NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.3) Exercise 10.3

Q.1

$\begin{array}{l}\text{Find the angle between two vectors}\overrightarrow{\mathrm{a}}\text{and}\overrightarrow{\mathrm{b}}\text{with}\\ \text{magnitude}\sqrt{\text{3}}\text{and\hspace{0.17em}\hspace{0.17em}2, respectively having}\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\sqrt{\text{6}}.\end{array}$

Ans

$\begin{array}{l}\text{It is given that,}\\ \left|\overrightarrow{\mathrm{a}}\right|=\sqrt{3},\left|\overrightarrow{\mathrm{b}}\right|=2\text{and}\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\sqrt{6}\\ \text{Now, we know that}\\ \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|\mathrm{cos\theta }\\ \Rightarrow \sqrt{6}=\sqrt{3}.2\mathrm{cos\theta }\\ \Rightarrow \mathrm{cos\theta }=\frac{\sqrt{6}}{\sqrt{3}.2}\\ =\frac{1}{\sqrt{2}}\end{array}$

$\begin{array}{l}\Rightarrow \mathrm{cos\theta }=\cos \frac{\mathrm{\pi }}{4}\\ \Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{4}\\ \text{Hence, the angle between the given vectors\hspace{0.17em}}\overrightarrow{\mathrm{a}}\text{and}\overrightarrow{\mathrm{b}}\text{is}\frac{\mathrm{\pi }}{4}.\end{array}$

Q.2

$\text{Find the angle between the vectors}\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}\text{and}3\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}+\widehat{\mathrm{k}}.$

Ans

$\begin{array}{l}\text{The given vectors are\hspace{0.17em}}\overrightarrow{\mathrm{a}}=\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}\text{and}\overrightarrow{\mathrm{b}}=3\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}+\widehat{\mathrm{k}}.\\ \left|\overrightarrow{\mathrm{a}}\right|=|\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}|\\ =\sqrt{1^2+{(-2)}^2+3^2}\\ =\sqrt{1+4+9}\\ =\sqrt{14}\\ \left|\overrightarrow{\mathrm{b}}\right|=|3\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}+\widehat{\mathrm{k}}|\\ =\sqrt{3^2+{(-2)}^2+1^2}\\ =\sqrt{9+4+1}\\ =\sqrt{14}\\ \text{Now,}\\ \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=(\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}).(3\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}+\widehat{\mathrm{k}})\\ =3+4+3=10\end{array}$

$\begin{array}{l}\text{Also, we know that}\\ \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\left|\overrightarrow{\mathrm{a}}\right|.\left|\overrightarrow{\mathrm{b}}\right|\mathrm{cos\theta }\\ 10=\sqrt{14}.\sqrt{14}\mathrm{cos\theta }\\ \mathrm{cos\theta }=\frac{10}{14}\\ =\frac{5}{7}\\ \mathrm{\theta }={\cos }^{-1}\left(\frac{5}{7}\right)\end{array}$

Q.3

$\text{Find the projection of the vector}\widehat{\mathrm{i}}-\widehat{\mathrm{j}}\text{on the vector}\widehat{\mathrm{i}}+\widehat{\mathrm{j}}.$

Ans

$\begin{array}{l}\text{Let}\overrightarrow{\mathrm{a}}=\widehat{\mathrm{i}}-\widehat{\mathrm{j}}\text{and}\overrightarrow{\mathrm{b}}=\widehat{\mathrm{i}}+\widehat{\mathrm{j}}\\ \text{Now, projection of vector}\overrightarrow{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}on}\overrightarrow{\mathrm{b}}\text{\hspace{0.17em}\hspace{0.17em}is given by,}\\ \frac{1}{\left|\overrightarrow{\mathrm{b}}\right|}(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}})=\frac{1}{|\widehat{\mathrm{i}}+\widehat{\mathrm{j}}|}(\widehat{\mathrm{i}}+\widehat{\mathrm{j}}).(\widehat{\mathrm{i}}-\widehat{\mathrm{j}})\\ =\frac{1}{\sqrt{1^2+1^2}}(1-1)=0\\ \text{Hence, the projection of vector}\overrightarrow{\mathrm{a}}\text{\hspace{0.17em}on\hspace{0.17em}\hspace{0.17em}}\overrightarrow{\mathrm{b}}\text{is 0.}\end{array}$

