\hat{j}+\hat{k})(\hat{i}+\hat{j}-2\hat{k})$
= (2)(1) + (-3)(1) + (1)(-2)
= 2 – 3 – 2
= -3
| $\vec{a}$ | = $|2\hat{i}-3\hat{j}+\hat{k}|$
= $\sqrt{(2)^2+(-3)^2+(-1)^2}$
= √14
| $\vec{b}$ | =| $\hat{i}+\hat{j}-2\hat{k}$ |
= $\sqrt{(1)^2+(1)^2+(-2)^2}$
= √6
cos θ = $\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}$
Now, cos θ = -3/(√14×√6)
= -3/√84
θ = cos^-1(-3/√84)

(v) $\vec{a} =\hat{i}+2\hat{j}-\hat{k}$ and $\vec{b} =\hat{i}-\hat{j}+\hat{k}$

Solution:

Let the angle be θ between $\vec{a}$ and $\vec{b}$
Now, $\vec{a} . \vec{b}$
= $(\hat{i}+2\hat{j}-\hat{k})(\hat{i}-\hat{j}+\hat{k})$
= (1)(1) + (2)(-1) + (-1)(1)
= 1 – 2 – 1
= -2
| $\vec{a}$ | = | $\hat{i}+2\hat{j}-\hat{k}$ |
= $\sqrt{(1)^2+(2)^2+(-1)^2}$
= √6
| $\vec{b}$ | = | $\hat{i}-\hat{j}+\hat{k}$ |
= $\sqrt{(1)^2+(-1)^2+(1)^2}$
= √3
cos θ = $\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}$
Now, cos θ = -2/(√6×√3)
= -2/√18
= -2/3√2
θ = cos^-1(-√2 /3)

Question 6. Find the angles which the vectors $\vec{a} =\hat{i}-\hat{j}+\sqrt2\hat{k}$ makes with the coordinate axes.

Solution:

Components along x, y and z axis are $\hat{i},\hat{j}$ and $\hat{k}$ respectively.

Let the angle between $\vec{a}$ and $\hat{i}$ be θ₁

Now, $\vec{a} . \hat{i}$

= $(\hat{i}-\hat{j}-\sqrt2\hat{k})(\hat{i}-0\hat{j}+0\hat{k})$

= (1)(1) + (-1)(0) + (√2)(0)

= 1

$|\vec{a}| = |\hat{i}-\hat{j}+\sqrt2\hat{k}|$

= $\sqrt{(1)^2+(-1)^2+(√2)^2}$

= √4 = 2

$|\hat{i}| = |\hat{i}+0\hat{j}+0\hat{k}|$

= √1 = 1

cos θ₁ = $\frac{\vec{a}.\hat{i}}{|\vec{a}||\hat{i}|}$

Now, cos θ₁ = 1/(2×1)

= 1/2

θ₁ = cos^-1(1/2) = π/3

Let the angle between $\vec{a}$ and $\hat{j}$ be θ₂

Now, $\vec{a} . \hat{j}$

= $(\hat{i}-\hat{j}+\sqrt2\hat{k})(0\hat{i}+\hat{j}+0\hat{k})$

= (1)(0) + (-1)(1) + (√2)(0)

= -1

$|\hat{j}| = |0\hat{i}+\hat{j}+0\hat{k}|$

= √1 = 1

cos θ₂ = $\frac{\vec{a}.\hat{j}}{|\vec{a}||\hat{j}|}$

Now, cos θ₂ = -1/(2×1)

= -1/2

θ₂ = cos^-1(-1/2) = 2π/3

Let the angle between $\vec{a}$ and $\hat{k}$ be θ₃

Now, $\vec{a} . \hat{k}$

= $(\hat{i}-\hat{j}+\sqrt2\hat{k})(0\hat{i}+0\hat{j}+\hat{k})$

= (1)(0) + (-1)(0) + (√2)(1)

= √2

$|\hat{k}| = |0\hat{i}+0\hat{j}+\hat{k}|$

= √1 = 1

cos θ₃ = $\frac{\vec{a}.\hat{k}}{|\vec{a}||\hat{k}|}$

= 1/(√2)

= cos^-1(1/√2) = π/4

Question 7(i). Dot product of a vector with $\hat{i}+\hat{j}-3\hat{k}, \hat{i}+3\hat{j}-2\hat{k}$ and $2\hat{i}+\hat{j}+4\hat{k}$ are 0, 5 and 8respectively. Find the vector.

Solution:

Let $\vec{a} =\hat{i}+\hat{j}-3\hat{k}, \vec{b} =\hat{i}+3\hat{j}-2\hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}+4\hat{k}$ be three given vectors.

Let $\vec{r}=x\hat{i}+y\hat{j}+j\hat{k}$ be a vector such that its dot products with $\vec{a}, \vec{b}$ , and $\vec{c}$ are 0, 5 and 8 respectively. Then,

$\vec{r}. \vec{a} = 0$

⇒ $(x\hat{i}+y\hat{j}+j\hat{k})(\hat{i}+\hat{j}-3\hat{k})$ = 0

⇒ x + y – 3z = 0 ….(1)

$\vec{r}. \vec{b} = 5$

⇒ $(x\hat{i}+y\hat{j}+j\hat{k})(\hat{i}+3\hat{j}-2\hat{k})$ = 5

⇒ x + 3y – 2z = 5 …..(2)

$\vec{r}. \vec{c} = 8$

⇒ $(x\hat{i}+y\hat{j}+j\hat{k})(2\hat{i}+\hat{j}+4\hat{k})$ = 8

⇒ 2x + y + 4z = 8 …..(3)

