Ex 10.3 Class 12 Maths Ncert Solutions

Given, \( \vec{a} \cdot \vec{b}=\sqrt{6},|\overrightarrow{a}|=\sqrt{3},|\vec{b}|=2 \)
We know, \( \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \)
Putting the value of \( \vec{a} \cdot \vec{b},=|\vec{a}| \) and \( |\vec{b}| \)
\(
\therefore \sqrt{6}=\sqrt{3} \times 2 \times \cos \theta\)
\(\Rightarrow \cos \theta=\frac{\sqrt{6}}{\sqrt{3} \times 2}\)
\(\Rightarrow \cos \theta=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \theta=\cos ^{-1} \frac{1}{\sqrt{2}}\)
\(\Rightarrow \theta=\frac{\pi}{4}
\)
So, the angle between the vectors \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{4} \).

Given vectors are \( \hat{\imath}-2 \hat{\jmath}+3 \hat{k} \) and \( 3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \)
Let \( \overrightarrow{\mathrm{a}}=\hat{\imath}-2 \hat{\jmath}+3 \hat{k} \) and \( \overrightarrow{\mathrm{b}}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \)
\( |\overrightarrow{a}|=\sqrt{1^{2}+(-2)^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{35} \)
\( |\overrightarrow{b}|=\sqrt{3^{2}+(-2)^{2}+1^{2}}=\sqrt{9+4+1}=\sqrt{14} \)
Now \( \overrightarrow{a} \cdot \overrightarrow{b}=(\hat{\imath}-2 \hat{\jmath}+3 \hat{k}) \cdot(3 \hat{\imath}-2 \hat{\jmath}+\hat{k}) \)
\( \overrightarrow{a} \cdot \overrightarrow{b}=(\hat{\imath} \cdot 3 \hat{\imath}-\hat{\imath} \cdot 2 \hat{\jmath}+\hat{\imath} \cdot \hat{k}-2 \hat{\jmath} \cdot 3 \hat{\imath}-2 \hat{\jmath} \cdot 2 \hat{\jmath}+2 \hat{\jmath} \cdot \hat{k}+3 \hat{k} \cdot 3 \hat{\imath}-3 \hat{k} \cdot 2 \hat{\jmath}+3 \hat{k} \cdot \hat{k}) \)
\( \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=(3-0+0-0+4-0+0-0+3)=10[\hat{\imath} \hat{\jmath}=\hat{\jmath} \hat{k}=\hat{k} . \hat{\imath}=0] \)
\( \therefore \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=10[\hat{\imath} \cdot \hat{\imath}=\hat{\jmath} \cdot \hat{\jmath}=\hat{k} \cdot \hat{\mathrm{k}}=1] \text{ We know } \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta \)
\( \Rightarrow 10=\sqrt{14} \times \sqrt{14} \times \cos \theta \)
\( \Rightarrow 10=14 \cos \theta \)
\( \Rightarrow \cos \theta=\frac{10}{14 }\)
\( \Rightarrow \theta=\cos ^{-1}(\frac{5 }{7}) \)

Let \( \vec{a}=\hat{\imath}-\hat{\jmath} \) and \( \vec{b}=\hat{\imath}-\hat{\jmath} \)
Projection of \( \vec{a} \) and \( \vec{b} \) is given by \( \frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b}) \)
Now \( |\vec{b}|=|\hat{\imath}-\hat{\jmath}|=\sqrt{1^{2}+(-1)^{2}}=\sqrt{1+1}=\sqrt{2} \)
\( \vec{a} \cdot \vec{b}=(\hat{\imath}-\hat{\jmath}) \cdot(\hat{\imath}+\hat{\jmath}) \)
\( \Rightarrow \vec{a} \cdot \vec{b}=\hat{\imath} . \hat{\imath}+\hat{\imath} . \hat{\jmath}-\hat{\jmath} . \hat{\imath}+\hat{\jmath} . \hat{\jmath}[\hat{\imath} . \hat{\jmath}=\hat{\jmath} \cdot \hat{k}=\hat{k} . \hat{\imath}=0] \)
\( \Rightarrow \vec{a} \cdot \vec{b}=1+0-0-1 \quad [\hat{\imath} . \hat{\imath}=\hat{\jmath} \cdot \hat{\jmath}=\hat{k} . \hat{k}=1] \)
\( \Rightarrow \vec{a} \cdot \vec{b}=1-1=0 \)
\( \therefore \) Projection of \( \vec{a} \) on \( \vec{b}=\frac{1}{|\overrightarrow{b}|}(\vec{a} \cdot \vec{b}) \)
\(
=\frac{1}{\sqrt{2}} \times 0
\)
\( =0 \)
\( \therefore \) Projection of \( \overrightarrow{\mathrm{a}} \) on \( \overrightarrow{\mathrm{b}} \) is 0 .

