Ncert Solutions For Class 12 Maths Ex 10.3

Ncert Solutions For Class 12 Maths Chapter 10 Ex 10.3

Q.1 : Find the angle between two vectors \(\vec{ m }\) and \(\vec{ n }\) with magnitude \(\sqrt{ 3 }\)and 2 , respectively having \(\vec{ m }.\vec{ n } = \sqrt{ 6 }\)

Solution 1:

It is given that,

\(\left | \vec{ m } \right |= \sqrt{ 3 }\) ,

\(\left | \vec{ n } \right |\) = 2

And \(\vec{ m }.\vec{ n } = \sqrt{ 6 }\)

\(\vec{ m }.\vec{ n } = \left | \vec{ m } \right |\left | \vec{ n } \right | \cos \theta\)

Now, we know that

Therefore,

\(\boldsymbol{\Rightarrow }\) \(\sqrt{ 6 } = \sqrt{ 3 } \times 2 \times cos\theta\)

\(\boldsymbol{\Rightarrow }\) \(\cos \theta = \frac{ \sqrt{ 6 }}{ \sqrt{ 3 } \times 2 }\)

\(\boldsymbol{\Rightarrow }\) \(\cos \theta = \frac{ 1 }{ \sqrt{ 2 }}\)

\(\boldsymbol{\Rightarrow }\) \(\theta = \frac{ \pi }{ 4 }\)

Therefore, the angle between the given vectors \(\vec{ m }\) and \(\vec{ n }\) is \(\frac{ \pi }{ 4 }\)

Q 2 : Find the angle between the vectors \(\hat{ a } – 2 \hat{ b } + 3 \hat{ c } and 3 \hat{ a } – 2 \hat{ b } + \hat{ c }\)

Solution 2:

The given vectors are:

\(\vec{ m } = \hat{ a } – 2 \hat{ b } + 3 \hat{ c } and \vec{ n } = 3 \hat{ a } – 2 \hat{ b } + \hat{ c }\) \(\left |\vec{ m } \right | = \sqrt{ 1 ^{ 2 } + \left ( – 2 \right )^{ 2 } + 3 ^{ 2 }}\) \(\left |\vec{ m } \right | = \sqrt{ 1 + 4 + 9 }\) \(\left |\vec{ m } \right | = \sqrt{ 14 }\) \(\left |\vec{ n } \right | = \sqrt{ 3 ^{ 2 } + \left ( – 2 \right )^{ 2 } + 1 ^{ 2 }}\) \(\left |\vec{ n } \right | = \sqrt{ 9 + 4 + 1 }\) \(\left |\vec{ n } \right | = \sqrt{ 14 }\) \(Now, \vec{ m }.\vec{ n } = \left ( \hat{ a } – 2 \hat{ b } + 3 \hat{ c } \right )\left ( 3 \hat{ a } – 2 \hat{ b } + \hat{ c } \right )\) \(Now, \vec{ m }.\vec{ n } = 1.3 + \left ( – 2 \right )\left ( – 2 \right ) + 3.1\) \(Now, \vec{ m }.\vec{ n } = 3 + 4 + 3\) \(Now, \vec{ m }.\vec{ n } = 10\)

Also, we know that

\(\vec{ m }.\vec{ n } = \left | \vec{ m } \right |\left | \vec{ n } \right | \cos \theta\)

Therefore,

\(10 = \sqrt{ 14 } \sqrt{ 14 }\cos \theta\) \(\cos \theta = \frac{ 10 }{ 14 }\) \(\theta = \cos^{-1} \frac{ 5 }{ 7 }\)

Q 3. Find the projection of the vector \(\hat{ a } – \hat{ b }\) on the vector \(\hat{ a } + \hat{ b }\).

