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

RD Sharma’s Class 12 Maths Chapter 24, Scalar or Dot Product Exercise 24.1 solutions are given in this article. These solutions are well designed to help the students in grasping the solving techniques for a wide array of problems. Additionally, they incorporate handy shortcuts and real-world examples to facilitate quick comprehension of concepts and expedite the problem-solving learning curve.

Scalar or Dot Product refers to a mathematical operation between two vectors that results in a scalar quantity. It is calculated by multiplying the magnitudes of the vectors and the cosine of the angle between them.

Question 1. Find[Tex] \vec{a}.\vec{b}[/Tex] when

(i) [Tex]\vec{a} = \hat{i}-2 \hat{j}+ \hat{k}[/Tex] and [Tex]\vec{b} = 4 \hat{i} -4\hat{j} +7\hat{k}[/Tex]

Solution:

 = [Tex](\hat{i}-2 \hat{j}+ \hat{k})(\hat{i}-2 \hat{j}+ \hat{k})[/Tex]

= (1)(4) + (-2)(-4) + (1)(7)

= 4 + 8 + 7

= 19

(ii) [Tex]\vec{a} = \hat{j}+2 \hat{k}[/Tex] and [Tex]\vec{b} = 2\hat{i}+\hat{k}[/Tex]

Solution:

= [Tex](\hat{j}+2 \hat{k} )(2\hat{i}+\hat{k})[/Tex]

= (0)(2) + (1)(0) + (2)(1)

= 2

(iii) [Tex] \vec{a} = \hat{j}-\hat{k}  [/Tex]and [Tex] \vec{b} = 2\hat{i}+3\hat{j}-2 \hat{k} [/Tex]

Solution:

= [Tex](\hat{j}-\hat{k})(2\hat{i}+3\hat{j}-2 \hat{k} )[/Tex]

= (0)(2) + (1)(3) + (-1)(-2)

= 0 + 3 + 2

= 5

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

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

Solution:

[Tex]\vec{a}  [/Tex]and [Tex]\vec{b}  [/Tex]are perpendicular to each other

So [Tex]\vec{a} . \vec{b} = 0[/Tex]

⇒[Tex](λ\hat{i}+2\hat{j}+\hat{k})(4\hat{i}-9\hat{j}+2\hat{k}) = 0        [/Tex] 

⇒ λ(4) + (2)(-9) + (1)(2) = 0

⇒ 4λ – 18 + 2 = 0

⇒ 4λ = 16

⇒ λ = 4

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

Solution:

[Tex]\vec{a}  [/Tex]and [Tex]\vec{b}  [/Tex]are perpendicular to each other

so[Tex] \vec{a} . \vec{b}  [/Tex]= 0 

⇒ [Tex](λ\hat{i}+2\hat{j}+\hat{k})(5\hat{i}-9\hat{j}+2\hat{k})[/Tex]

⇒ λ(5) + (2)(-9) + (1)(2) = 0

⇒ 5λ – 18 + 2 = 0

⇒ 5λ = 16

⇒ λ = 16/5

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

Solution:

[Tex]\vec{a}  [/Tex]and [Tex]\vec{b}  [/Tex]are perpendicular to each other

so[Tex] \vec{a} . \vec{b}  [/Tex] = 0  

⇒[Tex] (2\hat{i}+3\hat{j}+4\hat{k})(3\hat{i}+2\hat{j}-λ\hat{k})  [/Tex] =0

⇒ (2)(3) + (3)(2) – (4)λ = 0

⇒ 6 + 6 – 4λ = 0

⇒ 4λ = 12

⇒ λ = 3

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

Solution:

[Tex]\vec{a}  [/Tex]and [Tex]\vec{b}  [/Tex]are perpendicular to each other

so [Tex]\vec{a} . \vec{b}=0[/Tex] 

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

⇒ λ(1) + (3)(-1) + (2)(3) = 0

⇒ λ – 3 + 6 = 0

⇒ λ = 3

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

Solution:

Let the angle be θ 

cos θ = [Tex]\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}[/Tex]

= 6 /(4×3) = 1/2

Therefore, θ = cos^-1(1/2)