Solving 1,2 and 3 we get x = 1, y = 2 and z = 1,

Hence, the required vector is $\vec{r}=\hat{i}+2\hat{j}+\hat{k}$

Question 8. If $\vec{a}$ and $\vec{b}$ are unit vectors inclined at an angle θ then prove that

(i) cos θ/2 = 1/2 $|\hat{a}+\hat{b}|$

Solution:

| $\hat{a}$ | = | $\hat{b}$ | = 1
| $\hat{a}+\hat{b}$ |² =( $\hat{a}+\hat{b}$ )²
= $(\hat{a})^2+(\hat{b})^2+2\hat{a}.\hat{b}$
= 1 + 1 + 2 $\hat{a}.\hat{b}$
= 2 + 2| $\hat{a}||\hat{b}$ |cos θ
= 2(1 + (1)(1)cos θ)
= 2(2cos² θ/2)
| $\hat{a}+\hat{b}$ |²= 4cos² θ/2
$\hat{a}+\hat{b}$ = 2 cos θ/2
cos θ/2 = 1/2| $\hat{a}+\hat{b}$ |

(ii) tan θ/2 = $\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}$

Solution:

$|\hat{a}| = |\hat{b}|$ = 1
$\frac{|\hat{a}-\hat{b}|^2}{|\hat{a}+\hat{b}|^2}$ = $\frac{(\hat{a}-\hat{b})^2}{(\hat{a}+\hat{b})^2}$
= $\frac{(\hat{a})^2+(\hat{b})^2-2\hat{a}.\hat{b}}{(\hat{a})^2+(\hat{b})^2+2\hat{a}.\hat{b}}$
= $\frac{2-2|\hat{a}||\hat{b}|cos θ }{2+2|\hat{a}||\hat{b}|cos θ }$
= $\frac{2(1-cos θ)}{2(1+cos θ)}$
= $\frac{2sin^2 θ/2}{2cos^2 θ/2}$
= tan² θ/2
Therefore, tan θ/2 = $\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}$

Question 9. If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is √3.

Solution:

Let $\hat{a}$ and $\hat{b}$ be two unit vectors

Then, $|\hat{a}| = |\hat{b}| = 1$

According to question:

$|\hat{a}+\hat{b}| = 1$

Taking square on both sides

⇒ $|\hat{a}+\hat{b}|^2 = (1)^2$

⇒ $(\hat{a})^2+(\hat{b})^2+2\hat{a}.\hat{b} = 1$

⇒ (1)²+(1)²+ $2\hat{a}.\hat{b}$ = 1

⇒ 2+ 2 $\hat{a}.\hat{b}$ = 1

⇒ 2 $\hat{a}.\hat{b}$ = -1

⇒ \hat{a}.\hat{b} =-1/2

Now, $|\hat{a}-\hat{b}|^2 = (\hat{a}-\hat{b})^2$

= $(\hat{a})^2 + (\hat{b})^2 - 2\hat{a}.\hat{b}$

= (1)² + (1)²– 2 (-1/2)

= 2 + 1 = 3

Therefore, $|\hat{a}-\hat{b}|^2$ = 3

$|\hat{a}-\hat{b}|$ =√3

Question 10. If $\vec{a},\vec{b},\vec{c}$ are three mutually perpendicular unit vectors, then prove that | $\vec{a}+\vec{b}+\vec{c}$ | =√3.

Solution:

Given $\vec{a},\vec{b},\vec{c}$ are mutually perpendicular so,

$\vec{a}.\vec{b}=\vec{b}.\vec{c}=\vec{c}.\vec{a} = 0$

$|\vec{a}| = |\vec{b}| = |\vec{c}|=1$

Now,

$|\vec{a}+\vec{b}+\vec{c}|^2$ = $(\vec{a}+\vec{b}+\vec{c})^2$

= $(\vec{a})^2+(\vec{b})^2+(\vec{c})^2+2\vec{a}\vec{b}+2\vec{b}\vec{c}+2\vec{c}\vec{a}$

= (1)² + (1)² +(1)² + 0

= 3

$|\vec{a}+\vec{b}+\vec{c}|$ = √3

Question 11. If $|\vec{a}+\vec{b}|$ = 60, $|\vec{a}-\vec{b}|$ = 40 and $|\vec{b}|$ = 46, find $|\vec{a}|$

Solution:

Given $|\vec{a}+\vec{b}|$ =60, $|\vec{a}-\vec{b}|$ = 40 and $|\vec{b}|$ = 46
We know that,
(a + b)²+ (a – b)² = 2(a² + b²)
⇒ $|\vec{a}+\vec{b}|^2+|\vec{a}-\vec{b}|^2 = 2(|\vec{a}|^2+|\vec{b}|^2)$
⇒ 60² + 40² = 2( $|\vec{a}|$ ² + 49²)
⇒ 3600 + 1600 = 2 $|\vec{a}|^2$ + 2401
⇒ $2|\vec{a}|$ = 968
⇒ $|\vec{a}|$ = √484 =22

Question 12. Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined with the coordinate axes.

Solution:

Let $\vec{a} = \hat{i}+\hat{j}+\hat{k}$
$|\vec{a}| =$ √(1+1+1) = √3
Let θ₁, θ₂, θ₃be the angle between the coordinate axes and the $\vec{a}$
cos θ₁ = $\frac{\vec{a}.\hat{i}}{|\vec{a}||\hat{i}|}$
= 1/√3
cos θ₂ = $\frac{\vec{a}.\hat{j}}{|\vec{a}||\hat{j}|}$
= 1/√3
cos θ₃ = $\frac{\vec{a}.\hat{k}}{|\vec{a}||\hat{k}|}$
= 1/√3
Since, cos θ₁= cos θ₂= cos θ₃
Therefore, Given vector is equally inclined with coordinate axis.