Solution 3:

Let, \(\hat{ i } = \hat{ a } – \hat{ b }\)

And \(\hat{ j } = \hat{ a } + \hat{ b }\)

Now, projection of vector \(\vec{ i }\) on \(\vec{ j }\) is given by,

\(\frac{ 1 }{ \left | \vec{ j} \right |} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 1 }{ \sqrt{ 1 + 1 }} \left \{ 1.1 + \left ( – 1 \right ) \left ( 1 \right ) \right \} = \frac{ 1 }{ \sqrt{ 2 }}\left ( 1 – 1 \right ) = 0\)

Hence the projection of vector \(\vec{ i }\) on \(\vec{ j }\) is 0

 

 

Q 4. Find the projection of the vector \(\hat{ a } + 3 \hat{ b } + 7 \hat{ c }\) on the vector \(7\hat{ a } – \hat{ b } + 8 \hat{ c }\)

Solution 4:

Let \(\hat{ i } = \hat{ a } + 3 \hat{ b } + 7 \hat{ c }\) and \(\hat{ j } = 7\hat{ a } – \hat{ b } + 8 \hat{ c }\)

Now, projection of vector \(\vec{ i }\) on \(\vec{ j }\) is given by,

\(\frac{ 1 }{ \vec{ j }} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 1 }{ \sqrt{ 7^{ 2 } + \left ( – 1 \right )^{ 2 } + 8 ^{ 2 }}} \left \{ 1 \left ( 7 \right ) + 3 \left ( – 1 \right ) + 7 \left ( 8 \right )\right \}\) \(\frac{ 1 }{ \vec{ j }} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 7 – 3 + 56 }{ \sqrt{ 49 + 1 + 64 }}\) \(\frac{ 1 }{ \vec{ j }} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 60 }{ \sqrt{ 114 }}\)

Q 5: Show that each of the given three vectors is a unit vector :

\(\frac{ 1 }{ 7 } \left ( 2 \hat{ a } + 3 \hat{ b } + 6 \hat{ c }\right ) , \frac{ 1 }{ 7 } \left ( 3 \hat{ a } – 6 \hat{ b } + 2 \hat{ c }\right ) , \frac{ 1 }{ 7 } \left ( 6 \hat{ a } + 2 \hat{ b } – 3 \hat{ c }\right )\)

Also, show that they are mutually perpendicular to each other.

Solution 5:

\(Let \vec{ i } = \frac{ 1 }{ 7 } \left ( 2 \hat{ a } + 3 \hat{ b } + 6 \hat{ c }\right ) = \frac{ 2 }{ 7 } \hat{ a } + \frac{ 3 }{ 7 } \hat{ b } + \frac{ 6 }{ 7 } \hat{ c } \\ Let \vec{ j } = \frac{ 1 }{ 7 } \left ( 3 \hat{ a } – 6 \hat{ b } + 2 \hat{ c }\right ) = \frac{ 3 }{ 7 } \hat{ a } – \frac{ 6 }{ 7 } \hat{ b } + \frac{ 2 }{ 7 } \hat{ c } \\ Let \vec{ k } = \frac{ 1 }{ 7 } \left ( 6 \hat{ a } + 2 \hat{ b } – 3 \hat{ c }\right ) = \frac{ 6 }{ 7 } \hat{ a } + \frac{ 2 }{ 7 } \hat{ b } – \frac{ 3 }{ 7 } \hat{ c }\)

\(\boldsymbol{\Rightarrow }\) \(\left | \vec{ i } \right | = \sqrt{ \left (\frac{ 2 }{ 7 }\right )^{ 2 } + \left (\frac{ 3 }{ 7 }\right )^{ 2 } + \left (\frac{ 6 }{ 7 }\right )^{ 2 }}\)

\(\left | \vec{ i } \right | = \sqrt{ \frac{ 4 }{ 49 } + \frac{ 9 }{ 49 } + \frac{ 36 }{ 49 }}\) \(\left | \vec{ i } \right | = 1\)