Let \( \overrightarrow{a}=\hat{\imath}+3 \hat{\jmath}+7 \hat{k} \) and \( \overrightarrow{b}=7 \hat{\imath}-\hat{\jmath}+8 \hat{k} \)
Projection of \( \overrightarrow{a} \) on \( \overrightarrow{b} \) is given by \( \frac{1}{|\overrightarrow{b}|}(\overrightarrow{a} \cdot \overrightarrow{b}) \)
Now \( |\overrightarrow{b}|=|7 \hat{\imath}-\hat{\jmath}+8 \hat{k}|=\sqrt{7^{2}+(-1)^{2}+8^{2}}=\sqrt{49+1+64}=\sqrt{114} \)
\( \overrightarrow{a} \cdot \overrightarrow{b}=(\hat{\imath}+3 \hat{\jmath}+7 \hat{k}) \cdot(7 \hat{\imath}-\hat{\jmath}+8 \hat{k}) \)
\( \Rightarrow \overrightarrow{a} \cdot \overrightarrow{b}=(\hat{\imath} .7 \hat{\imath}-\hat{\imath} . \hat{\jmath}+\hat{\imath} .8 \hat{k}-2 \hat{\jmath} .7 \hat{\imath}-3 \hat{\jmath} . \hat{\jmath}+3 \hat{\jmath} .8 \hat{k}+7 \hat{k} .7 \hat{\imath}-7 \hat{k} . \hat{\jmath}+7 \hat{k} .8 \hat{k}) \)
\( \Rightarrow \overrightarrow{a} \cdot \overrightarrow{b}=7-0+0+0-3+0+0-0+56[\hat{i} \cdot \hat{\jmath}=\hat{\jmath} \cdot \hat{k}=\hat{k} \cdot \hat{\imath}=0] \)
\( \Rightarrow \overrightarrow{a} \cdot \overrightarrow{b}=7-3+56=60 \)
\( [\hat{\imath} \cdot \hat{\imath}=\hat{\jmath} \cdot \hat{\jmath}=\hat{k} . \hat{k}=1] \)
\( \therefore \) Projection of \( \overrightarrow{a} \) on \( \overrightarrow{b}=\frac{1}{|\overrightarrow{b}|}(\overrightarrow{a} \cdot \overrightarrow{b}) \)
\( =\frac{1}{\sqrt{114}} \times 60 \)
\( =\frac{60}{\sqrt{114}} \)
\( \therefore \) Projection of \( \overrightarrow{a} \) on \( \overrightarrow{b} \) is \( \frac{60}{\sqrt{114}} \).