Q.4

$\begin{array}{l}\text{Find the projection of the vector}\widehat{\mathrm{i}}+\mathbf{3}\widehat{\mathrm{j}}+7\widehat{k}\text{on the vector}\\ 7\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+8\widehat{\mathrm{k}}.\end{array}$

Ans

$\begin{array}{l}\text{Let}\overrightarrow{\mathrm{a}}=\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+7\widehat{\mathrm{k}}\text{and}\overrightarrow{\mathrm{b}}=7\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+8\widehat{\mathrm{k}}\\ \text{Now, projection of vector}\overrightarrow{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}on}\overrightarrow{\mathrm{b}}\text{\hspace{0.17em}\hspace{0.17em}is given by,}\\ \frac{1}{\left|\overrightarrow{\mathrm{b}}\right|}(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}})=\frac{1}{|7\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+8\widehat{\mathrm{k}}|}(\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+7\widehat{\mathrm{k}}).(7\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+8\widehat{\mathrm{k}})\\ =\frac{1}{\sqrt{7^2+{(-1)}^2+8^2}}(7-3+56)\\ =\frac{1}{\sqrt{49+1+64}}\left(60\right)=\frac{1}{\sqrt{114}}\left(60\right)\\ \text{Hence, the projection of vector}\overrightarrow{\mathrm{a}}\text{\hspace{0.17em}on\hspace{0.17em}\hspace{0.17em}}\overrightarrow{\mathrm{b}}\text{is}\frac{\text{60}}{\sqrt{114}}\text{.}\end{array}$

Q.5

$\begin{array}{l}\text{Show that each of the given three vectors is a unit vector:}\\ \frac{1}{7}(2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+6\widehat{\mathrm{k}}),\frac{1}{7}(3\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}),\frac{1}{7}(6\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-3\widehat{\mathrm{k}})\\ \text{Also, show that they are mutually perpendicular to each}\\ \text{other.}\end{array}$

Ans

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\overrightarrow{\mathrm{a}}=\frac{1}{7}(2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+6\widehat{\mathrm{k}})\\ \overrightarrow{\mathrm{b}}=\frac{1}{7}(3\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}+2\widehat{\mathrm{k}})\\ \overrightarrow{\mathrm{c}}=\frac{1}{7}(6\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-3\widehat{\mathrm{k}})\\ \Rightarrow \left|\overrightarrow{\mathrm{a}}\right|=\left|\frac{1}{7}(2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+6\widehat{\mathrm{k}})\right|\end{array}$

$\begin{array}{l}=\frac{1}{7}\sqrt{2^2+3^2+6^2}\\ =\frac{1}{7}\sqrt{4+9+36}\\ =\frac{7}{7}=1\\ \left|\overrightarrow{\mathrm{b}}\right|=\left|\frac{1}{7}(3\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}+2\widehat{\mathrm{k}})\right|\\ =\frac{1}{7}\sqrt{3^2+{(-6)}^2+2^2}\\ =\frac{1}{7}\sqrt{9+36+4}\\ =\frac{7}{7}=1\\ \text{and}\\ \left|\overrightarrow{\mathrm{c}}\right|=\left|\frac{1}{7}(6\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-3\widehat{\mathrm{k}})\right|\\ =\frac{1}{7}\sqrt{6^2+2^2+{(-3)}^2}\\ =\frac{1}{7}\sqrt{36+4+9}\\ =\frac{7}{7}=1\\ \text{Thus, each of the given three vectors is a unit vector.}\\ \text{Now,}\\ \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\frac{1}{7}(2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+6\widehat{\mathrm{k}}).\frac{1}{7}(3\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}+2\widehat{\mathrm{k}})\\ =\frac{1}{49}(6-18+12)=0\end{array}$

$\begin{array}{l}\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}=\frac{1}{7}(3\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}).\frac{1}{7}(6\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-3\widehat{\mathrm{k}})\\ =\frac{1}{49}(18-12-6)=0\\ \overrightarrow{\mathrm{c}}.\overrightarrow{\mathrm{a}}=\frac{1}{7}(6\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-3\widehat{\mathrm{k}}).\frac{1}{7}(2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+6\widehat{\mathrm{k}})\\ =\frac{1}{49}(12+6-18)=0\\ \text{Hence, the given three vectors are mutually perpendicular}\\ \text{to each other.}\end{array}$