\(\boldsymbol{\Rightarrow }\) \(\left | \vec{ j } \right | = \sqrt{ \left (\frac{ 3 }{ 7 }\right )^{ 2 } + \left (\frac{ – 6 }{ 7 }\right )^{ 2 } + \left (\frac{ 2 }{ 7 }\right )^{ 2 }}\)

\(\left | \vec{ j } \right | = \sqrt{ \frac{ 9 }{ 49 } + \frac{ 36 }{ 49 } + \frac{ 4 }{ 49 }}\) \(\left | \vec{ j } \right | = 1\)

\(\boldsymbol{\Rightarrow }\) \(\left | \vec{ k } \right | = \sqrt{ \left (\frac{ 6 }{ 7 }\right )^{ 2 } + \left (\frac{ 2 }{ 7 }\right )^{ 2 } + \left (\frac{ – 3 }{ 7 }\right )^{ 2 }}\)

\(\left | \vec{ k } \right | = \sqrt{ \frac{ 36 }{ 49 } + \frac{ 4 }{ 49 } + \frac{ 9 }{ 49 }}\) \(\left | \vec{ k } \right | = 1\)

Thus, each of the given three vectors is a unit vector.

\(\boldsymbol{\Rightarrow }\) \(\vec{ i }.\vec{ j } = \frac{ 2 }{ 7 }\times \frac{ 3 }{ 7 } + \frac{ 3 }{ 7 } \times \left ( \frac{ -6 }{ 7 } \right ) + \frac{ 6 }{ 7 } \times \frac{2}{7}\)

\(\vec{ i }.\vec{ j } = \frac{6}{49} – \frac{18}{49} + \frac{12}{49}\) \(\vec{ i }.\vec{ j } = 0\)

\(\boldsymbol{\Rightarrow }\) \(\vec{ j }.\vec{ k } = \frac{3}{ 7 }\times \frac{ 6 }{ 7 } + \frac{ -6 }{ 7 } \times \left ( \frac{ 2 }{ 7 } \right ) + \frac{ 2 }{ 7 } \times \frac{ -3}{7}\)

\(\vec{ j }.\vec{ k } = \frac{18}{49} – \frac{12}{49} – \frac{6}{49}\) \(\vec{ j }.\vec{ k } = 0\)

\(\boldsymbol{\Rightarrow }\) \(\vec{ k }.\vec{ i } = \frac{6}{ 7 }\times \frac{ 2 }{ 7 } + \frac{ 2 }{ 7 } \times \left ( \frac{ 3 }{ 7 } \right ) + \frac{ -3 }{ 7 } \times \frac{ 6}{7}\)

\(\vec{ k }.\vec{ i } = \frac{12}{49} – \frac{6}{49} – \frac{18}{49}\) \(\vec{ k }.\vec{ i } = 0\)

Hence, the given three vectors are mutually perpendicular to each other.

Q 6: Find:

\(\left | \vec{m} \right | and \left | \vec{n} \right |\), if \(\left ( \vec{m} + \vec{n} \right ).\left ( \vec{m} + \vec{n} \right ) = 8\) and \(\left | \vec{m} \right | = 8\left | \vec{n} \right |\)

Solution 6:

\(\left ( \vec{m} + \vec{n} \right ).\left ( \vec{m} – \vec{n} \right ) = 8\)

\(\boldsymbol{\Rightarrow }\) \(\vec{ m}.\vec{ m} – \vec{ m}\vec{n} + \vec{ n}\vec{ m} – \vec{n}.\vec{ n} = 8\)

\(\boldsymbol{\Rightarrow }\) \(\left | \vec{ m} \right |^{ 2} – \left | \vec{ n} \right |^{ 2} = 8\)

\(\boldsymbol{\Rightarrow }\) \(8 \left | \vec{ n} \right |^{2} – \left | n \right |^{ 2} = 8\) . . . . . . . . . \(\left [ \left | \vec{ m } = 8\left | \vec{ n} \right | \right | \right ]\)