Question 13. Show that the vectors $\vec{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}), \vec{b}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}),\vec{c}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})$ are mutually perpendicular unit vectors.

Solution:

Given, $\vec{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})$

$\vec{b}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})$

$\vec{c}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})$

$|\vec{a}|$ = (1/7)√(2²+ 3²+ 6²) = (1/7)(√49) = 1

$|\vec{b}|$ = (1/7)√(3²+ (-6)²+ 2²) = (1/7)(√49) = 1

$|\vec{c}|$ = (1/7)√(6²+ 2²+ (-3)²) = (1/7)(√49) = 1

Now, $\vec{a}.\vec{b} =$ 1/49[3 × 2 – 3 × 6 + 6 × 2]

= 1/49[6 – 18 + 12] = 0

$\vec{b}.\vec{c} =$ 1/49[3 × 6 – 6 × 2 – 2 × 3]

= 1/49[18 – 12 – 6] = 0

Since, $\vec{a}.\vec{b} = \vec{b}.\vec{c}=0$ they are mutually perpendicular unit vectors.

Question 14. For any two vectors $\vec{a}$ and $\vec{b}$ , Show that $(\vec{a}+\vec{b}).(\vec{a}-\vec{b})=0\Leftrightarrow|\vec{a}|=|\vec{b}|.$

Solution:

To prove $(\vec{a}+\vec{b}).(\vec{a}-\vec{b})=0\Leftrightarrow|\vec{a}|=|\vec{b}|$
⇒ $(\vec{a}+\vec{b}).(\vec{a}-\vec{b})=0$
⇒ $|\vec{a}|^2-|\vec{b}|^2=0$
⇒ $|\vec{a}|=|\vec{b}|$
Hence Proved

Question 15. If $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ , $\vec{b}=\hat{i}+\hat{j}-2\hat{k}$ and $\vec{c}=\hat{i}+3\hat{j}-\hat{k}$ , find such that $\vec{a}$ is perpendicular to $λ\vec{b}+\vec{c}$ .

Solution:

Given: $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$

$\vec{b}=\hat{i}+\hat{j}-2\hat{k}$

$\vec{c}=\hat{i}+3\hat{j}-\hat{k}$

According to question

$\vec{a}(λ\vec{b}+\vec{c})=0$

⇒ $(2\hat{i}-\hat{j}+\hat{k})[λ(\hat{i}+\hat{j}-2\hat{k})+(\hat{i}+3\hat{j}-\hat{k})]=0$

⇒ $(2\hat{i}-\hat{j}+\hat{k})(λ\hat{i}+λ\hat{j}-2λ\hat{k}+\hat{i}+3\hat{j}-\hat{k})=0$

⇒ 2(λ+1) – (λ+3) -2λ-1 = 0

⇒ 2λ + 2 -λ – 3 – 2λ – 1 = 0

⇒ -λ = 2

⇒ λ = -2

Question 16. If $\vec{p}=5\hat{i}+λ\hat{j}-3\hat{k}$ and $\vec{q}=\hat{i}+3\hat{j}-5\hat{k}$ , then find the value of λ so that $\vec{p}+\vec{q}$ and $\vec{p}-\vec{q}$ are perpendicular vectors.

Solution:

Given, $\vec{p}=5\hat{i}+λ\hat{j}-3\hat{k}$

$\vec{q}=\hat{i}+3\hat{j}-5\hat{k}$

According to question

$(\vec{p}+\vec{q})(\vec{p}-\vec{q})=0$

⇒ $|\vec{p}|^2-|\vec{q}|^2=0$

⇒ $|\vec{p}|^2=|\vec{q}|^2$

⇒ $\sqrt{5^2+λ^2+(-3)^2}=\sqrt{1^2+3^2+(-5)^2}$

⇒ 25 + λ²+ 9 = 1 + 9 + 25

⇒ λ² = 1

⇒ λ = 1

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

Question 33. Find the angle between the two vectors $\vec{a}$ and $\vec{b}$ , if

(i) $|\vec{a}|$ =√3, $|\vec{b}|$ = 2 and $\vec{a}.\vec{b}$ = √6

Solution:

We know, $\vec{a}\vec{b} = |\vec{a}||\vec{b}|cos\theta$
⇒ √6 = 2√3 cos θ
⇒ cos θ = 1/√2
⇒ θ = cos^-1(1/√2)
⇒ θ = π/4

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

Solution:

We know, $\vec{a}\vec{b} = |\vec{a}||\vec{b}|cos\theta$
⇒ 1 = 3×3 cos θ
⇒ cos θ = 1/9
⇒ θ = cos^-1(1/9)

Question 34. Express the vector $\vec{a}=5\hat{i}-2\hat{j}+5\hat{k}$ as the sum of two vectors such that one is parallel to the vector $\vec{b}=3\hat{i}+\hat{k}$ and other is perpendicular to $\vec{b}$

Solution:

Given, $\vec{a}=5\hat{i}-2\hat{j}+5\hat{k}$

$\vec{b}=3\hat{i}+\hat{k}$

Let the two vectors be $\vec{a_1}, \vec{a_2}$

Now, $\vec{a}=\vec{a_1}+\vec{a_2}$ ….(1)

Assuming $\vec{a_1}$ is parallel to $\vec{b}$

Then, $\vec{a_1}=λ\vec{b}$ ……(2)

$\vec{a_2}$ is perpendicular to $\vec{b}$

Then, $\vec{a_2}.\vec{b}=0$ ……(3)

From eq(1)

$\vec{a_1}=λ\vec{b}$

⇒ $\vec{a_2} = \vec{a}-λ\vec{b}$

⇒ $\vec{a_2} =5\hat{i}-2\hat{j}+5\hat{k} -λ(3\hat{i}+\hat{k})$

⇒ $\vec{a_2} =(5-3λ)\hat{i}-2\hat{j}+(5-λ)\hat{k}$

From eq(3)