Let \( \overrightarrow{a}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k})=\frac{2}{7} \hat{\imath}+\frac{3}{7} \hat{\jmath}+\frac{6}{7} \hat{k} \)
\(
\overrightarrow{\mathrm{b}}=\frac{1}{7}(3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k})=\frac{3}{7} \hat{\imath}-\frac{6}{7} \hat{\jmath}+\frac{2}{7} \hat{k}\)
\(\overrightarrow{\mathrm{c}}=\frac{1}{7}(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k})=\frac{6}{7} \hat{\imath}+\frac{2}{7} \hat{\jmath}-\frac{3}{7} \hat{k}
\)
Now we need to find out the magnitude of \( \overrightarrow{a}, \overrightarrow{b} \& \overrightarrow{c} \)
\(
|\overrightarrow{\mathrm{a}}|=\sqrt{\left(\frac{2}{7}\right)^{2}+\left(\frac{3}{7}\right)^{2}+\left(\frac{6}{7}\right)^{2}}=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=\sqrt{\frac{49}{49}}=1\)
\(|\overrightarrow{\mathrm{b}}|=\sqrt{\left(\frac{3}{7}\right)^{2}+\left(-\frac{6}{7}\right)^{2}+\left(\frac{2}{7}\right)^{2}}=\sqrt{\frac{9}{49}+\frac{36}{49}+\frac{4}{49}}=\sqrt{\frac{49}{49}}=1\)
\(|\overrightarrow{\mathrm{c}}|=\sqrt{\left(\frac{6}{7}\right)^{2}+\left(\frac{2}{7}\right)^{2}+\left(-\frac{3}{7}\right)^{2}}=\sqrt{\frac{36}{49}+\frac{4}{49}+\frac{9}{49}}=\sqrt{\frac{49}{49}}=1
\)
Since, \( |\overrightarrow{a}|=1,|\overrightarrow{b}|=1 \) and \( |\overrightarrow{c}|=1 \)
\( \therefore \) the three given vectors are unit vectors.
To show that each of three vectors are mutually perpendicular to each other
We have to show \( \overrightarrow{a} \cdot \overrightarrow{b}=0, \overrightarrow{b} \cdot \overrightarrow{c}=0 \) and \( \overrightarrow{c} \cdot \overrightarrow{a}=0 \)
\(
\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}= \left(\frac{2}{7} \hat{\imath}+\frac{3}{7} \hat{\jmath}+\frac{6}{7} \hat{k}\right)\left(\frac{3}{7} \hat{\imath}-\frac{6}{7} \hat{\jmath}+\frac{2}{7} \hat{k}\right)\)
\(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}= \frac{2}{7} \hat{\imath} \cdot \frac{3}{7} \hat{\jmath}-\frac{2}{7} \hat{\imath} \cdot \frac{6}{7} \hat{k}+\frac{2}{7} \hat{\imath} \cdot \frac{2}{7} \hat{k}+\frac{3}{7} \hat{\jmath} \cdot \frac{3}{7} \hat{\imath}-\frac{3}{7} \hat{\jmath} \cdot \frac{6}{7} \hat{\jmath}+\frac{3}{7} \hat{\jmath} \cdot \frac{2}{7} \hat{k}+\frac{6}{7} \hat{k} \frac{3}{7} \hat{\imath}\)
\(-\frac{6}{7} \hat{k} \cdot \frac{6}{7} \hat{\jmath}+\frac{6}{7} \hat{k} \cdot \frac{2}{7} \hat{k}\)
\(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}= \frac{6}{7}-0+0+0-\frac{18}{7}+0+0-0+\frac{12}{7}
\)
\( \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\frac{18}{7}-\frac{18}{7}=0[\hat{i} \cdot \hat{\jmath}=\hat{\jmath} \cdot \hat{\mathrm{k}}=\hat{k} \cdot \hat{\mathrm{i}}=0] \)
\( [\hat{\imath} \cdot \hat{\imath}=\hat{\jmath} \cdot \hat{\jmath}=\hat{k} \cdot \hat{k}=1] \)
\( \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\left(\frac{3}{7} \hat{\imath}-\frac{6}{7} \hat{\jmath}+\frac{2}{7} \hat{k}\right)\left(\frac{6}{7} \hat{\imath}+\frac{2}{7} \hat{\jmath}-\frac{3}{7} \hat{k}\right) \)
\( \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\frac{3}{7} \hat{\imath} \cdot \frac{6}{7} \hat{\imath}+\frac{3}{7} \hat{\imath} \cdot \frac{2}{7} \hat{\jmath}-\frac{3}{7} \hat{\imath} \cdot \frac{3}{7} \hat{k}-\frac{6}{7} \hat{\jmath} \cdot \frac{6}{7} \hat{\imath}+\frac{6}{7} \hat{\jmath} \cdot \frac{2}{7} \hat{\jmath}-\frac{6}{7} \hat{\jmath} \frac{3}{7} \hat{k}+\frac{2}{7} \hat{k} \cdot \frac{6}{7} \hat{\imath} \)
\(
+\frac{2}{7} \hat{k} \cdot \frac{2}{7} \hat{\jmath}-\frac{2}{7} \hat{k} \cdot \frac{3}{7} \hat{k}
\)
\( \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\frac{18}{7}+0-0-0-\frac{12}{7}+0+0+0+\frac{6}{7} \)
\( \overrightarrow{b} \cdot \overrightarrow{c}=\frac{18}{7}-\frac{18}{7}=0 \)
\( \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=\left(\frac{6}{7} \hat{\imath}+\frac{2}{7} \hat{\jmath}-\frac{3}{7} \hat{k}\right)\left(\frac{2}{7} \hat{\imath}+\frac{3}{7} \hat{\jmath}+\frac{6}{7} \hat{k}\right) \)
\( \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=\frac{6}{7} \hat{\imath} \cdot \frac{2}{7} \hat{\imath}+\frac{6}{7} \hat{\imath} \cdot \frac{3}{7} \hat{\jmath}+\frac{6}{7} \hat{\imath} \cdot \frac{6}{7} \hat{k}+\frac{2}{7} \hat{\jmath} \cdot \frac{2}{7} \hat{\imath}+\frac{2}{7} \hat{\jmath} \cdot \frac{3}{7} \hat{\jmath}+\frac{2}{7} \hat{\jmath} \cdot \frac{6}{7} \hat{k}-\frac{3}{7} \hat{k} \frac{2}{7} \hat{\imath} \)
\(
+\frac{2}{7} \hat{k} \cdot \frac{2}{7} \hat{\jmath}-\frac{2}{7} \hat{k} \cdot \frac{3}{7} \hat{k}
\)
\( \overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}}=\frac{12}{7}+0+0+0+\frac{6}{7}+0-0-0-\frac{18}{7} \)
\( \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=\frac{18}{7}-\frac{18}{7}=0 \)
Since, \( \overrightarrow{a} \cdot \overrightarrow{b}=0, \overrightarrow{b} \cdot \overrightarrow{c}=0 \) and \( \overrightarrow{c} \cdot \overrightarrow{a}=0 \)
\( \therefore \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}} \) and \( \overrightarrow{\mathrm{c}} \) are mutually perpendicular to each other.