\(\boldsymbol{\Rightarrow }\) \(64 \left | \vec{n} \right |^{2} – \left | \vec{n} \right |^{2} = 8\)

\(\boldsymbol{\Rightarrow }\) \(63 \left | \vec{ n } \right |^{2} = 8\)

\(\boldsymbol{\Rightarrow }\) \(\left | \vec{n} \right |^{2} = \frac{ 8}{ 63}\)

\(\boldsymbol{\Rightarrow }\) \(\left | \vec{n} \right |^{2} = \sqrt{\frac{ 8 }{ 63 }}\) [ magnitude of a vector is non-negative]

\(\boldsymbol{\Rightarrow }\) \(\left | \vec{n} \right |^{2} = \frac{ 2 \sqrt{2}}{ 3 \sqrt{7}}\)

\(\left | \vec{m} \right | = 8 \left | \vec{n} \right |\) \(\left | \vec{m} \right | = \frac{ 8 \times 2\sqrt{2}}{ 3\sqrt{7}}\) \(\left | \vec{m} \right | = \frac{ 16 \sqrt{2}}{ 3 \sqrt{7}}\)

 

 

Q 7: Find the product of the following : \(\left ( 3 \vec{m} – 5 \vec{n}\right ).\left ( 2\vec{m} + 7\vec{n}\right )\)

Solution 7:

\(\left ( 3 \vec{m} – 5 \vec{n}\right ).\left ( 2\vec{m} + 7\vec{n}\right )\) \(= 3 \vec{m}.2 \vec{m} + 3 \vec{m}.7 \vec{n} – 5 \vec{n}.2 \vec{m} – 5 \vec{n}.7 \vec{n}\) \(= 6 \vec{m}. \vec{m} + 21 \vec{m}. \vec{n} – 10 \vec{n}. \vec{m} – 35 \vec{n}. \vec{n}\) \(= 6 \left | \vec{m} \right |^{2} + 11 \vec{m}.\vec{n} – 35 \left | \vec{n} \right |^{2}\)

Q 8: Find the magnitude of two vectors \(\vec{m} and \vec{n}\), having the same magnitude and such that the angle between them is 60° and their scalar product is \(\frac{ 1 }{ 2 }\)

Solution 8:

Let θ be the angle between the vectors \(\vec{m} and \vec{n}\).

As given in the question:

 

\(\left | \vec{m} \right | = \left | \vec{n} \right |, \vec{m}.\vec{n} = \frac{1}{2} and \theta = 60 ^{\circ}\). . . . . . . . . . . . . . . . . . . . . . . ( 1 )

We know that:

\(\vec{ m }.\vec{ n } = \left | \vec{ m } \right |\left | \vec{ n } \right | \cos \theta\)

Therefore,

\(\frac{1}{2} = \left | \vec{m} \right |\left | \vec{m} \right | cos 60 ^{\circ}\) . . . . . . . . . . . . . . . . . . . [using ( 1 )]

\(\boldsymbol{\Rightarrow }\) \(\frac{1}{2} = \left | \vec{m} \right |^{2} \times \frac{1}{2}\)

\(\boldsymbol{\Rightarrow }\) \(\left | \vec{m} \right |^{2} = 1\)

\(\boldsymbol{\Rightarrow }\) \(\left | \vec{m} \right |^{2} = \left | \vec{n} \right |^{2} = 1\)

Q 9: Find:

\(\left | \vec{y} \right |\) , if for a unit vector \(\vec{b}\) , \(\left (\vec{y } – \vec{b} \right ).\left (\vec{y } + \vec{b} \right ) = 12\)

Solution 9:

\(\left (\vec{y } – \vec{b} \right ).\left (\vec{y } + \vec{b} \right ) = 12\)