$\vec{a_2}.\vec{b}=0$

⇒ $((5-3λ)\hat{i}-2\hat{j}+(5-λ)\hat{k})(3\hat{i}+\hat{k})$

⇒ (5-3λ)3+(5-λ)=0

⇒ 15-9λ+5-λ=0

⇒ -10λ = -20

⇒ λ=2

From eq(2)

$\vec{a_1}=6\hat{i}+2\hat{k}$

$\vec{a_2}=-\hat{i}-2\hat{j}+3\hat{k}$

$\vec{a}=(6\hat{i}+2\hat{k})+(-\hat{i}-2\hat{j}+3\hat{k})$

Question 35. If $\vec{a}$ and $\vec{b}$ are two vectors of the same magnitude inclined at an angle of 30° such that $\vec{a}.\vec{b}$ = 3, find $|\vec{a}|, |\vec{b}|.$

Solution:

Given that two vectors of the same magnitude inclined at an angle of 30°, and $\vec{a}.\vec{b} = 3$
To find $|\vec{a}|, |\vec{b}|$
We know, $\vec{a}\vec{b} = |\vec{a}||\vec{b}|cos\theta$
⇒ 3 = $|\vec{a}||\vec{a}|cos θ$
⇒ 3 = $|\vec{a}|^2 cos 30°$
⇒ 3 = $|\vec{a}|^2$ (√3/2)
⇒ $|\vec{a}|^2$ = 6/√3
⇒ $|\vec{a}|=|\vec{b}|=\sqrt{2\sqrt{3}}$

Question 36. Express $2\hat{i}-\hat{j}+3\hat{k}$ as the sum of a vector parallel and a vector perpendicular to $2\hat{i}+4\hat{j}-2\hat{k}.$

Solution:

Assuming $\vec{a}=2\hat{i}-\hat{j}+3\hat{k}$

$\vec{b}=2\hat{i}+4\hat{j}-2\hat{k}$

Let the two vectors be $\vec{a_1}, \vec{a_2}$

Now, $\vec{a}=\vec{a_1}+\vec{a_2}$

or $\vec{a_2}=\vec{a}-\vec{a_1}$ ….(1)

Assuming $\vec{a_1}$ is parallel to $\vec{b}$

then, $\vec{a_1}=λ\vec{b}$ …(2)

$\vec{a_2}$ is perpendicular to $\vec{b}$

then, $\vec{a_2}.\vec{b}=0$ ……(3)

Putting eq(2) in eq(1), we get

$\vec{a_1}=λ\vec{b}$

⇒ $\vec{a_2} = \vec{a}-λ\vec{b}$

⇒ $\vec{a_2} =2\hat{i}-\hat{j}+3\hat{k} -λ(2\hat{i}+4\vec{j}-2\hat{k})$

⇒ $\vec{a_2} =(2-2λ)\hat{i}-(1+4λ)\hat{j}+(3+2λ)\hat{k}$

From eq(3)

$\vec{a_2}.\vec{b}=0$

⇒ $((2-2λ)\hat{i}-(1+4λ)\hat{j}+(3+2λ)\hat{k})(2\hat{i}+4\hat{j}-2\hat{k})=0$

⇒ (2 – 2λ)2 – (1 + 4λ)4 – (3 + 2λ)2 = 0

⇒ 4 – 4λ – 4 – 16λ – 6 – 4λ = 0

⇒ 24λ = -6

⇒ λ = -6/24

From eq(2)

$\vec{a_1}=-1/4(2\hat{i}+4\hat{j}-2\hat{k})$

$\vec{a_1}=-1/2\hat{i}-1\hat{j}+1/2\hat{k}$

$\vec{a_2}=(2+1/2)\hat{i}-0\hat{j}+(3-1/2)\hat{k}$

$\vec{a_2}=5/2\hat{i}+5/2\hat{k}$

$\vec{a}=\vec{a_1}+\vec{a_2}$

$\vec{a}=(-1/2\hat{i}-1\hat{j}+1/2\hat{k})+(5/2\hat{i}+5/2\hat{k})$

Question 37. Decompose the vector $6\hat{i}-3\hat{j}-6\hat{k}$ into vectors which are parallel and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}.$

Solution:

Let $\vec{a}=6\hat{i}-3\hat{j}-6\hat{k}$ and $\vec{m}=\hat{i}+\hat{j}+\hat{k}$

Let $\vec{b}$ be a vector parallel to $\hat{i}+\hat{j}+\hat{k}.$

Therefore, $\vec{b} =λ(\hat{i}+\hat{j}+\hat{k})$

$\vec{a}$ to be decomposed into two vectors

$\vec{a}=\vec{b}+\vec{c}$

⇒ $\vec{c}=\vec{a}-\vec{b}$

⇒ $\vec{c}=(6-λ)\hat{i}+(-3-λ)\hat{j}+(-6-λ)\vec{k}$

Now, $\vec{c}$ is perpendicular to $\vec{m}$

or $\vec{c}.\vec{m}=0$

⇒ $((6-λ)\hat{i}+(-3-λ)\hat{j}+(-6-λ)\vec{k})(\hat{i}+\hat{j}+\hat{k})=0$

⇒ 6 – λ – 3 – λ – 6 – λ = 0

⇒ λ = -1

Therefore, the required vectors are $\vec{b} =-\hat{i}-\hat{j}-\hat{k}$ and $\vec{c}=7\hat{i}-2\hat{j}-5\vec{k}$

Question 38. Let $\vec{a}=5\hat{i}-\hat{j}+7\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+λ\hat{k}$ . Find λ such that $\vec{a}+\vec{b}$ is orthogonal to $\vec{a}-\vec{b}$