Q.6

$\text{Find}\left|\overrightarrow{\text{a}}\right|\text{and}\left|\overrightarrow{\text{b}}\right|\text{, if}\left(\overrightarrow{\text{a}}+\overrightarrow{\text{b}}\right)\text{.}\left(\overrightarrow{\text{a}}–\overrightarrow{\text{b}}\right)=\text{8 and}\left|\overrightarrow{\text{a}}\right|=\text{8}\left|\overrightarrow{\text{b}}\right|\text{.}$

Ans

$\begin{array}{l}\text{We have,}\\ (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}).(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})=8\text{and}\left|\overrightarrow{\mathrm{a}}\right|=8\left|\overrightarrow{\mathrm{b}}\right|\\ \text{So,}\\ (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}).(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})=8\\ \Rightarrow \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{b}}=8\\ \Rightarrow {\left|\overrightarrow{\mathrm{a}}\right|}^2-{\left|\overrightarrow{\mathrm{b}}\right|}^2=8[\because \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{a}}]\\ \Rightarrow {\left(8\left|\overrightarrow{\mathrm{b}}\right|\right)}^2-{\left|\overrightarrow{\mathrm{b}}\right|}^2=8\\ \Rightarrow 64{\left|\overrightarrow{\mathrm{b}}\right|}^2-{\left|\overrightarrow{\mathrm{b}}\right|}^2=8\\ \end{array}$

Q.7

$\text{Evaluate the product}(3\overrightarrow{\mathrm{a}}-5\overrightarrow{\mathrm{b}}).(2\overrightarrow{\mathrm{a}}+7\overrightarrow{\mathrm{b}}).$

Ans

$\begin{array}{l}(3\overrightarrow{\mathrm{a}}-5\overrightarrow{\mathrm{b}}).(2\overrightarrow{\mathrm{a}}+7\overrightarrow{\mathrm{b}})=6\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{a}}+21\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}-10\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{a}}-35\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{b}}\\ =6{\left|\overrightarrow{\mathrm{a}}\right|}^2+21\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}-10\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{a}}-35{\left|\overrightarrow{\mathrm{b}}\right|}^2\\ =6{\left|\overrightarrow{\mathrm{a}}\right|}^2+11\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}-35{\left|\overrightarrow{\mathrm{b}}\right|}^2[\because \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{a}}]\end{array}$

Q.8

$\begin{array}{l}\text{Find the magnitude of the vectors}\overrightarrow{\text{a}}\text{and}\overrightarrow{\text{b}}\text{, having the same}\\ \text{magnitude and such that the angle between them is 60°}\\ \text{and their scalar product is}\frac{1}{2}\text{.}\end{array}$

Ans

$\begin{array}{l}\text{Let}\mathrm{\theta }\text{be the angle between the vectors}\overrightarrow{\text{a}}\text{and}\overrightarrow{\text{b}}\text{.\hspace{0.17em}Then,}\\ \overrightarrow{\mathrm{a}}\text{\hspace{0.17em}.\hspace{0.17em}}\overrightarrow{\mathrm{b}}=\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|\mathrm{cos\theta }\\ \frac{1}{2}=\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{a}}\right|\cos 60°[\because \left|\overrightarrow{\mathrm{a}}\right|=\left|\overrightarrow{\mathrm{b}}\right|]\end{array}$

$\begin{array}{l}\frac{1}{2}={\left|\overrightarrow{\mathrm{a}}\right|}^2\times \frac{1}{2}\\ 1={\left|\overrightarrow{\mathrm{a}}\right|}^2\\ \Rightarrow \left|\overrightarrow{\mathrm{a}}\right|=1\\ \text{So,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left|\overrightarrow{\mathrm{b}}\right|=\left|\overrightarrow{\mathrm{a}}\right|=1\end{array}$

Q.9

$\text{Find}\left|\overrightarrow{\mathrm{x}}\right|,\text{if for a unit vector}\overrightarrow{\mathrm{a}},(\overrightarrow{\mathrm{x}}-\overrightarrow{\mathrm{a}}).(\overrightarrow{\mathrm{x}}+\overrightarrow{\mathrm{a}})=12.$