\(\boldsymbol{\Rightarrow }\) \(\vec{y}.\vec{y} + \vec{y}.\vec{b} – \vec{b}.\vec{y} – \vec{b}.\vec{b} = 12\)

\(\boldsymbol{\Rightarrow }\) \(\left |\vec{y} \right |^{ 2 } – \left |\vec{b} \right |^{ 2 } = 12\)

\(\boldsymbol{\Rightarrow }\) \(\left |\vec{y} \right |^{ 2 } – 1 = 12 \left [ \left |\vec{b} \right | = 1 as \vec{b} is a unit vector \right ]\)

\(\boldsymbol{\Rightarrow }\) \(\left |\vec{y} \right |^{ 2 } = 13\)

Therefore,

\(\left | \vec{ y } \right | = \sqrt{ 13 }\)

 

 

Q 10: If \(\vec{i} = 2\hat{a} + 2\hat{b} + 3\hat{c}\) ,

\(\vec{j} = -\hat{a} + 2\hat{b} + \hat{c}\)

and \(\vec{k} = 3\hat{a} + \hat{b}\) are such that \(\vec{i} + \lambda \vec{j}\) is perpendicular to \(\vec{k}\) then find the value of λ

Solution 10:

The given vectors are \(\vec{i} = 2\hat{a} + 2\hat{b} + 3\hat{c}\) , \(\vec{j} = -\hat{a} + 2\hat{b} + \hat{c}\) and \(\vec{k} = 3\hat{a} + \hat{b}\)

Now,

\(\vec{i} + \lambda \vec{j} = \left ( 2 \hat{a} + 2 \hat{b} + 3 \hat{c} \right ) + \lambda \left ( – \hat{a} + 2 \hat{b} + \hat{c} \right ) = ( 2 – \lambda ) \hat{a} + ( 2 + 2 \lambda )\hat{b} + \left ( 3 + \lambda \right )\hat{c}\)

If \(\left (\vec{i} + \lambda \vec{j} \right )\) is perpendicular to \(\vec{k}\) ,then

\(\left (\vec{i} + \lambda \vec{j} \right ).\vec{k} = 0\)

\(\boldsymbol{\Rightarrow }\) \(\left [\left ( 2 – \lambda \right )\hat{a} + \left ( 2 + 2\lambda \right )\hat{b} + \left ( 3 + \lambda \right )\hat{c} \right ]\left ( 3\hat{a} + \hat{b} \right ) = 0\)

\(\boldsymbol{\Rightarrow }\) \(\left [\left ( 2 – \lambda \right )3 + \left ( 2 + 2\lambda \right )1 + \left ( 3 + \lambda \right )0 \right ] = 0\)

\(\boldsymbol{\Rightarrow }\) \(6 – 3\lambda + 2 + 2\lambda = 0\)

\(\boldsymbol{\Rightarrow }\) \(– \lambda + 8 = 0\)

\(\boldsymbol{\Rightarrow }\) \(\lambda = 8\)

Hence, the required value of λ is 8.

Q 11: Show that:

\(\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a}\) is perpendicular to \(\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a}\) , for any two non zero vectors \(\vec{a} and \vec{b}\)

Solution 11:

\(\boldsymbol{\Rightarrow }\) \(\left (\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a} \right ).\left (\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a} \right )\)

\(\left (\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a} \right ).\left (\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a} \right )\) = \(\left |\vec{a} \right |^{2} \vec{b}.\vec{b} – \left |\vec{a} \right |\left |\vec{b} \right |\vec{b}.\vec{a} + \left |\vec{b} \right |\left |\vec{a} \right |\vec{a}.\vec{b} – \left |\vec{b} \right |^{2} \vec{a}.\vec{a}\)

\(= \left | \vec{a} \right |^{ 2 } \left |\vec{b} \right |^{ 2 }\)

\(\left (\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a} \right ).\left (\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a} \right )\) = 0

Hence, \(\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a}\) is perpendicular to \(\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a}\)

Q 12: If \(\vec{a}.\vec{a} = 0 and \vec{a}.\vec{b} = 0\), then what can be concluded about the vector \(\vec{b}\) ?