Solution:

Given, $\vec{a}=5\hat{i}-\hat{j}+7\hat{k}$

$\vec{b}=\hat{i}-\hat{j}+λ\hat{k}$

According to question

$(\vec{a}+\vec{b})(\vec{a}-\vec{b})=0$

⇒ $|\vec{a}|^2-|\vec{b}|^2=0$

⇒ $|\vec{a}|^2=|\vec{b}|^2$

⇒ $\sqrt{5^2+(-1)^2+7^2}=\sqrt{1^2+(-1)^2+λ^2}$

⇒ 25 + 1 + 49 = 1 + 1 + λ²

⇒ λ²= 73

⇒ λ = √73

Question 39. If $\vec{a}.\vec{a}=0$ and $\vec{a}.\vec{b}=0$ , what can you conclude about the vector $\vec{b}$ ?

Solution:

Given, $\vec{a}.\vec{a}=0$ , $\vec{a}.\vec{b}=0$

$|\vec{a}| = 0$

Now, $\vec{a}.\vec{b}=0$

We conclude that $\vec{a}=0$ or $\vec{b}=0$ or θ = 90°

Thus, $\vec{b}$ can be any arbitrary vector.

Question 40. If $\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$ , then prove that it is perpendicular to both $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$

Solution:

Given $\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$
$\vec{c}.\vec{a}=0$ ….(1)
$\vec{c}.\vec{b}=0$ ….(2)
To prove $\vec{c}.(\vec{a}+\vec{b})=0$ and $\vec{c}.(\vec{a}-\vec{b})=0$
Now, $\vec{c}.(\vec{a}+\vec{b})$
⇒ $\vec{c}.\vec{a}+\vec{c}.\vec{b}=0$ [From eq(1) and (2)]
Again, $\vec{c}.(\vec{a}-\vec{b})$
⇒ $\vec{c}.\vec{a}-\vec{c}.\vec{b}=0$ [From eq(1) and (2)]
Hence Proved

Question 41. If $|\vec{a}|= a$ and $|\vec{b}|= b$ , prove that $(\frac{\vec{a}}{a^2}-\frac{\vec{b}}{b^2})^2= (\frac{\vec{a}-\vec{b}}{ab})^2$

Solution:

Given, $|\vec{a}| = a$ and $|\vec{b}| = b,$

To prove

$(\frac{\vec{a}}{a^2}-\frac{\vec{b}}{b^2})^2= (\frac{\vec{a}-\vec{b}}{ab})^2.$

Taking LHS

$(\frac{\vec{a}}{a^2}-\frac{\vec{b}}{b^2})^2$

= $\frac{\vec{a}.\vec{a}}{a^4}+\frac{\vec{b}.\vec{b}}{b^4}-2\frac{\vec{a}.\vec{b}}{a^2b^2}$

= $\frac{a^2}{a^4}+\frac{b^2}{b^4}-2\frac{\vec{a}.\vec{b}}{a^2b^2}$

= $\frac{1}{a^2}+\frac{1}{b^2}-2\frac{\vec{a}.\vec{b}}{a^2b^2}$

Taking RHS

$(\frac{\vec{a}-\vec{b}}{ab})^2 \vec{d}.\vec{a}=0$

= $(\frac{a^2+b^2-2\vec{a}\vec{b}}{a^2b^2})$

= $\frac{1}{a^2}+\frac{1}{b^2}-2\frac{\vec{a}.\vec{b}}{a^2b^2}$

LHS = RHS

Hence Proved

Question 42. If $\vec{a},\vec{b},\vec{c}$ are three non- coplanar vectors such that $\vec{d}.\vec{a}=\vec{d}.\vec{b}=\vec{d}.\vec{c} =0$ then show that $\vec{d}$ is the null vector.

Solution:

Given that $\vec{d}.\vec{a}=0$
So either $\vec{d} = 0$ or $\vec{d}⊥\vec{a}=0$
Similarly, $\vec{d}.\vec{b} = 0$
Either $\vec{d}= 0$ or $\vec{d}⊥\vec{b}=0$
Also, $\vec{d}.\vec{c} =0$
So $\vec{d} = 0$ or $\vec{d}⊥\vec{c}=0$
But $\vec{d}$ can’t be perpendicular to $\vec{a},\vec{b}$ and $\vec{c}$ because $\vec{a},\vec{b},\vec{c}$ are non-coplanar.
So $\vec{d}$ = 0 or $\vec{d}$ is a null vector

Question 43. If a vector $\vec{a}$ is perpendicular to two non- collinear vectors $\vec{b}$ and $\vec{c}$ , then is $\vec{a}$ perpendicular to every vector in the plane of $\vec{b}$ and $\vec{c}$

Solution:

Given that $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$

$\vec{a}.\vec{b}=0 , \vec{a}.\vec{c}=0$

Let $\vec{r}$ be any vector in the plane of $\vec{b}$ and $\vec{c}$ and $\vec{r}$ is the linear combination of $\vec{b}$ and $\vec{c}$

$\vec{r} =x\vec{b}+y\vec{c}$ [x, y are scalars]

Now $, \vec{a}.\vec{r}$

⇒ $\vec{a}.\vec{r} =\vec{a}(x\vec{b}+y\vec{c})$

⇒ $\vec{a}.\vec{r} =x(\vec{a}.\vec{b})+y(\vec{a}.\vec{c})$

⇒ $\vec{a}.\vec{r} =x0+y0$

⇒ $\vec{a}.\vec{r} = 0$

Therefore, $\vec{a}$ is perpendicular to $\vec{r}$ i.e. $\vec{a}$ is perpendicular to every vector.