\( \operatorname{Let}(\overrightarrow{a}+\overrightarrow{b}),(\overrightarrow{a}-\overrightarrow{b})=8 \)
Given, \( |\overrightarrow{a}|=8|\overrightarrow{b}||\overrightarrow{a}|=8|\overrightarrow{b}| \)
\(
\text { Now, }(\overrightarrow{a}+\overrightarrow{b}),(\overrightarrow{a}-\overrightarrow{b})=8\)
\(\Rightarrow \overrightarrow{a} \cdot \overrightarrow{a}-\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{a}-\overrightarrow{b} \cdot \overrightarrow{b}=8\)
\(\Rightarrow|\overrightarrow{a}|^{2}-|\overrightarrow{b}|^{2}=8[\text { since, } \overrightarrow{a} \cdot \overrightarrow{b}=\overrightarrow{b} \cdot \overrightarrow{a}]\)
\(\Rightarrow(8|\overrightarrow{b}|)^{2}-|\overrightarrow{b}|^{2}=8\)
\(\Rightarrow 64|\overrightarrow{b}|^{2}-|\overrightarrow{b}|^{2}=8\)
\(\Rightarrow 63|\overrightarrow{b}|^{2}=8\)
\(\Rightarrow|\overrightarrow{b}|^{2}=\frac{8}{63}\)
\(\Rightarrow|\overrightarrow{b}|=\sqrt{\frac{8}{63}}=\frac{2 \sqrt{2}}{3 \sqrt{7}}\)
\(\therefore|\overrightarrow{b}|=\frac{2 \sqrt{2}}{3 \sqrt{7}}\)
\(|\overrightarrow{a}|=8|\overrightarrow{b}|\)
\(\Rightarrow|\overrightarrow{a}|=8 \times \frac{2 \sqrt{2}}{3 \sqrt{7}}\)
\(\therefore|\overrightarrow{a}|=\frac{16 \sqrt{2}}{3 \sqrt{7}}
\)

To find \( (3 \overrightarrow{a}-5 \overrightarrow{b}) \cdot(2 \overrightarrow{a}+7 \overrightarrow{b}) \)
\(
=3 \overrightarrow{a} \cdot 2 \overrightarrow{a}+3 \overrightarrow{a} \cdot 7 \overrightarrow{b}-5 \overrightarrow{b} \cdot 2 \overrightarrow{a}-5 \overrightarrow{b} \cdot 7 \overrightarrow{b}\)
\(=6 \overrightarrow{a} \cdot \overrightarrow{a}+21 \overrightarrow{a} \cdot \overrightarrow{b}-10 \overrightarrow{a} \cdot \overrightarrow{b}-35 \overrightarrow{b} \cdot \overrightarrow{b} \quad \text{[Since,} \overrightarrow{a} \cdot \overrightarrow{b}=\overrightarrow{b} \cdot \overrightarrow{a}] \)
\( =6|\overrightarrow{a}|^{2}+11 \overrightarrow{a} \cdot \overrightarrow{b}-35|\overrightarrow{b}|^{2} \)

Let \( \theta \) be the angle between \( \overrightarrow{a} \) and \( \overrightarrow{b} \)
Given, magnitude of vector \( \overrightarrow{a}= \) magnitude of vector \( \overrightarrow{b} \)
\( |\overrightarrow{a}|=|\overrightarrow{b}| \) and \( \theta=60^{\circ} \)
Scalar product of \( \overrightarrow{a} \) and \( \overrightarrow{b}=\frac{1}{2} \)
i.e. \( \overrightarrow{a} \cdot \overrightarrow{b}=\frac{1}{2} \)
We know \( \overrightarrow{a} \cdot \overrightarrow{b}=|\overrightarrow{a}|=|\overrightarrow{b}| \cos \theta \)
\(
\Rightarrow \overrightarrow{a} \cdot \overrightarrow{b}=|\overrightarrow{a}|^{2} \cos \theta\)
\(\Rightarrow \frac{1}{2}=|\overrightarrow{a}|^{2} \cos 60^{\circ}\)
\(\Rightarrow \frac{1}{2}=|\overrightarrow{a}|^{2} \times \frac{1}{2}
\)
\( \Rightarrow|\overrightarrow{a}|^{2}=1 \) [magnitude of vector is positive]
So, \( |\overrightarrow{b}|=1 \)
Hence, \( |\overrightarrow{a}|=|\overrightarrow{b}|=1 \)

Given vector \( \overrightarrow{a} \) is a unit vector
\(
\therefore \ \mid \overrightarrow{a} \mid=1\)
\((\overrightarrow{x}-\overrightarrow{a}) \cdot(\overrightarrow{x}+\overrightarrow{a})=12\)
\(\Rightarrow \overrightarrow{x} \cdot \overrightarrow{x}+\overrightarrow{x} \cdot \overrightarrow{a}-\overrightarrow{a} \cdot \overrightarrow{x}+\overrightarrow{a} \cdot \overrightarrow{a}=12\)
\(\Rightarrow|\overrightarrow{x}|^{2}+\overrightarrow{x} \cdot \overrightarrow{a}-\overrightarrow{x} \cdot \overrightarrow{a}-|\overrightarrow{a}|^{2}=12\)
\( [ \) Since, \( \overrightarrow{x} \cdot \overrightarrow{a}=\overrightarrow{a} \cdot \overrightarrow{x}] \)
\(\Rightarrow|\overrightarrow{x}|^{2}-|\overrightarrow{a}|^{2}=12\)
\(\Rightarrow|\overrightarrow{x}|^{2}-1^{2}=12\)
\(\Rightarrow|\overrightarrow{x}|^{2}=12+1=13\)
\(\Rightarrow|\overrightarrow{x}|^{2}=\sqrt{13}\)
\(\therefore|\overrightarrow{x}|^{2}=\sqrt{13}
\)