Solution:

It is given that \(\vec{a}.\vec{a} = 0 and \vec{a}.\vec{b} = 0\)

Now,

that \(\vec{a}.\vec{a} = 0\)

\(\boldsymbol{\Rightarrow }\) \(\left |\vec{a} \right |^{ 2 } = 0\)

\(\boldsymbol{\Rightarrow }\) \(\left |\vec{a} \right |= 0\)

Therefore, \(\vec{a}\) is a zero vector

Hence, vector \(\vec{b}\) satisfying \(\vec{a}.\vec{b} = 0\) can be any vector

Q 13:

If \(\vec{a} , \vec{b} , \vec{c}\) are unit vectors such that \(\vec{a} + \vec{b} + \vec{c}\) = 0, find the value of \(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}\)

Solution 13:

\(\left |\vec{a} + \vec{b }+ \vec{c} \right |^{ 2 } = \left (\vec{a} + \vec{b} + \vec{c} \right )\left (\vec{a} + \vec{b} + \vec{c} \right )\)

\(\left |\vec{a} + \vec{b }+ \vec{c} \right |^{ 2 }\) = \(\left |\vec{a} \right | ^{2}+ \left |\vec{b} \right |^{ 2 } +\left | \vec{c} \right |^{ 2 } + 2 \left ( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right )\)

\(\boldsymbol{\Rightarrow }\) \(0 = 1 + 1 + 1 + 2 \left ( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right )\)

\(\boldsymbol{\Rightarrow }\) \(\left ( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right ) = \frac{ – 3 }{ 2 }\)

Q 14: If either vector \(\vec{a} = \vec{0} or \vec{b} = \vec{0 }\) , then \(\vec{a}.\vec{b} = 0\). But the converse need not be true. Justify your answer with an example.

Solution 14:

Consider \(\vec{a} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k} and \vec{b} = 3 \hat{i} + 3 \hat{j } – 6 \hat{k}\)

Then,

\(\vec{a}.\vec{b} = 2.3 + 4.3 + 3 \left ( – 6 \right ) = 6 + 12 – 18 = 0\)

We now observe that :

\(\left |\vec{a} \right | = \sqrt{ 2 ^{2} + 4 ^{ 2 } + 3 ^{2}} = \sqrt{ 29 }\)

Therefore,

\(\vec{a} \neq \vec{0}\) \(\left |\vec{ b } \right | = \sqrt{ 3 ^{2} + 3 ^{ 2 } + ( -6 ) ^{2}} = \sqrt{ 54 }\)

Therefore,

\(\vec{b} \neq \vec{0}\)

Hence, the converse of the given statement need not be true

Q 15: If the vertices A, B, C of a triangle ABC are ( 1 , 2 , 3 ) , ( – 1 , 0 , 0 ) , ( 0 , 1 , 2 ) , respectively, then find ABC. [ ABC is the angle between the vectors \(\vec{BA} and \vec{BC}\) ]

Solution 15:

The vertices of ∆ABC are given as A ( 1 , 2 , 3 ) , B ( – 1 , 0 , 0 ) , and C ( 0 , 1 , 2 ). Also, it is given that ∠ ABC is the angle between the vectors \(\vec{BA} and \vec{BC}\)

\(\vec{BA} = \left \{ 1 – \left ( – 1 \right ) \right \}\hat{ i} + \left ( 2 – 0 \right ) \hat{j} + \left ( 3 – 0 \right ) \hat{k} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}\) \(\vec{BC} = \left \{ 0 – \left ( – 1 \right ) \right \}\hat{ i} + \left ( 1 – 0 \right ) \hat{j} + \left ( 2 – 0 \right ) \hat{k} = \hat{i} + \hat{j} + 2 \hat{k}\)

