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

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

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

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

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Last Updated : 28 Mar, 2021

Class 12 RD Sharma Solutions - Chapter 23 Algebra of Vectors - Exercise 23.9

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

Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solutions of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors, and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solutions of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors, and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

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

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

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.