Given \( \overrightarrow{\mathrm{a}}=2 \hat{\imath}-2 \hat{\jmath}+3 \hat{k}, \overrightarrow{\mathrm{b}}=-\hat{\imath}+2 \hat{\jmath}+\hat{k} \) and \( \overrightarrow{\mathrm{c}}=3 \hat{\imath}+\hat{\jmath} \)
\(
\overrightarrow{a}+\lambda \overrightarrow{b}=(2 \hat{\imath}-2 \hat{\jmath}+3 \hat{k})+\lambda(-\hat{\imath}+2 \hat{\jmath}+\hat{k})\)
\(\Rightarrow \overrightarrow{a}+\lambda \overrightarrow{b}=(2-\lambda) \hat{\imath}-(2+\lambda) \hat{\jmath}+(3+\lambda) \hat{k}
\)
It is given \( \overrightarrow{a}+\lambda \overrightarrow{b} \) is perpendicular to \( \overrightarrow{c} \) then
\(
(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}=0\)
\(((2-\lambda) \hat{\imath}-(2+2 \lambda) \hat{\jmath}+(3+\lambda) \hat{k}) \cdot(3 \hat{\imath}+\hat{\jmath})=0\)
\(\Rightarrow(2-\lambda) \hat{\imath} \cdot 3 \hat{\imath}+(2-\lambda) \hat{\imath} \cdot \hat{\jmath}+(2+2 \lambda) \hat{\jmath} \cdot 3 \hat{\imath}+(2+2 \lambda) \hat{\jmath} \cdot \hat{\jmath}+\)
\(\quad(3+\lambda) \hat{k} \cdot 3 \hat{\imath}+(3+\lambda) \hat{k} \cdot \hat{\jmath}=0\)
\(\Rightarrow(2-\lambda) \times 3+0+0+2+2 \lambda+0+0=0[\hat{\imath} \cdot \hat{\jmath}=\hat{\jmath} \cdot \hat{k}=\hat{k} \cdot \hat{\imath}=0]
\)
\(
\Rightarrow 6-3 \lambda+2+2 \lambda=0\)
\(\Rightarrow 8-\lambda=0\)
\(\therefore \lambda=8
\)

To show \( |\overrightarrow{a}| \overrightarrow{b}+|\overrightarrow{b}| \overrightarrow{a} \) is perpendicular to \( |\overrightarrow{a}| \overrightarrow{b}-|\overrightarrow{b}| \overrightarrow{a} \) for any two non zero vectors \( \overrightarrow{a} \) and \( \overrightarrow{b} \), we need to show dot product of \( |\overrightarrow{a}| \overrightarrow{b}+|\overrightarrow{b}| \overrightarrow{a} \) and \( |\overrightarrow{a}| \overrightarrow{b}- \) \( |\overrightarrow{b}| \overrightarrow{a} \) is zero.
\(
(|\overrightarrow{a}| \overrightarrow{b}+|\overrightarrow{b}| \overrightarrow{a}) \cdot(|\overrightarrow{a}| \overrightarrow{b}-|\overrightarrow{b}| \overrightarrow{a})\)
\(= (|\overrightarrow{a}| \overrightarrow{b}) \cdot(|\overrightarrow{a}| \overrightarrow{b})-(|\overrightarrow{a}| \overrightarrow{b}) \cdot(|\overrightarrow{b}| \overrightarrow{a})+(|\overrightarrow{b}| \overrightarrow{a}) \cdot(|\overrightarrow{a}| \overrightarrow{b})-(|\overrightarrow{b}| \overrightarrow{a}) \cdot(|\overrightarrow{b}| \overrightarrow{a})\)
\(= |\overrightarrow{a}|^{2} \overrightarrow{b} \cdot \overrightarrow{b}-|\overrightarrow{b}|^{2} \overrightarrow{a} \cdot \overrightarrow{a}\)
\(= |\overrightarrow{a}|^{2}|\overrightarrow{b}|^{2}-|\overrightarrow{b}|^{2}|\overrightarrow{a}|^{2} \quad[\text { Since }, \overrightarrow{a} \cdot \overrightarrow{b}=\overrightarrow{b} \cdot \overrightarrow{a}]\)
\(= 0
\)
Since, dot product of \( |\overrightarrow{a}| \overrightarrow{b}+|\overrightarrow{b}| \overrightarrow{a} \) and \( |\overrightarrow{a}| \overrightarrow{b}-|\overrightarrow{b}| \overrightarrow{a} \) is zero.
\( \therefore|\overrightarrow{a}| \overrightarrow{b}+|\overrightarrow{b}| \overrightarrow{a} \) is perpendicular to \( |\overrightarrow{a}| \overrightarrow{b}-|\overrightarrow{b}| \overrightarrow{a} \)