Therefore,

\(\vec{BA }. \vec{BC} = \left ( 2 \hat{i} + 2 \hat{j} + 3 \hat{k} \right ).\left ( \hat{i} + \hat{j} + 2 \hat{k} \right ) = 2 \times 1 + 2 \times 1 + 3 \times 2 = 2 + 2 + 6 = 10\) \(\left |\vec{BA } \right | = \sqrt{ 2 ^{ 2 } + 2 ^{2} + 3^{ 2 }} = \sqrt{ 4 + 4 + 9 } = \sqrt{ 17 }\) \(\left |\vec{BC } \right | = \sqrt{ 1 + 1 + 2 ^{ 2 }} = \sqrt{ 6 }\)

Now, it is known that :

\(\vec{BA} . \vec{BC} = \left |\vec{BA} \right |\left |\vec{BC} \right | cos \left ( \angle ABC \right )\)

Therefore,

\(10 = \sqrt{ 17 } \times \sqrt{ 6 } cos \left ( \angle ABC \right )\)

\(\boldsymbol{\Rightarrow }\) \(cos \left ( \angle ABC \right ) = \frac{ 10 }{ \sqrt{ 17 } \times \sqrt{6}}\)

\(\boldsymbol{\Rightarrow }\) \(\angle ABC = \cos^{-1}\left ( \frac{ 10 }{ \sqrt{ 102 }} \right )\)

 

 

Q 16: Show that the points A ( 1 , 2 , 7 ) , B ( 2 , 6 , 3 ) and C ( 3 , 10 , – 1 ) are collinear.

Solution 16:

The given points are A ( 1 , 2 , 7 ) , B ( 2 , 6 , 3 ) , and C ( 3 , 10 , – 1 ).

Therefore,

\(\vec{AB} = \left ( 2 – 1 \right ) \hat{i} + \left ( 6 – 2 \right ) \hat{j} + \left ( 3 – 7 \right ) \hat{k} = \hat{i} + 4 \hat{j} – 4 \hat{k}\) \(\vec{BC} = \left ( 3 – 2 \right ) \hat{i} + \left ( 10 – 6 \right ) \hat{j} + \left ( – 1 – 3 \right ) \hat{k} = \hat{i} + 4 \hat{j} – 4 \hat{k}\) \(\vec{AC} = \left ( 3 – 1 \right ) \hat{i} + \left ( 10 – 2 \right ) \hat{j} + \left ( – 1 – 7 \right ) \hat{k} = 2 \hat{i} + 8 \hat{j} – 8 \hat{k}\)

\(\boldsymbol{\Rightarrow }\) \(\left |\vec{AB} \right | = \sqrt{ 1 ^{2} + 4 ^{2} + \left ( – 4 \right )^{ 2 }} = \sqrt{ 1 + 16 + 16 } = \sqrt{ 33 }\)

\(\boldsymbol{\Rightarrow }\) \(\left |\vec{BC} \right | = \sqrt{ 1 ^{2} + 4 ^{2} + \left ( – 4 \right )^{ 2 }} = \sqrt{ 1 + 16 + 16 } = \sqrt{ 33 }\)

\(\boldsymbol{\Rightarrow }\) \(\left |\vec{AC} \right | = \sqrt{ 2 ^{2} + 8 ^{2} + 8 ^{2} } = \sqrt{ 4 + 64 + 64 } = \sqrt{ 132 }\) = \(2 \sqrt{33}\)

Therefore,

\(\left |\vec{AC} \right | = \left |\vec{AB} \right | + \left |\vec{BC} \right |\)

Hence, the given points A, B, and C are collinear.

Q 17: Show that the vectors \(2 \hat{i} + \hat{j} + \hat{k} , \hat{i} – 3 \hat{j } – 5 \hat{k} and 3 \hat{i} – 4 \hat{j} – 4 \hat{k}\) form the vertices of a right angled triangle.