Ans

$\begin{array}{l}\text{Since,}\\ (\overrightarrow{\mathrm{x}}-\overrightarrow{\mathrm{a}}).(\overrightarrow{\mathrm{x}}+\overrightarrow{\mathrm{a}})=12\\ \overrightarrow{\mathrm{x}}.\overrightarrow{\mathrm{x}}+\overrightarrow{\mathrm{x}}.\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{x}}-\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{a}}=12\\ {\left|\overrightarrow{\mathrm{x}}\right|}^2-{\left|\overrightarrow{\mathrm{a}}\right|}^2=12[\because \overrightarrow{\mathrm{x}}.\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{x}}]\\ \Rightarrow {\left|\overrightarrow{\mathrm{x}}\right|}^2-1^2=12[\overrightarrow{\mathrm{a}}\text{is a unit vector so,}\left|\overrightarrow{\mathrm{a}}\right|=1]\\ \Rightarrow {\left|\overrightarrow{\mathrm{x}}\right|}^2=12+1\\ \Rightarrow \left|\overrightarrow{\mathrm{x}}\right|=\sqrt{13}\end{array}$

Q.10

$\begin{array}{l}\text{If}\overrightarrow{\mathrm{a}}=2\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}},\overrightarrow{\mathrm{b}}=-\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\text{and}\overrightarrow{\mathrm{c}}=3\widehat{\mathrm{i}}+\widehat{\mathrm{j}}\text{are such that}\\ \overrightarrow{\mathrm{a}}+\mathrm{\lambda }\overrightarrow{\mathrm{b}}\text{is perpendicular to}\overrightarrow{\mathrm{c}},\text{then find the value of}\mathrm{\lambda }.\end{array}$

Ans

$\begin{array}{l}\text{The given vectors are:}\\ \overrightarrow{\mathrm{a}}=2\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}},\overrightarrow{\mathrm{b}}=-\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\text{and}\overrightarrow{\mathrm{c}}=3\widehat{\mathrm{i}}+\widehat{\mathrm{j}}\end{array}$

$\begin{array}{l}\therefore \overrightarrow{\mathrm{a}}+\mathrm{\lambda }\overrightarrow{\mathrm{b}}=(2\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}})+\mathrm{\lambda }(-\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+\widehat{\mathrm{k}})\\ =(2-\mathrm{\lambda })\widehat{\mathrm{i}}+(2+2\mathrm{\lambda })\widehat{\mathrm{j}}+(3+\mathrm{\lambda })\widehat{\mathrm{k}}\\ \text{Since,\hspace{0.17em}}(\overrightarrow{\mathrm{a}}+\mathrm{\lambda }\overrightarrow{\mathrm{b}})\text{is perpendicular to}\overrightarrow{\mathrm{c}}\text{, so}\\ (\overrightarrow{\mathrm{a}}+\mathrm{\lambda }\overrightarrow{\mathrm{b}}).\overrightarrow{\mathrm{c}}=0\\ \Rightarrow \{(2-\mathrm{\lambda })\widehat{\mathrm{i}}+(2+2\mathrm{\lambda })\widehat{\mathrm{j}}+(3+\mathrm{\lambda })\widehat{\mathrm{k}}\}.(3\widehat{\mathrm{i}}+\widehat{\mathrm{j}})=0\\ \Rightarrow 3(2-\mathrm{\lambda })+(2+2\mathrm{\lambda })=0\\ \Rightarrow 6-3\mathrm{\lambda }+2+2\mathrm{\lambda }=0\\ \Rightarrow 8-\mathrm{\lambda }=0\\ \Rightarrow \mathrm{\lambda }=8\\ \text{Therefore, the required value of}\mathrm{\lambda }\text{is 8.}\end{array}$

Q.11

\begin{array}{l}\text{Show that}\left|\overrightarrow{a}\right|\overrightarrow{b}+\left|\overrightarrow{b}\right|\overrightarrow{a}\text{is perpendicular to}\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{b}\right|\overrightarrow{a},\text{for any}\\ \text{two nonzero vectors}\overrightarrow{a}\text{and}\overrightarrow{b}.\end{array}