Given, \( \overrightarrow{a} \cdot \overrightarrow{a}=0 \)
\(
\Rightarrow|\vec{a}|^{2}=0\)
\(\Rightarrow|\overrightarrow{a}|=0
\)
\( \therefore \overrightarrow{\mathrm{a}} \) is a zero vector.
Hence, \( \overrightarrow{b} \) in \( \overrightarrow{a} \cdot \overrightarrow{b}=0 \) can be any vector.

Given \( \overrightarrow{a}, \overrightarrow{b} \) and \( \overrightarrow{c} \) are unit vectors and \( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0 \)
\(
|\overrightarrow{a}|=1,|\overrightarrow{b}|=1 \text { and }|\overrightarrow{c}|=1
\)
Now
\(
\overrightarrow{a} \cdot(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{a} \cdot \overrightarrow{0}\)
\(\Rightarrow \overrightarrow{a} \cdot \overrightarrow{a}+\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{a} \cdot \overrightarrow{c}=0\)
\(\Rightarrow 1+\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{a} \cdot \overrightarrow{c}=0\ldots(i))\)
\(\overrightarrow{b} \cdot(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{b} \cdot \overrightarrow{0}\)
\(\Rightarrow \overrightarrow{b} \cdot \overrightarrow{a}+\overrightarrow{b} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}=0\)
\(\Rightarrow 1+\overrightarrow{b} \cdot \overrightarrow{a}+\overrightarrow{b} \cdot \overrightarrow{c}=0\ldots(ii)\)
\( \overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{c}} \cdot \overrightarrow{0} \)
\(
\Rightarrow \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}=0\)
\(\Rightarrow 1+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}}=0\ldots(iii)\)
Adding (i), (ii) and (iii) we get
\(
1+\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{a} \cdot \overrightarrow{c}+1+\overrightarrow{b} \cdot \overrightarrow{a}+\overrightarrow{b} \cdot \overrightarrow{c}+1+\overrightarrow{c} \cdot \overrightarrow{a}+\overrightarrow{c} \cdot \overrightarrow{b}=0\)
\(\Rightarrow 3+\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{c} \cdot \overrightarrow{a}+\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}+\overrightarrow{b} \cdot \overrightarrow{c}=0 {[\text { Since, } \overrightarrow{a} \cdot \overrightarrow{b}=\overrightarrow{b} \cdot \overrightarrow{a}]}\)
\(\Rightarrow 3+2(\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a})=0 {[\text { Since, } \overrightarrow{b} \cdot \overrightarrow{c}=\overrightarrow{c} \cdot \overrightarrow{b}]}\)
\(\Rightarrow 2(\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a})=-3 {[\text { Since, } \overrightarrow{a} \cdot \overrightarrow{c}=\overrightarrow{c} \cdot \overrightarrow{a}]}\)
\(\therefore \overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}=-\frac{3}{2}
\)

If either \( \overrightarrow{a}=0 \) Or \( \overrightarrow{b}=0 \) then \( \overrightarrow{a} \cdot \overrightarrow{b}=0 \)
Now let \( \overrightarrow{a}=2 \hat{\imath}+3 \hat{\jmath}+6 \hat{\mathrm{k}} \) and \( \overrightarrow{b}=3 \hat{\imath}-6 \hat{\jmath}+2 \hat{\mathrm{k}} \)
\(
\overrightarrow{a} \cdot \overrightarrow{b}=(2 \hat{\mathrm{\imath}}+3 \hat{\mathrm{\jmath}}+6 \hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{\imath}}-6 \hat{\mathrm{\jmath}}+2 \hat{\mathrm{k}})\)
\(\Rightarrow \overrightarrow{a} \cdot \overrightarrow{b}=2 \hat{\mathrm{\imath}} \cdot 3 \hat{\imath}-2 \hat{\mathrm{\imath}} \cdot 6 \hat{\jmath}+2 \hat{\mathrm{\imath}} \cdot 2 \hat{\mathrm{k}}+3 \hat{\jmath} \cdot 3 \hat{\imath}-3 \hat{\jmath} \cdot 6 \hat{\jmath}+3 \hat{\jmath} \cdot 2 \hat{\mathrm{k}}+6 \hat{\mathrm{k}} \cdot 3 \hat{\mathrm{\imath}}\)
\( -6 \hat{\mathrm{k}} \cdot 6 \hat{\jmath}+6 \hat{\mathrm{k}} \cdot 2 \hat{\mathrm{k}}
\)
\(\Rightarrow \overrightarrow{a} \cdot \overrightarrow{b}=6-0+0+0-18+0+0-0+12 \quad[\hat{\imath} . \hat{\jmath}=\hat{\mathrm{\jmath}} \cdot \hat{\mathrm{k}}=\hat{\mathrm{k}} \cdot \hat{\mathrm{i}}=0]\)
\(\Rightarrow \overrightarrow{a} \cdot \overrightarrow{b}=18-18=0[\hat{\mathrm{\imath}} \cdot \hat{\mathrm{\imath}}=\hat{\mathrm{\jmath}} \cdot \hat{\jmath}=\hat{\mathrm{k}} \cdot \hat{\mathrm{k}}=1]
\)
So, \( \overrightarrow{a} \cdot \overrightarrow{b}=0 \)
\(
|\overrightarrow{a}|=\sqrt{2^{2}+3^{2}+6^{2}}=\sqrt{4+9+36}=\sqrt{49}=7\)
\(\therefore|\overrightarrow{a}| \neq 0\)
\(|\overrightarrow{b}|=\sqrt{3^{2}+(-6)^{2}+2^{2}}=\sqrt{9+36+4}=\sqrt{49}=7\)
\(\therefore|\overrightarrow{b}| \neq 0\)
Since \( \overrightarrow{a} . \overrightarrow{b}=0 \overrightarrow{a} \neq 0, \overrightarrow{b} \neq 0 \)
Hence, the converse of given statement need not be true.