= π/3

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

Solution:

[Tex](\vec{a}-2\vec{b})[/Tex] = [Tex] (\hat{i}-\hat{j})-2(-\hat{j}+2\hat{k})[/Tex]

=[Tex] \hat{i}-\hat{j}+2\hat{j}-4\hat{k}[/Tex]

=[Tex] \hat{i}+\hat{j}-4\hat{k}        [/Tex] 

[Tex] (\vec{a}+\vec{b})[/Tex] = [Tex](\hat{i}-\hat{j})+(-\hat{j}+2\hat{k})[/Tex]

= [Tex]\hat{i}-\hat{j}-\hat{j}+2\hat{k}[/Tex]

= [Tex]\hat{i}-2\hat{j}+2\hat{k}[/Tex]

Now, [Tex](\vec{a}-2\vec{b}).(\vec{a}+\vec{b}) [/Tex]

= [Tex](\hat{i}+\hat{j}-4\hat{k})(\hat{i}-2\hat{j}+2\hat{k})[/Tex]

= (1)(1) + (1)(-2) + (-4)(2)

= 1 – 2 – 8

= -9

Therefore, [Tex](\vec{a}-2\vec{b}).(\vec{a}+\vec{b})[/Tex] = -9

Question 5. Find the angle between the vectors [Tex]\vec{a}  [/Tex]and [Tex]\vec{b}  [/Tex]where :

(i) [Tex]\vec{a} = \hat{i}-\hat{j}  [/Tex]and [Tex]\vec{b} = \hat{j}+\hat{k}[/Tex]

Solution:

 Let the angle be θ between [Tex]\vec{a}  [/Tex]and [Tex]\vec{b}[/Tex]

cos θ = [Tex]\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}[/Tex]

Now, [Tex]\vec{a} . \vec{b}[/Tex]

= [Tex](\hat{i}-\hat{j})(\hat{j}+\hat{k})[/Tex]

= (1)(0) + (-1)(1) + (0)(1)

= 0 – 1 + 0 = -1

|[Tex]\vec{a}  [/Tex]|= |[Tex]\hat{i}-\hat{j}  [/Tex]|

= [Tex]\sqrt{(1)^2+(-1)^2}[/Tex]

= √2

[Tex]|\vec{b}|[/Tex] = |[Tex]\hat{j}+\hat{k}[/Tex]|

= [Tex]\sqrt{(1)^2+(1)^2}[/Tex]

= √2

Now, cos θ = -1/(√2×√2)

= -1/2

θ = cos^-1(-1/2)

= 2π/3

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

Solution:

Let the angle be θ between [Tex] \vec{a}[/Tex] and [Tex]\vec{b}[/Tex]

Now, [Tex] \vec{a} . \vec{b}[/Tex]

=[Tex](3\hat{i}-2\hat{j}-6\hat{k})(4\hat{i}-\hat{j}+8\hat{k})[/Tex]

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

= 12 + 2 – 48

= -34

|[Tex]\vec{a}[/Tex]| = |[Tex]3\hat{i}-2\hat{j}-6\hat{k}[/Tex]|

= [Tex]\sqrt{(3)^2+(-2)^2+(-6)^2}[/Tex]

= √49 = 7

[Tex]|\vec{b}| = |4\hat{i}-\hat{j}+8\hat{k}|[/Tex]

= [Tex]\sqrt{(4)^2+(-1)^2+(8)^2}[/Tex]

= √81 = 9

cos θ = [Tex]\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}[/Tex]

Now, cos θ = -34/(7×9)

= -34/63

θ = cos^-1(-34/63)

(iii) [Tex]\vec{a} =2\hat{i}-\hat{j}+2\hat{k}[/Tex] and [Tex]\vec{b} =4\hat{i}+4\hat{j}-2\hat{k}[/Tex]

Solution:

Let the angle be θ between [Tex]\vec{a}[/Tex] and [Tex]\vec{b}[/Tex]

Now, [Tex] \vec{a} . \vec{b}[/Tex]

=[Tex](2\hat{i}-\hat{j}+2\hat{k})(4\hat{i}+4\hat{j}-2\hat{k})[/Tex]

