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

  • Last Updated : 28 Mar, 2021

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, θ3 be 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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