Ans

\begin{array}{l}\left(\left|\overrightarrow{a}\right|\overrightarrow{b}+\left|\overrightarrow{b}\right|\overrightarrow{a}\right).\left(\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{b}\right|\overrightarrow{a}\right)={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2-\left|\overrightarrow{a}\right|\overrightarrow{b}\left|\overrightarrow{b}\right|\overrightarrow{a}+\left|\overrightarrow{b}\right|\overrightarrow{a}\left|\overrightarrow{a}\right|\overrightarrow{b}-{\left|\overrightarrow{b}\right|}^2{\left|\overrightarrow{a}\right|}^2\\ =-\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\overrightarrow{b}\overrightarrow{a}+\left|\overrightarrow{b}\right|\left|\overrightarrow{a}\right|\overrightarrow{a}\overrightarrow{b}\\ =0\left[\because \overrightarrow{b}\overrightarrow{a}=\overrightarrow{a}\overrightarrow{b}\right]\\ \text{So,}\left(\left|\overrightarrow{a}\right|\overrightarrow{b}+\left|\overrightarrow{b}\right|\overrightarrow{a}\right)\text{and}\left(\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{b}\right|\overrightarrow{a}\right)\text{are perpendicular to each other}\text{.}\end{array}

Q.12

$\begin{array}{l}\text{If}\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{a}}=0\text{and}\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\text{0, then what can be concluded about}\\ \text{the vector}\overrightarrow{\mathrm{b}}\text{?}\end{array}$

Ans

$\begin{array}{l}\text{Since, it is given that}\\ \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{a}}=0\\ \Rightarrow {\left|\overrightarrow{\mathrm{a}}\right|}^2=0\\ \Rightarrow \left|\overrightarrow{\mathrm{a}}\right|=0\\ \therefore \overrightarrow{\mathrm{a}}\text{is a zero vector.}\\ \text{Since, vector}\overrightarrow{\mathrm{b}}\text{satisfying}\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=0\text{can be any vector.}\end{array}$

Q.13

$\begin{array}{l}\text{If}\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}\text{are unit vectors such that}\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0},\text{find the}\\ \text{value of}\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}.\overrightarrow{\mathrm{a}}.\end{array}$

Ans

$\begin{array}{l}\text{Given:}\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}\text{are unit vectors such that}\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}\\ \text{Then,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{[\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}]}^2=0\\ {\left|\overrightarrow{\mathrm{a}}\right|}^2+{\left|\overrightarrow{\mathrm{b}}\right|}^2+{\left|\overrightarrow{\mathrm{c}}\right|}^2+2[\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}.\overrightarrow{\mathrm{a}}]=0\\ 1+1+1+2[\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}.\overrightarrow{\mathrm{a}}]=0\\ \Rightarrow \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}.\overrightarrow{\mathrm{a}}=-\frac{3}{2}\end{array}$

Q.14

$\begin{array}{l}\text{If either vector}\overrightarrow{\mathrm{a}}=0\text{or}\overrightarrow{\mathrm{b}}=0,\text{then}\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=0.\text{But the converse}\\ \text{need not be true. Justify your answer with an example.}\end{array}$

Ans

$\begin{array}{l}\text{Let}\overrightarrow{\mathrm{a}}=3\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}\text{and}\overrightarrow{\mathrm{b}}=2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}-6\widehat{\mathrm{k}}\\ \text{Then,}\\ \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=(3\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+3\widehat{\mathrm{k}})(2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}-6\widehat{\mathrm{k}})\end{array}$

$\begin{array}{l}=6+12-18\\ =0\\ \text{And}\\ \left|\overrightarrow{\mathrm{a}}\right|=|3\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}|\\ =\sqrt{3^2+4^2+3^2}\\ =\sqrt{9+16+9}\\ =\sqrt{34}\ne 0\\ \left|\overrightarrow{\mathrm{b}}\right|=|2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}-6\widehat{\mathrm{k}}|\\ =\sqrt{2^2+3^2+{(-6)}^2}\\ =\sqrt{4+9+36}\\ =\sqrt{49}\\ =7\ne 0\\ \text{Hence, the converse of the given statement need not be true.}\end{array}$

Q.15

$\begin{array}{l}\text{If the vertices A, B, C of a triangle ABC are}\left(\text{1,2,3}\right)\text{,}\left(\text{-1,0,0}\right)\text{,}\\ \left(\text{0,1,2}\right)\text{respectively, then find}\angle \text{ABC.}\\ [\angle \text{ABC is the angle between the vectors}\overrightarrow{\text{BA}}\text{and}\overrightarrow{\text{BC}}].\end{array}$