= (2)(4) + (-1)(4) + (2)(-2)

= 8 – 4 – 4 = 0

|[Tex]\vec{a}[/Tex]| = |[Tex]2\hat{i}-\hat{j}+2\hat{k}[/Tex]|

= [Tex]\sqrt{(2)^2+(-1)^2+(2)^2}[/Tex]

= √9 = 3

|[Tex]\vec{b}[/Tex]| = |[Tex]4\hat{i}+4\hat{j}-2\hat{k}[/Tex]|

=[Tex] \sqrt{(4)^2+(4)^2+(-2)^2}[/Tex]

= √36 = 6

Now, cos θ = [Tex]\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}        [/Tex]  

cos θ = 0/(3×6) = 0

θ = cos^-1(0)

θ = π/2

(iv) [Tex]\vec{a} =2\hat{i}-3\hat{j}+\hat{k}[/Tex] and [Tex]\vec{b} =\hat{i}+\hat{j}-2\hat{k}[/Tex]

Solution:

Let the angle be θ between [Tex]\vec{a}[/Tex] and [Tex]\vec{b}[/Tex]

Now, [Tex]\vec{a} . \vec{b}[/Tex]

=[Tex](2\hat{i}-3\hat{j}+\hat{k})(\hat{i}+\hat{j}-2\hat{k})[/Tex]

= (2)(1) + (-3)(1) + (1)(-2)

= 2 – 3 – 2

= -3

|[Tex]\vec{a}[/Tex]| = [Tex]|2\hat{i}-3\hat{j}+\hat{k}|[/Tex]

=[Tex] \sqrt{(2)^2+(-3)^2+(-1)^2}[/Tex]

= √14 

|[Tex]\vec{b}[/Tex]| =|[Tex]\hat{i}+\hat{j}-2\hat{k}[/Tex]|

=[Tex] \sqrt{(1)^2+(1)^2+(-2)^2}[/Tex]

= √6

cos θ = [Tex] \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}[/Tex]

Now, cos θ = -3/(√14×√6)

= -3/√84

θ = cos^-1(-3/√84)

(v) [Tex]\vec{a} =\hat{i}+2\hat{j}-\hat{k}[/Tex] and [Tex]\vec{b} =\hat{i}-\hat{j}+\hat{k}[/Tex]

Solution:

Let the angle be θ between [Tex]\vec{a}[/Tex] and[Tex] \vec{b}[/Tex]

Now,[Tex] \vec{a} . \vec{b}[/Tex]

=[Tex](\hat{i}+2\hat{j}-\hat{k})(\hat{i}-\hat{j}+\hat{k})[/Tex]

= (1)(1) + (2)(-1) + (-1)(1)

= 1 – 2 – 1

= -2

|[Tex]\vec{a}[/Tex]| = |[Tex]\hat{i}+2\hat{j}-\hat{k}[/Tex]|

=[Tex] \sqrt{(1)^2+(2)^2+(-1)^2}[/Tex]

= √6

|[Tex]\vec{b}[/Tex]| = |[Tex]\hat{i}-\hat{j}+\hat{k}[/Tex]|

=[Tex] \sqrt{(1)^2+(-1)^2+(1)^2}[/Tex]

= √3 

cos θ = [Tex]\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}[/Tex]

Now, cos θ = -2/(√6×√3)

= -2/√18

= -2/3√2

θ = cos^-1(-√2 /3)

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

Solution:

Components along x, y and z axis are [Tex] \hat{i},\hat{j}[/Tex] and [Tex]\hat{k}[/Tex] respectively.

