Skip to content
Related Articles

Related Articles

Class 12 RD Sharma Solutions – Chapter 24 Scalar or Dot Product – Exercise 24.1 | Set 2
  • Last Updated : 28 Mar, 2021

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 vales in eq(1) to find the value of a3

(1/√2)2 + (1/2)2 + a32 = 1  ….(1)

a32 = 1/4

a3 = 1/2

Now we find the value of θ 

\vec{a}k=a_3

|\vec{a}||k|cos\theta=a_3

(1)(1)cosθ = 1/2

cosθ = 1/2

cosθ = π/3

and components of \vec{a}

\vec{a}=\frac{1}{\sqrt{2}}i + \frac{1}{2}j+\frac{1}{2}k

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

Solution:

Given, (3\vec{a}-5\vec{b}).(2\vec{a}+7\vec{b})

=6|\vec{a}|^2+21\vec{a}.\vec{b}-10\vec{a}.\vec{b}-35|\vec{b}|^2

=6|\vec{a}|^2+11\vec{a}.\vec{b}-35|\vec{b}|^2       

= 6(2)2 + 11(1) – 35(1)2 

= 24 + 11 – 35

= 35 – 35 = 0

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

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

Solution:

Given, (\vec{x}-\vec{a})(\vec{x}+\vec{a})=8      

|\vec{x}|^2-\vec{x}.\vec{a}+\vec{a}\vec{x}+|\vec{a}|^2=8

⇒ |\vec{x}|^2-|\vec{a}|^2=8

⇒ |\vec{x}|^2-1=8

⇒ |\vec{x}|^2=9

|\vec{x}|=3

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

Solution:

Given, (\vec{x}-\vec{a})(\vec{x}+\vec{a})=12

⇒ |\vec{x}|^2-\vec{x}.\vec{a}+\vec{a}\vec{x}+|\vec{a}|^2=12

⇒ |\vec{x}|^2-|\vec{a}|^2=12

⇒ |\vec{x}|^2-1=12

⇒ |\vec{x}|^2=13

⇒ |\vec{x}|  =√13

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

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

Solution:

Given, (\vec{a}+\vec{b})(\vec{a}-\vec{b})  = 12

⇒  |\vec{a}|^2-\vec{a}.\vec{b}+\vec{a}\vec{b}-|\vec{b}|^2=12

⇒  |\vec{a}|^2-|\vec{b}|^2   = 12

⇒ 4|\vec{b}|^2-|\vec{b}|^2   = 12

⇒ 3|\vec{b}|^2  = 12

⇒ |\vec{b}|  = 2

So, 

|\vec{a}|  = 4

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

Solution:

Given, (\vec{a}+\vec{b})(\vec{a}-\vec{b})  = 8

⇒ |\vec{a}|^2-\vec{a}.\vec{b}+\vec{a}\vec{b}-|\vec{b}|^2=8

⇒ |\vec{a}|^2-|\vec{b}|^2=8

 64|\vec{b}|^2-|\vec{b}|^2=8

⇒ 63|\vec{b}|^2=8

⇒ |\vec{b}|  = √(8/63)

So, 

|\vec{a}|  = 8√(8/63)

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

Solution:

Given, (\vec{a}+\vec{b})(\vec{a}-\vec{b})= 3

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

⇒ |\vec{a}|^2-|\vec{b}|^2 = 3

⇒ 4|\vec{b}|^2-|\vec{b}|^2=3

⇒ 3|\vec{b}|^2  = 3

⇒ |\vec{b}|  = 1

So, 

|\vec{a}|  = 2

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

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

Solution:

We have, |\vec{a}-\vec{b}|

 |\vec{a}-\vec{b}|^2 =|\vec{a}|^2-2\vec{a}.\vec{b}+|\vec{b}|^2

 |\vec{a}-\vec{b}|^2  = 22 – 2 × 8 + 52

⇒ |\vec{a}-\vec{b}|^2  = 4 – 16 + 25

⇒ |\vec{a}-\vec{b}|^2  = 13

 |\vec{a}-\vec{b}|  = √13

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

Solution:

We have, |\vec{a}-\vec{b}|

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

 |\vec{a}-\vec{b}|^2   = 32 – 2 × 1 + 42

⇒ |\vec{a}-\vec{b}|^2   = 9 – 2 + 16

⇒ |\vec{a}-\vec{b}|^2  = 23

 |\vec{a}-\vec{b}|   = √23

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

Solution:

We have,  |\vec{a}-\vec{b}|

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

 |\vec{a}-\vec{b}|^2  = 22 – 2 × 4 + 32

⇒ |\vec{a}-\vec{b}|^2  = 4 – 8 + 9

 |\vec{a}-\vec{b}|^2  = 5

⇒ |\vec{a}-\vec{b}|   = √5

My Personal Notes arrow_drop_up
Recommended Articles
Page :