Ans

$\begin{array}{l}\text{The vertices of ΔABC are given as A (1, 2, 3), B (–1, 0, 0),}\\ \text{and C (0, 1, 2).}\end{array}$

$\begin{array}{l}\text{Also, it is given that}\angle \text{ABC is the angle between the vectors\hspace{0.17em}}\overrightarrow{\text{BA}}\\ \text{and\hspace{0.17em}\hspace{0.17em}}\overrightarrow{\text{BC}}\text{.}\\ \therefore \overrightarrow{\text{BA}}\text{=}\left(\text{1 + 1}\right)\widehat{\text{i}}\text{+}\left(\text{2 – 0}\right)\widehat{\text{j}}\text{+}\left(\text{3 – 0}\right)\widehat{\text{k}}\\ \text{\hspace{0.17em}= 2}\widehat{\text{i}}\text{+ 2}\widehat{\text{j}}\text{+ 3\hspace{0.17em}}\widehat{\text{k}}\\ \overrightarrow{\text{BC}}\text{=}\left(\text{0 + 1}\right)\widehat{\text{i}}\text{+}\left(\text{1 – 0}\right)\widehat{\text{j}}\text{+}\left(\text{2 – 0}\right)\widehat{\text{k}}\\ \text{\hspace{0.17em}=}\widehat{\text{i}}\text{+}\widehat{\text{j}}\text{+ 2}\widehat{\text{k}}\\ \therefore \overrightarrow{\text{BA}}\cdot \overrightarrow{\text{BC}}\\ \text{\hspace{0.17em}=}\left(\text{2}\widehat{\text{i}}\text{+ 2}\widehat{\text{j}}\text{+ 3\hspace{0.17em}}\widehat{\text{k}}\right)\cdot \left(\widehat{\text{i}}\text{+}\widehat{\text{j}}\text{+ 2}\widehat{\text{k}}\right)\\ \text{\hspace{0.17em}= 2 + 2 + 6}\\ \text{\hspace{0.17em}= 10}\\ \left|\overrightarrow{\text{BA}}\right|\text{=}\left|\text{2}\widehat{\text{i}}\text{+ 2}\widehat{\text{j}}\text{+ 3\hspace{0.17em}}\widehat{\text{k}}\right|\\ \text{\hspace{0.17em}=}\sqrt{{\text{2}}^{\text{2}}{\text{+ 2}}^{\text{2}}{\text{+ 3}}^{\text{2}}}\\ \text{\hspace{0.17em}=}\sqrt{\text{4 + 4 + 9}}\\ \text{\hspace{0.17em}=}\sqrt{\text{17}}\\ \left|\overrightarrow{\text{BC}}\right|\text{=}\left|\widehat{\text{i}}\text{+}\widehat{\text{j}}\text{+ 2}\widehat{\text{k}}\right|\\ \text{\hspace{0.17em}=}\sqrt{{\text{1}}^{\text{2}}{\text{+ 1}}^{\text{2}}{\text{+ 2}}^{\text{2}}}\\ \text{\hspace{0.17em}=}\sqrt{\text{1 + 1 + 4}}\\ \text{\hspace{0.17em}=}\sqrt{\text{6}}\\ \text{Now, it is known that:\hspace{0.17em}}\\ \overrightarrow{\text{BA}}\cdot \overrightarrow{\text{BC}}\text{=}\left|\overrightarrow{\text{BA}}\right|\cdot \left|\overrightarrow{\text{BC}}\right|\text{cos}(\angle \text{ABC})\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}10 =}\sqrt{\text{17}}\sqrt{\text{6}}\text{cos}(\angle \text{ABC)}\end{array}$

$\begin{array}{l}\text{cos}(\angle \text{ABC})\text{=}\frac{\text{10}}{\sqrt{\text{102}}}\\ \angle {\text{ABC = cos}}^{\text{–1}}\left(\frac{\text{10}}{\sqrt{\text{102}}}\right)\end{array}$

