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Class 12 RD Sharma Solutions – Chapter 24 Scalar or Dot Product – Exercise 24.1 | Set 3
  • Last Updated : 28 Mar, 2021

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

⇒ x2 – 2 – 2 > 0

⇒ x2 > 4

x ∈ (2, -2)

Case II: When angle between \vec{b}   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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