Given points are \( \mathrm{A}(1,2,3), \mathrm{B}(-1,0,0) \) and \( \mathrm{C}(0,1,2) \)
\( \angle \mathrm{ABC} \) is the angle between vectors \( \overrightarrow{\mathrm{BA}} \) and \( \overrightarrow{\mathrm{BC}} \)
\(\overrightarrow{\mathrm{BA}}=\{1-(-1)\} \hat{\imath}+(2-0) \hat{\jmath}+(3-0) \hat{\mathrm{k}}=2 \hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{BC}}=\{0-(-1)\} \hat{\imath}+(1-0) \hat{\jmath}+(2-0) \hat{\mathrm{k}}=\hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=(2 \hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}}) \cdot(\hat{\imath}+\hat{\jmath}+2 \hat{k})\)
\(\Rightarrow \overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=2 \hat{\imath} \cdot \hat{\imath}+2 \hat{\imath} \cdot \hat{\jmath}+2 \hat{\imath} \cdot 2 \hat{\mathrm{k}}+2 \hat{\jmath} \cdot \hat{\imath}+2 \hat{\jmath} \cdot \hat{\jmath}+2 \hat{\jmath} \cdot 2 \hat{\mathrm{k}}+3 \hat{\mathrm{k}} \cdot \hat{\imath}+3 \hat{k} \cdot \hat{\jmath}+3 \hat{k} \cdot 2 \hat{k}\)
\(\Rightarrow \overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=2+0+0+0+2+0+0+0+6[\hat{\imath} \cdot \hat{\jmath}=\hat{\jmath} \cdot \hat{k}=\hat{k} \cdot \hat{\imath}=0]\)
\(\Rightarrow \overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=10 \quad[\hat{\mathrm{i}} \cdot \hat{\imath}=\hat{\jmath} \cdot \hat{\jmath}=\hat{\mathrm{k}} \cdot \hat{\mathrm{k}}=1]\)
\(|\overrightarrow{\mathrm{BA}}|=|2 \hat{\imath}+2 \hat{\jmath}+3 \hat{k}|=\sqrt{2^{2}+2^{2}+3^{2}}=\sqrt{4+4+9}=\sqrt{17}\)
\(|\overrightarrow{\mathrm{BC}}|=|\hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}|=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{1+1+4}=\sqrt{6}\)
We know
\(\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=|\overrightarrow{\mathrm{BA}}||\overrightarrow{\mathrm{BC}}| \cos \angle \mathrm{ABC}\)
\(\Rightarrow 10=\sqrt{17} \times \sqrt{6} \times \cos \angle \mathrm{ABC}\)
\(\Rightarrow \cos \angle \mathrm{ABC}=\frac{10}{\sqrt{17} \times \sqrt{6}}\)
\(\Rightarrow \cos \angle \mathrm{ABC}=\frac{10}{\sqrt{102}}\)
\(\Rightarrow \angle \mathrm{ABC}=\cos ^{-1}(\frac{10}{ \sqrt{102}} )\)