$\begin{array}{l}\left|\overrightarrow{\text{AC}}\right|\text{=}\left|\text{2}\widehat{\text{i}}\text{+ 8}\widehat{\text{j}}\text{– 8}\widehat{\text{k}}\right|\\ \text{=}\sqrt{{\text{2}}^{\text{2}}{\text{+ 8}}^{\text{2}}\text{+}{\left(\text{–8}\right)}^{\text{2}}}\\ \text{=}\sqrt{\text{4 + 64 + 64}}\\ \text{=}\sqrt{\text{132}}\text{= 2}\sqrt{\text{33}}\\ \therefore \left|\overrightarrow{\text{AC}}\right|\text{=}\left|\overrightarrow{\text{AB}}\right|\text{+}\left|\overrightarrow{\text{BC}}\right|\\ \text{Hence, the given points A, B, and C are collinear.}\end{array}$

Q.16

$\begin{array}{l}\text{Show that the vectors}2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}},\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}-5\widehat{\mathrm{k}}\text{and}3\widehat{\mathrm{i}}-4\widehat{\mathrm{j}}-4\widehat{\mathrm{k}}\\ \text{form the vertices of a right angled triangle.}\end{array}$

Ans

$\begin{array}{l}\text{Let vectors}2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}},\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}-5\widehat{\mathrm{k}}\mathrm{a}\mathrm{nd}3\widehat{\mathrm{i}}-4\widehat{\mathrm{j}}-4\widehat{\mathrm{k}}\text{be position vectors}\\ \text{of points A, B and C respectively.}\\ \text{i.e.,}\overrightarrow{\mathrm{OA}}=2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}},\overrightarrow{\mathrm{OB}}=\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}-5\widehat{\mathrm{k}},\overrightarrow{\mathrm{OC}}=3\widehat{\mathrm{i}}-4\widehat{\mathrm{j}}-4\widehat{\mathrm{k}}\\ \therefore \overrightarrow{\mathrm{AB}}=(1-2)\widehat{\mathrm{i}}+(-3+1)\widehat{\mathrm{j}}+(-5-1)\widehat{\mathrm{k}}=-\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}-6\widehat{\mathrm{k}}\\ \overrightarrow{\mathrm{BC}}=(3-1)\widehat{\mathrm{i}}+(-4+3)\widehat{\mathrm{j}}+(-4+5)\widehat{\mathrm{k}}=2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\\ \overrightarrow{\mathrm{CA}}=(2-3)\widehat{\mathrm{i}}+(-1+4)\widehat{\mathrm{j}}+(1+4)\widehat{\mathrm{k}}=-\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+5\widehat{\mathrm{k}}\\ \left|\overrightarrow{\mathrm{AB}}\right|=|-\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}-6\widehat{\mathrm{k}}|\\ =\sqrt{{(-1)}^2+{(-2)}^2+{(-6)}^2}\\ =\sqrt{1+4+36}\\ =\sqrt{41}\\ \left|\overrightarrow{\mathrm{BC}}\right|=|2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}}|\end{array}$

$\begin{array}{l}=\sqrt{2^2+{(-1)}^2+1^2}\\ =\sqrt{4+1+1}\\ =\sqrt{6}\\ \left|\overrightarrow{\mathrm{CA}}\right|=|-\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+5\widehat{\mathrm{k}}|\\ =\sqrt{{(-1)}^2+3^2+5^2}\\ =\sqrt{1+9+25}\\ =\sqrt{35}\\ \therefore {\left|\overrightarrow{\mathrm{BC}}\right|}^2+{\left|\overrightarrow{\mathrm{CA}}\right|}^2={\left(\sqrt{6}\right)}^2+{\left(\sqrt{35}\right)}^2\end{array}$

$\begin{array}{l}=6+35\\ =41\\ ={\left|\overrightarrow{\mathrm{AB}}\right|}^2\\ \text{Hence,}\mathrm{\Delta }\text{ABC is a right-angled triangle.}\end{array}$

Q.17

$\begin{array}{l}\text{Show that the vectors}2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}},\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}-5\widehat{\mathrm{k}}\text{and}3\widehat{\mathrm{i}}-4\widehat{\mathrm{j}}-4\widehat{\mathrm{k}}\\ \text{form the vertices of a right angled triangle.}\end{array}$