Solution 17:

Let vectors \(2 \hat{i} + \hat{j} + \hat{k} , \hat{i} – 3 \hat{j } – 5 \hat{k} and 3 \hat{i} – 4 \hat{j} – 4 \hat{k}\) be position vectors of points A, B, and C respectively.

i.e , \( \vec{OA} = 2 \hat{i} – \hat{j} + \hat{k} , OB = \hat{i} – 3 \hat{j} – 5 \hat{k} and OC = 3 \hat{i} – 4 \hat{j } – 4 \hat{k}\)

Therefore,

\(\vec{AB} = \left ( 1 – 2 \right ) \hat{i} + \left ( – 3 + 1 \right ) \hat{j} + \left ( – 5 – 1 \right ) \hat{k} = – \hat{i} – 2 \hat{j} – 6 \hat{k}\) \(\vec{BC} = \left ( 3 – 1 \right ) \hat{i} + \left ( – 4 + 3 \right ) \hat{j} + \left ( – 4 + 5 \right ) \hat{k} = – 2\hat{i} – \hat{j} – \hat{k}\) \(\vec{AC} = \left ( 2 – 3 \right ) \hat{i} + \left ( – 1 + 4 \right ) \hat{j} + \left ( 1 + 4 \right ) \hat{k} = – \hat{i} + 3 \hat{j} + 5 \hat{k}\)

\(\boldsymbol{\Rightarrow }\) \(\left |\vec{AB} \right | = \sqrt{ \left ( – 1 \right )^{ 2 } + \left ( – 2 \right )^{ 2 } + \left ( – 6 \right ) ^{ 2 }} = \sqrt{ 1 + 4 + 36 } = \sqrt{ 41 }\)

\(\boldsymbol{\Rightarrow }\) \(\left |\vec{BC} \right | = \sqrt{ \left ( 2 \right )^{ 2 } + \left ( – 1 \right )^{ 2 } + \left ( 1 \right ) ^{ 2 }} = \sqrt{ 1 + 4 + 1 } = \sqrt{ 6 }\)

\(\boldsymbol{\Rightarrow }\) \(\left |\vec{AC} \right | = \sqrt{ \left ( – 1 \right )^{ 2 } + \left ( 3 \right )^{ 2 } + \left ( 5 \right ) ^{ 2 }} = \sqrt{ 1 + 9 + 25 } = \sqrt{ 35 }\)

Therefore,

\(\left |\vec{BC} \right |^{ 2 } + \left |\vec{AC} \right | ^{2} = 6 + 35 = 41 = \left |\vec{AB} \right | ^{ 2 }\)

Hence, ∆ ABC is a right – angled triangle.

Q 18:

If \(\vec{a}\) is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ \(\vec{a}\) is unit vector if

(A) λ = 1

(B) λ = – 1

(C) \(a = \left |\lambda \right |\)

(D) \(a = \frac{ 1 }{ \left | \lambda \right |}\)

Solution 18:

Vector \(\lambda \vec{a}\) is a unit vector if \(\left |\lambda \vec{a} \right | = 1\)

Now,

\(\left |\lambda \vec{a} \right | = 1\)

\(\boldsymbol{\Rightarrow }\) \(\left |\lambda \right | \left |\vec{a} \right | = 1\)

\(\boldsymbol{\Rightarrow }\) \(\left |\vec{a} \right | = \frac{ 1 }{ \left |\lambda \right | }\) . . . . . . . . . . . . [ λ ≠ 0 ]

\(\boldsymbol{\Rightarrow }\) \(a = \frac{ 1 }{ \left |\lambda \right | }\) . . . . . . . . . . . . . . . . . . [\(\left |\vec{a} \right | = a\) ]

Therefore, vector \(\lambda \vec{a}\) is a unit vector if \(a = \frac{ 1 }{ \left | \lambda \right |}\)

The correct answer is D.

 

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