Given points are A \( (1,2,7), B(2,6,3) \) and \( C(3,10,-1) \)
So,
\(\overrightarrow{\mathrm{AB}}=(2-1) \hat{\imath}+(6-2) \hat{\jmath}+(3-7) \hat{\mathrm{k}}=\hat{\imath}+4 \hat{\jmath}-4 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{BC}}=(3-2) \hat{\imath}+(10-6) \hat{\jmath}+(-1-3) \hat{\mathrm{k}}=\hat{\imath}+4 \hat{\jmath}-4 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{AC}}=(3-1) \hat{\imath}+(10-2) \hat{\jmath}+(-1-7) \hat{\mathrm{k}}=2 \hat{\imath}+8 \hat{\jmath}-8 \hat{\mathrm{k}}\)
\(|\overrightarrow{\mathrm{AB}}|=\sqrt{1^{2}+4^{2}+(-4)^{2}}=\sqrt{1+16+16}=\sqrt{33}\)
\(|\overrightarrow{\mathrm{BC}}|=\sqrt{1^{2}+4^{2}+(-4)^{2}}=\sqrt{1+16+16}=\sqrt{33}\)
\(|\overrightarrow{\mathrm{AC}}|=\sqrt{2^{2}+8^{2}+(-8)^{2}}=\sqrt{4+64+64}=\sqrt{132}=2 \sqrt{33}\)
Since, \( |\overrightarrow{A C}|=|\overrightarrow{A B}|+|\overrightarrow{B C}| \)
\( \therefore \) The given points \( \mathrm{A}, \mathrm{B} \) and C are collinear.

Let vectors \( 2 \hat{\imath}-\hat{\jmath}+\hat{k}, \hat{\imath}-3 \hat{\jmath}-5 \hat{k} \) and \( 3 \hat{\imath}-4 \hat{\jmath}-4 \hat{k} \) be position vectors of points \( \mathrm{A}, \mathrm{B} \) and C respectively.
Then, \( \overrightarrow{O A}=2 \hat{\imath}-\hat{\jmath}+\hat{k}, \overrightarrow{O B}=\hat{\imath}-3 \hat{\jmath}-5 \hat{k} \) and \( \overrightarrow{O C}=3 \hat{\imath}-4 \hat{\jmath}-4 \hat{k} \)
Now vectors \( \overrightarrow{A B}, \overrightarrow{B C} \) and \( \overrightarrow{A C} \) represent sides of \( \triangle A B C \)
\(\overrightarrow{\mathrm{AB}}=(1-2) \hat{\imath}+(-3-(-1)) \hat{\jmath}+(-5-1) \hat{\mathrm{k}}=-\hat{\imath}-2 \hat{\jmath}-6 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{BC}}=(3-1) \hat{\imath}+(-4-(-5)) \hat{\jmath}+(-4-(-5)) \hat{\mathrm{k}}=2 \hat{\imath}-\hat{\jmath}+\hat{k}\)
\(\overrightarrow{\mathrm{AC}}=(3-2) \hat{\imath}+(-4-(-1)) \hat{\jmath}+(-4-1) \hat{\mathrm{k}}=\hat{\imath}-3 \hat{\jmath}-5 \hat{k}\)
\(|\overrightarrow{\mathrm{AB}}|=\sqrt{(-1)^{2}+(-2)^{2}+(-6)^{2}}=\sqrt{1+4+36}=\sqrt{41}\)
\(|\overrightarrow{\mathrm{AB}}|^{2}=41\)
\(|\overrightarrow{\mathrm{BC}}|=\sqrt{2^{2}+(-1)^{2}+1^{2}}=\sqrt{4+1+1}=\sqrt{6}\)
\(|\overrightarrow{\mathrm{BC}}|^{2}=6\)
\(|\overrightarrow{\mathrm{AC}}|=\sqrt{1^{2}+(-3)^{2}+(-5)^{2}}=\sqrt{1+9+25}=\sqrt{35}\)
\(|\overrightarrow{\mathrm{AC}}|^{2}=35\)
Now \( |\overrightarrow{\mathrm{BC}}|^{2}+|\overrightarrow{\mathrm{AC}}|^{2}=35+6 \)
\(\Rightarrow|\overrightarrow{\mathrm{BC}}|^{2}+|\overrightarrow{\mathrm{AC}}|^{2}=41\)
And \( |\overrightarrow{\mathrm{AC}}|^{2}=41 \)
\(\therefore|\overrightarrow{\mathrm{BC}}|^{2}+|\overrightarrow{\mathrm{AC}}|^{2}=|\overrightarrow{\mathrm{AB}}|^{2}\)
From Pythagoras theorem we know
If \( \overrightarrow{\mathrm{BC}}|^{2}+|\overrightarrow{\mathrm{AC}}|^{2}=|\overrightarrow{\mathrm{AB}}|^{2} \) then \( \Delta \mathrm{ABC} \) is a right-angled triangle
\( \therefore \Delta \mathrm{ABC} \) is a right-angled triangle.

Given \( \lambda \overrightarrow{a} \) is a unit vector.
And \( |\overrightarrow{a}|=\mathrm{a} \)
\( |\lambda \overrightarrow{a}|=1 \)
\(\Rightarrow|\lambda||\overrightarrow{a}|=1\)
Since \( |\overrightarrow{a}|=a \)
\(\therefore \mathrm{a}=\frac{1}{|\lambda|}\)
\( \lambda \overrightarrow{a} \) is a unit vector if \( \mathrm{a}=\frac{1}{|\lambda|} \)