Let the angle between [Tex] \vec{a}[/Tex] and [Tex] \hat{i}[/Tex] be θ₁

Now, [Tex]\vec{a} . \hat{i}[/Tex]

= [Tex](\hat{i}-\hat{j}-\sqrt2\hat{k})(\hat{i}-0\hat{j}+0\hat{k})[/Tex]

= (1)(1) + (-1)(0) + (√2)(0) 

= 1

[Tex]|\vec{a}| = |\hat{i}-\hat{j}+\sqrt2\hat{k}|[/Tex]

= [Tex]\sqrt{(1)^2+(-1)^2+(√2)^2}[/Tex]

= √4 = 2

[Tex]|\hat{i}| = |\hat{i}+0\hat{j}+0\hat{k}|[/Tex]

= √1 = 1

cos θ₁ = [Tex]\frac{\vec{a}.\hat{i}}{|\vec{a}||\hat{i}|}[/Tex]

Now, cos θ₁ = 1/(2×1)

= 1/2

θ₁ = cos^-1(1/2) = π/3

Let the angle between [Tex]\vec{a}[/Tex] and [Tex] \hat{j}[/Tex] be θ₂

Now, [Tex]\vec{a} . \hat{j}[/Tex]

=[Tex] (\hat{i}-\hat{j}+\sqrt2\hat{k})(0\hat{i}+\hat{j}+0\hat{k})[/Tex]

= (1)(0) + (-1)(1) + (√2)(0)  

= -1

[Tex]|\hat{j}| = |0\hat{i}+\hat{j}+0\hat{k}|[/Tex]

= √1 = 1

cos θ₂ =[Tex] \frac{\vec{a}.\hat{j}}{|\vec{a}||\hat{j}|}[/Tex]

Now, cos θ₂ = -1/(2×1)

= -1/2

θ₂ = cos^-1(-1/2) = 2π/3

 Let the angle between [Tex]\vec{a}[/Tex] and [Tex]\hat{k}[/Tex] be θ₃

Now, [Tex]\vec{a} . \hat{k}[/Tex]

=[Tex] (\hat{i}-\hat{j}+\sqrt2\hat{k})(0\hat{i}+0\hat{j}+\hat{k})[/Tex]

= (1)(0) + (-1)(0) + (√2)(1)  

= √2

[Tex]|\hat{k}| = |0\hat{i}+0\hat{j}+\hat{k}|[/Tex]

= √1 = 1

cos θ₃ = [Tex] \frac{\vec{a}.\hat{k}}{|\vec{a}||\hat{k}|}[/Tex]

= 1/(√2)

= cos^-1(1/√2) = π/4

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

Solution:

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

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

[Tex] \vec{r}. \vec{a} = 0[/Tex]

⇒ [Tex](x\hat{i}+y\hat{j}+j\hat{k})(\hat{i}+\hat{j}-3\hat{k})      [/Tex] = 0

⇒ x + y – 3z = 0        ….(1)

[Tex] \vec{r}. \vec{b} = 5[/Tex]

⇒[Tex] (x\hat{i}+y\hat{j}+j\hat{k})(\hat{i}+3\hat{j}-2\hat{k})      [/Tex] = 5

⇒ x + 3y – 2z = 5     …..(2)

[Tex] \vec{r}. \vec{c} = 8[/Tex]

⇒ [Tex](x\hat{i}+y\hat{j}+j\hat{k})(2\hat{i}+\hat{j}+4\hat{k})      [/Tex] = 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[Tex] \vec{r}=\hat{i}+2\hat{j}+\hat{k}[/Tex]

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

(i) cos θ/2 = 1/2[Tex]|\hat{a}+\hat{b}|[/Tex]

Solution:

|[Tex]\hat{a}[/Tex]| = |[Tex]\hat{b}[/Tex]| = 1

|[Tex]\hat{a}+\hat{b}[/Tex]|² =([Tex]\hat{a}+\hat{b}[/Tex])² 

= [Tex](\hat{a})^2+(\hat{b})^2+2\hat{a}.\hat{b}[/Tex]

= 1 + 1 + 2[Tex]\hat{a}.\hat{b} [/Tex]

= 2 + 2|[Tex]\hat{a}||\hat{b}[/Tex]|cos θ 

= 2(1 + (1)(1)cos θ)

= 2(2cos² θ/2)

|[Tex]\hat{a}+\hat{b}[/Tex]|² = 4cos² θ/2

[Tex]\hat{a}+\hat{b}[/Tex] = 2 cos θ/2

cos θ/2 = 1/2|[Tex]\hat{a}+\hat{b}[/Tex]|

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

Solution:

[Tex]|\hat{a}| = |\hat{b}|[/Tex] = 1

[Tex]\frac{|\hat{a}-\hat{b}|^2}{|\hat{a}+\hat{b}|^2}[/Tex]= [Tex]\frac{(\hat{a}-\hat{b})^2}{(\hat{a}+\hat{b})^2} [/Tex]

=[Tex]\frac{(\hat{a})^2+(\hat{b})^2-2\hat{a}.\hat{b}}{(\hat{a})^2+(\hat{b})^2+2\hat{a}.\hat{b}}[/Tex]

= [Tex]\frac{2-2|\hat{a}||\hat{b}|cos θ }{2+2|\hat{a}||\hat{b}|cos θ }[/Tex]

=[Tex]\frac{2(1-cos θ)}{2(1+cos θ)}[/Tex]

= [Tex]\frac{2sin^2 θ/2}{2cos^2 θ/2}[/Tex]

= tan² θ/2

Therefore, tan θ/2 =[Tex]\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|} [/Tex]

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

Solution:

Let [Tex]\hat{a}[/Tex] and [Tex]\hat{b}[/Tex] be two unit vectors

Then, [Tex] |\hat{a}| = |\hat{b}| = 1[/Tex]

According to question:

[Tex]|\hat{a}+\hat{b}| = 1[/Tex]

Taking square on both sides

⇒[Tex]|\hat{a}+\hat{b}|^2 = (1)^2[/Tex]

⇒[Tex](\hat{a})^2+(\hat{b})^2+2\hat{a}.\hat{b} = 1[/Tex]

⇒ (1)²+(1)²+[Tex]2\hat{a}.\hat{b}[/Tex] = 1

⇒ 2+ 2[Tex]\hat{a}.\hat{b}[/Tex] = 1

⇒ 2[Tex]\hat{a}.\hat{b}[/Tex]= -1

⇒ \hat{a}.\hat{b} =-1/2

Now, [Tex]|\hat{a}-\hat{b}|^2 = (\hat{a}-\hat{b})^2[/Tex]

= [Tex](\hat{a})^2 + (\hat{b})^2 – 2\hat{a}.\hat{b}[/Tex]

= (1)² + (1)²  – 2 (-1/2)

= 2 + 1 = 3

Therefore, [Tex]|\hat{a}-\hat{b}|^2[/Tex] = 3 

[Tex]|\hat{a}-\hat{b}|[/Tex]=√3

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

Solution:

Given [Tex]\vec{a},\vec{b},\vec{c}  [/Tex] are mutually perpendicular so,

[Tex]\vec{a}.\vec{b}=\vec{b}.\vec{c}=\vec{c}.\vec{a} = 0[/Tex]

[Tex]|\vec{a}| = |\vec{b}| = |\vec{c}|=1     [/Tex] 

Now, 

[Tex]|\vec{a}+\vec{b}+\vec{c}|^2  [/Tex] = [Tex](\vec{a}+\vec{b}+\vec{c})^2[/Tex]

=[Tex](\vec{a})^2+(\vec{b})^2+(\vec{c})^2+2\vec{a}\vec{b}+2\vec{b}\vec{c}+2\vec{c}\vec{a}[/Tex]

= (1)² + (1)² +(1)² + 0

= 3

 [Tex]|\vec{a}+\vec{b}+\vec{c}|  [/Tex] = √3

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

Solution:

Given [Tex]|\vec{a}+\vec{b}|[/Tex]=60, [Tex]|\vec{a}-\vec{b}|[/Tex] = 40 and [Tex] |\vec{b}|[/Tex]= 46

We know that, 

(a + b)² + (a – b)² = 2(a² + b²)

⇒ [Tex]|\vec{a}+\vec{b}|^2+|\vec{a}-\vec{b}|^2 = 2(|\vec{a}|^2+|\vec{b}|^2)[/Tex]

⇒ 60² + 40² = 2([Tex]|\vec{a}| [/Tex]² + 49²)

⇒ 3600 + 1600 = 2[Tex]|\vec{a}|^2   [/Tex]+ 2401 

⇒ [Tex]2|\vec{a}|   [/Tex]= 968

⇒ [Tex]|\vec{a}|   [/Tex]= √484 =22

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

Solution:

Let [Tex]\vec{a} = \hat{i}+\hat{j}+\hat{k}[/Tex]

[Tex]|\vec{a}| =[/Tex]√(1+1+1) = √3

Let θ₁, θ₂, θ₃ be the angle between the coordinate axes and the [Tex]\vec{a}[/Tex]

cos θ₁ = [Tex]\frac{\vec{a}.\hat{i}}{|\vec{a}||\hat{i}|}[/Tex]

= 1/√3

cos θ₂ = [Tex]\frac{\vec{a}.\hat{j}}{|\vec{a}||\hat{j}|}[/Tex]

= 1/√3

cos θ₃ = [Tex]\frac{\vec{a}.\hat{k}}{|\vec{a}||\hat{k}|}[/Tex]

= 1/√3

Since, cos θ₁ = cos θ₂ = cos θ₃ 

Therefore, Given vector is equally inclined with coordinate axis.

Question 13. Show that the vectors [Tex]\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})[/Tex] are mutually perpendicular unit vectors.

Solution: 

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

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

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

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

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

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

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

= 1/49[6 – 18 + 12] = 0 

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

= 1/49[18 – 12 – 6] = 0

Since, [Tex]\vec{a}.\vec{b} = \vec{b}.\vec{c}=0[/Tex] they are mutually perpendicular unit vectors.

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

Solution:

To prove [Tex](\vec{a}+\vec{b}).(\vec{a}-\vec{b})=0\Leftrightarrow|\vec{a}|=|\vec{b}|[/Tex]

⇒[Tex](\vec{a}+\vec{b}).(\vec{a}-\vec{b})=0[/Tex]

⇒[Tex]|\vec{a}|^2-|\vec{b}|^2=0[/Tex]

⇒[Tex]|\vec{a}|=|\vec{b}|   [/Tex] 

Hence Proved

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

Solution:

Given: [Tex]\vec{a}=2\hat{i}-\hat{j}+\hat{k}[/Tex]

[Tex]\vec{b}=\hat{i}+\hat{j}-2\hat{k} [/Tex] 

[Tex]\vec{c}=\hat{i}+3\hat{j}-\hat{k}[/Tex]

According to question

[Tex]\vec{a}(λ\vec{b}+\vec{c})=0[/Tex]

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

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

⇒ 2(λ+1) – (λ+3) -2λ-1 = 0

⇒ 2λ + 2 -λ – 3 – 2λ – 1 = 0

⇒ -λ = 2

⇒ λ = -2

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

Solution:

Given, [Tex]\vec{p}=5\hat{i}+λ\hat{j}-3\hat{k} [/Tex]  

[Tex]\vec{q}=\hat{i}+3\hat{j}-5\hat{k}[/Tex]

According to question

[Tex](\vec{p}+\vec{q})(\vec{p}-\vec{q})=0[/Tex]

⇒[Tex]|\vec{p}|^2-|\vec{q}|^2=0  [/Tex]

⇒ [Tex]|\vec{p}|^2=|\vec{q}|^2[/Tex]

⇒ [Tex]\sqrt{5^2+λ^2+(-3)^2}=\sqrt{1^2+3^2+(-5)^2}[/Tex]

⇒ 25 + λ² + 9 = 1 + 9 + 25

⇒ λ² = 1

⇒ λ = 1

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The scalar or dot product is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. It is denoted as ?⋅?

Can the dot product be negative?

Yes, the dot product can be negative. This happens when the angle between the two vectors is greater than 90 degrees and less than 270 degrees. In such cases, cosθ is negative.

What is the difference between the dot product and the cross product?

The dot product returns a scalar, whereas the cross product returns a vector. The cross product is only defined in three dimensions and results in a vector perpendicular to the plane containing the two input vectors, following the right-hand rule.

What is the difference between the dot product and the cross product?

The dot product returns a scalar, whereas the cross product returns a vector. The cross product is only defined in three dimensions and results in a vector perpendicular to the plane containing the two input vectors, following the right-hand rule.