Ans

$\begin{array}{l}\text{Let vectors}2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}},\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}-5\widehat{\mathrm{k}}\mathrm{a}\mathrm{nd}3\widehat{\mathrm{i}}-4\widehat{\mathrm{j}}-4\widehat{\mathrm{k}}\text{be position vectors}\\ \text{of points A, B and C respectively.}\\ \text{i.e.,}\overrightarrow{\mathrm{OA}}=2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}},\overrightarrow{\mathrm{OB}}=\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}-5\widehat{\mathrm{k}},\overrightarrow{\mathrm{OC}}=3\widehat{\mathrm{i}}-4\widehat{\mathrm{j}}-4\widehat{\mathrm{k}}\\ \therefore \overrightarrow{\mathrm{AB}}=(1-2)\widehat{\mathrm{i}}+(-3+1)\widehat{\mathrm{j}}+(-5-1)\widehat{\mathrm{k}}=-\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}-6\widehat{\mathrm{k}}\\ \overrightarrow{\mathrm{BC}}=(3-1)\widehat{\mathrm{i}}+(-4+3)\widehat{\mathrm{j}}+(-4+5)\widehat{\mathrm{k}}=2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\\ \overrightarrow{\mathrm{CA}}=(2-3)\widehat{\mathrm{i}}+(-1+4)\widehat{\mathrm{j}}+(1+4)\widehat{\mathrm{k}}=-\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+5\widehat{\mathrm{k}}\\ \left|\overrightarrow{\mathrm{AB}}\right|=|-\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}-6\widehat{\mathrm{k}}|\\ =\sqrt{{(-1)}^2+{(-2)}^2+{(-6)}^2}\\ =\sqrt{1+4+36}\\ =\sqrt{41}\\ \left|\overrightarrow{\mathrm{BC}}\right|=|2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}}|\end{array}$

$\begin{array}{l}=\sqrt{2^2+{(-1)}^2+1^2}\\ =\sqrt{4+1+1}\\ =\sqrt{6}\\ \left|\overrightarrow{\mathrm{CA}}\right|=|-\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+5\widehat{\mathrm{k}}|\\ =\sqrt{{(-1)}^2+3^2+5^2}\\ =\sqrt{1+9+25}\\ =\sqrt{35}\\ \therefore {\left|\overrightarrow{\mathrm{BC}}\right|}^2+{\left|\overrightarrow{\mathrm{CA}}\right|}^2={\left(\sqrt{6}\right)}^2+{\left(\sqrt{35}\right)}^2\end{array}$

$\begin{array}{l}=6+35\\ =41\\ ={\left|\overrightarrow{\mathrm{AB}}\right|}^2\\ \text{Hence,}\mathrm{\Delta }\text{ABC is a right-angled triangle.}\end{array}$

Q.18

$\begin{array}{l}\text{If}\overrightarrow{\text{a}}\text{is a non zero vector of magnitude ‘a’ and λ a non zero}\\ \text{scalar, then λ\hspace{0.17em}}\overrightarrow{\text{a}}\text{is unit vector if}\\ \left(\text{A}\right)\text{\hspace{0.17em}\hspace{0.17em}λ = 1}\\ \left(\text{B}\right)\text{\hspace{0.17em}\hspace{0.17em}λ = –1}\\ \left(\text{C}\right)\text{\hspace{0.17em}a =}\left|\text{λ}\right|\\ \left(\text{D}\right)\text{a =}\frac{\text{1}}{\left|\text{λ}\right|}\end{array}$

Ans

$\begin{array}{l}\text{Vector λ}\overrightarrow{\text{a}}\text{\hspace{0.17em}\hspace{0.17em}is a unit vector if}\left|\text{λ}\overrightarrow{\text{a}}\right|\text{= 1.}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}\left|\text{λ}\overrightarrow{\text{a}}\right|\text{= 1}\\ \Rightarrow \left|\text{λ}\right|\left|\overrightarrow{\text{a}}\right|\text{= 1}\end{array}$

$\begin{array}{l}\Rightarrow \left|\overrightarrow{\text{a}}\right|\text{=}\frac{\text{1}}{\left|\text{λ}\right|}[\text{λ}\ne \text{0}]\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a =}\frac{\text{1}}{\left|\text{λ}\right|}\left[\left|\overrightarrow{\text{a}}\right|\text{= a}\right]\\ \text{Hence, λ}\overrightarrow{\text{a}}\text{\hspace{0.17em}vector is a unit vector if\hspace{0.17em}a =}\frac{\text{1}}{\left|\text{λ}\right|}\text{.}\\ \text{The correct answer is D.}\end{array}$