Question 44. If $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ , how that the angle θ between the vectors $\vec{b}$ and $\vec{c}$ is given by cos θ = $\frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|}$

Solution:

Given that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$
⇒ $\vec{a}=-(\vec{b}+\vec{c})$
⇒ $(\vec{a})^2=(\vec{b}+\vec{c})^2$
⇒ $\vec{a}.\vec{a}=(\vec{b}+\vec{c})(\vec{b}+\vec{c})$
⇒ $|\vec{a}|^2=|\vec{b}|^2+|\vec{b}||\vec{c}|+|\vec{b}||\vec{c}|+|\vec{c}|^2$
⇒ $|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2=2|\vec{b}||\vec{c}|$
⇒ $|\vec{b}||\vec{c}|=(|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2)/2$
⇒ cos θ = $\frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|}.$

Question 45. Let $\vec{u},\vec{v}$ and $\vec{w}$ be vector such $\vec{u}+\vec{v}+\vec{w}=\vec{0}$ . $|\vec{u}|$ = 3, $|\vec{v}|$ = 4 and $|\vec{w}|$ = 5, then find $\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u}$

Solution:

Given that $\vec{u},\vec{v}$ and $\vec{w}$ are vectors such that $\vec{u}+\vec{v}+\vec{w}=\vec{0}$ . $|\vec{u}|$ = 3, $|\vec{v}|$ = 4 and $|\vec{w}|$ =5,

To find $\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u}$

Taking

$\vec{u}+\vec{v}+\vec{w}=\vec{0}$

Squaring on both side, we get

⇒ $(\vec{u}+\vec{v}+\vec{w})^2=\vec{0}$

⇒ $|\vec{u}|^2+|\vec{v}|^2+|\vec{w}|^2+2(\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u})=0$

⇒ $2(\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u})=-(|\vec{u}|^2+|\vec{v}|^2+|\vec{w}|^2)$

⇒ $2(\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u})=-(3^2+4^2+5^2)$

⇒ $2(\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u})=-50$

Therefore, $\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u}=-25$

Question 46. Let $\vec{a}=x^2\hat{i}+2\hat{j}-2\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k},$ and $\vec{c}=x^2\hat{i}+5\hat{j}-4\hat{k}$ be three vectors. Find the values of x for which the angle between $\vec{a}$ and $\vec{b}$ is acute and the angle between $\vec{b}$ and $\vec{c}$ is obtuse.

Solution:

Given $\vec{a}=x^2\hat{i}+2\hat{j}-2\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}, \vec{c}=x^2\hat{i}+5\hat{j}-4\hat{k}$

Case I: When angle between $\vec{a}$ and $\vec{b}$ is acute:-

$\vec{a}.\vec{b}$ >0

⇒ $(x^2\hat{i}+2\hat{j}-2\hat{k})(\hat{i}-\hat{j}+\hat{k})>0″ height=”32″ width=”341″>⇒ x2 – 2 – 2 > 0⇒ x2 > 4x ∈ (2, -2)Case II: When angle between <img decoding=$ and $\vec{c}$ is obtuse:-

$\vec{b}.\vec{c}<0$

⇒ $(\hat{i}-\hat{j}+\hat{k})(x^2\hat{i}+5\hat{j}-4\hat{k})<0$

⇒ x²– 5 – 4 < 0

⇒ x²< 9

x ∈ (3, -3)

Therefore, x ∈ (-3, -2)∪(2, 3)

Question 47. Find the value of x and y if the vectors $\vec{a}=3\hat{i}+x\hat{j}-\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}+y\hat{k}$ are mutually perpendicular vectors of equal magnitude.

Solution:

Given $\vec{a}=3\hat{i}+x\hat{j}-\hat{k}, \vec{b}=2\hat{i}+\hat{j}+y\hat{k}$ are mutually perpendicular vectors of equal magnitude.

$|\vec{a}|^2=|\vec{b}|^2$

⇒ 3²+ x²+ (-1)² = 2²+ 1²+ y²

⇒ x²+10 = y²+5

⇒ x²– y²+ 5 = 0 ….(1)

Now, $\vec{a}.\vec{b} = 0$

⇒ 6 + x – y = 0

⇒ y = x + 6 …..(2)

From eq(1)

x²– (x + 6)²+ 5 = 0

⇒ x² – (x² + 36 – 12x) + 5 = 0

⇒ -12x – 31 = 0

⇒ x = -31/12

Now, y = -31/12 + 6

y = 41/12

Question 48. If $\vec{a}$ and $\vec{b}$ are two non-coplanar unit vectors such that $|\vec{a}+\vec{b}| =√3$ , find $(2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b}).$

Solution:

Given that $\vec{a}$ and $\vec{b}$ are two non-coplanar unit vectors such that $|\vec{a}+\vec{b}| =√3$

To find $(2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b}).$

Now,

$|\vec{a}+\vec{b}|^2 =3$

$⇒|\vec{a}|^2+|\vec{b}|^2+2\vec{a}.\vec{b}=3$

$⇒ 1+1+2\vec{a}.\vec{b}=3$

$⇒ \vec{a}.\vec{b}=1/2$

Now, $(2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b})$

= $6|\vec{a}|^2-13\vec{a}.\vec{b}-5|\vec{b}|^2$

= 6 – 13(1/2) – 5

= 1 – 13/2

= -11/2

Question 49. If $\vec{a},\vec{b}$ are two vectors such that | $\vec{a}+\vec{b}$ | = $|\vec{b}|$ , then prove that $\vec{a}+2\vec{b}$ is perpendicular to $\vec{a}.$

Solution:

To prove

$(\vec{a}+2\vec{b})\vec{a}= 0$

Now,

$\vec{a}+\vec{b} = \vec{b}$

Squaring on both side, we get

$|\vec{a}+\vec{b}|^2=|\vec{b}|^2$

⇒ $(\vec{a}+\vec{b})(\vec{a}+\vec{b}) = \vec{b}.\vec{b}$

⇒ $\vec{a}\vec{a}+\vec{a}\vec{b}+\vec{b}\vec{a}+\vec{b}\vec{b} = \vec{b}.\vec{b}$

⇒ $\vec{a}\vec{a}+2\vec{a}\vec{b} = 0$

⇒ $\vec{a}(\vec{a}+2\vec{b}) = 0$

Therefore, $\vec{a}+2\vec{b}$ is perpendicular to $\vec{a}$

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RD Sharma Class 12 Ex 24.2 Solutions Chapter 24 Scalar or Dot Product

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Question 1. Find when

(i) and

Question 2. For what value of λ are the vector and perpendicular to each other? where:

(i) and

(ii) and

(iii) and

(iv) and

Question 3. If and are two vectors such that ||=4, || = 3 and = 6. Find the angle between and

Question 4. If and , find .

Question 5. Find the angle between the vectors and where :

(i) and

(ii) and

(iii) and

(iv) and

(v) and

Question 6. Find the angles which the vectors makes with the coordinate axes.

Question 7(i). Dot product of a vector with and are 0, 5 and 8respectively. Find the vector.

Question 8. If and are unit vectors inclined at an angle θ then prove that

(i) cos θ/2 = 1/2

(ii) tan θ/2 =

Question 9. If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is √3.

Question 10. If are three mutually perpendicular unit vectors, then prove that || =√3.

Question 11. If = 60, = 40 and = 46, find

Question 12. Show that the vector is equally inclined with the coordinate axes.

Question 13. Show that the vectors are mutually perpendicular unit vectors.

Question 14. For any two vectors and , Show that

Question 15. If , and , find such that is perpendicular to .

Question 16. If and , then find the value of λ so that and are perpendicular vectors.

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

Question 33. Find the angle between the two vectors and , if

(i) =√3, = 2 and = √6

(ii) = 3, = 3 and = 1

Question 34. Express the vector as the sum of two vectors such that one is parallel to the vector and other is perpendicular to

Question 35. If and are two vectors of the same magnitude inclined at an angle of 30° such that = 3, find

Question 36. Express as the sum of a vector parallel and a vector perpendicular to

Question 37. Decompose the vector into vectors which are parallel and perpendicular to the vector

Question 38. Let and . Find λ such that is orthogonal to

Question 39. If and , what can you conclude about the vector ?

Question 40. If is perpendicular to both and , then prove that it is perpendicular to both and

Question 41. If and , prove that

Question 42. If are three non- coplanar vectors such that then show that is the null vector.

Question 43. If a vector is perpendicular to two non- collinear vectors and , then is perpendicular to every vector in the plane of and

Question 44. If , how that the angle θ between the vectors and is given by cos θ =

Question 45. Let and be vector such . = 3, = 4 and = 5, then find

Question 46. Let and be three vectors. Find the values of x for which the angle between and is acute and the angle between and is obtuse.

Question 47. Find the value of x and y if the vectorsand are mutually perpendicular vectors of equal magnitude.

Question 48. If and are two non-coplanar unit vectors such that , find

Question 49. If are two vectors such that || = , then prove that is perpendicular to

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Question 1. Find $\vec{a}.\vec{b}$ when

(i) $\vec{a} = \hat{i}-2 \hat{j}+ \hat{k}$ and $\vec{b} = 4 \hat{i} -4\hat{j} +7\hat{k}$

Question 2. For what value of λ are the vector $\vec{a}$ and $\vec{b}$ perpendicular to each other? where:

(i) $\vec{a} = λ\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b} =4\hat{i}-9\hat{j}+2\hat{k}$

(ii) $\vec{a} = λ\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b} =5\hat{i}-9\hat{j}+2\hat{k}$

(iii) $\vec{a} = 2\hat{i}+3\hat{j}+4\hat{k}$ and $\vec{b} =3\hat{i}+2\hat{j}-λ\hat{k}$

(iv) $\vec{a} = λ\hat{i}+3\hat{j}+2\hat{k}$ and $\vec{b} =\hat{i}-\hat{j}+3\hat{k}$

Question 3. If $\vec{a}$ and $\vec{b}$ are two vectors such that | $\vec{a}$ |=4, | $\vec{b}$ | = 3 and $\vec{a}.\vec{b}$ = 6. Find the angle between $\vec{a}$ and $\vec{b}$

Question 4. If $\vec{a} = \hat{i}-\hat{j}$ and $\vec{b} =-\hat{j}+2\hat{k}$ , find $(\vec{a}-2\vec{b}). (\vec{a}+\vec{b})$ .

Question 5. Find the angle between the vectors $\vec{a}$ and $\vec{b}$ where :

(i) $\vec{a} = \hat{i}-\hat{j}$ and $\vec{b} = \hat{j}+\hat{k}$

(ii) $\vec{a} =3\hat{i}-2\hat{j}-6\hat{k}$ and $\vec{b} =4\hat{i}-\hat{j}+8\hat{k}$

(iii) $\vec{a} =2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b} =4\hat{i}+4\hat{j}-2\hat{k}$

(iv) $\vec{a} =2\hat{i}-3\hat{j}+\hat{k}$ and $\vec{b} =\hat{i}+\hat{j}-2\hat{k}$

(v) $\vec{a} =\hat{i}+2\hat{j}-\hat{k}$ and $\vec{b} =\hat{i}-\hat{j}+\hat{k}$

Question 6. Find the angles which the vectors $\vec{a} =\hat{i}-\hat{j}+\sqrt2\hat{k}$ makes with the coordinate axes.

Question 7(i). Dot product of a vector with $\hat{i}+\hat{j}-3\hat{k}, \hat{i}+3\hat{j}-2\hat{k}$ and $2\hat{i}+\hat{j}+4\hat{k}$ are 0, 5 and 8respectively. Find the vector.

Question 8. If $\vec{a}$ and $\vec{b}$ are unit vectors inclined at an angle θ then prove that

(i) cos θ/2 = 1/2 $|\hat{a}+\hat{b}|$

(ii) tan θ/2 = $\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}$

Question 10. If $\vec{a},\vec{b},\vec{c}$ are three mutually perpendicular unit vectors, then prove that | $\vec{a}+\vec{b}+\vec{c}$ | =√3.

Question 11. If $|\vec{a}+\vec{b}|$ = 60, $|\vec{a}-\vec{b}|$ = 40 and $|\vec{b}|$ = 46, find $|\vec{a}|$

Question 12. Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined with the coordinate axes.

Question 13. Show that the vectors $\vec{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}), \vec{b}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}),\vec{c}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})$ are mutually perpendicular unit vectors.

Question 14. For any two vectors $\vec{a}$ and $\vec{b}$ , Show that $(\vec{a}+\vec{b}).(\vec{a}-\vec{b})=0\Leftrightarrow|\vec{a}|=|\vec{b}|.$

Question 15. If $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ , $\vec{b}=\hat{i}+\hat{j}-2\hat{k}$ and $\vec{c}=\hat{i}+3\hat{j}-\hat{k}$ , find such that $\vec{a}$ is perpendicular to $λ\vec{b}+\vec{c}$ .

Question 16. If $\vec{p}=5\hat{i}+λ\hat{j}-3\hat{k}$ and $\vec{q}=\hat{i}+3\hat{j}-5\hat{k}$ , then find the value of λ so that $\vec{p}+\vec{q}$ and $\vec{p}-\vec{q}$ are perpendicular vectors.

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

Question 33. Find the angle between the two vectors $\vec{a}$ and $\vec{b}$ , if

(i) $|\vec{a}|$ =√3, $|\vec{b}|$ = 2 and $\vec{a}.\vec{b}$ = √6

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

Question 34. Express the vector $\vec{a}=5\hat{i}-2\hat{j}+5\hat{k}$ as the sum of two vectors such that one is parallel to the vector $\vec{b}=3\hat{i}+\hat{k}$ and other is perpendicular to $\vec{b}$

Question 35. If $\vec{a}$ and $\vec{b}$ are two vectors of the same magnitude inclined at an angle of 30° such that $\vec{a}.\vec{b}$ = 3, find $|\vec{a}|, |\vec{b}|.$

Question 36. Express $2\hat{i}-\hat{j}+3\hat{k}$ as the sum of a vector parallel and a vector perpendicular to $2\hat{i}+4\hat{j}-2\hat{k}.$

Question 37. Decompose the vector $6\hat{i}-3\hat{j}-6\hat{k}$ into vectors which are parallel and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}.$

Question 38. Let $\vec{a}=5\hat{i}-\hat{j}+7\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+λ\hat{k}$ . Find λ such that $\vec{a}+\vec{b}$ is orthogonal to $\vec{a}-\vec{b}$

Question 39. If $\vec{a}.\vec{a}=0$ and $\vec{a}.\vec{b}=0$ , what can you conclude about the vector $\vec{b}$ ?

Question 40. If $\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$ , then prove that it is perpendicular to both $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$

Question 41. If $|\vec{a}|= a$ and $|\vec{b}|= b$ , prove that $(\frac{\vec{a}}{a^2}-\frac{\vec{b}}{b^2})^2= (\frac{\vec{a}-\vec{b}}{ab})^2$

Question 42. If $\vec{a},\vec{b},\vec{c}$ are three non- coplanar vectors such that $\vec{d}.\vec{a}=\vec{d}.\vec{b}=\vec{d}.\vec{c} =0$ then show that $\vec{d}$ is the null vector.

Question 43. If a vector $\vec{a}$ is perpendicular to two non- collinear vectors $\vec{b}$ and $\vec{c}$ , then is $\vec{a}$ perpendicular to every vector in the plane of $\vec{b}$ and $\vec{c}$

Question 44. If $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ , how that the angle θ between the vectors $\vec{b}$ and $\vec{c}$ is given by cos θ = $\frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|}$

Question 45. Let $\vec{u},\vec{v}$ and $\vec{w}$ be vector such $\vec{u}+\vec{v}+\vec{w}=\vec{0}$ . $|\vec{u}|$ = 3, $|\vec{v}|$ = 4 and $|\vec{w}|$ = 5, then find $\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u}$

Question 46. Let $\vec{a}=x^2\hat{i}+2\hat{j}-2\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k},$ and $\vec{c}=x^2\hat{i}+5\hat{j}-4\hat{k}$ be three vectors. Find the values of x for which the angle between $\vec{a}$ and $\vec{b}$ is acute and the angle between $\vec{b}$ and $\vec{c}$ is obtuse.

Question 47. Find the value of x and y if the vectors $\vec{a}=3\hat{i}+x\hat{j}-\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}+y\hat{k}$ are mutually perpendicular vectors of equal magnitude.

Question 48. If $\vec{a}$ and $\vec{b}$ are two non-coplanar unit vectors such that $|\vec{a}+\vec{b}| =√3$ , find $(2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b}).$

Question 49. If $\vec{a},\vec{b}$ are two vectors such that | $\vec{a}+\vec{b}$ | = $|\vec{b}|$ , then prove that $\vec{a}+2\vec{b}$ is perpendicular to $\vec{a}.$