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

Exercise 10.3

Q1. $\begin{array}{c}\text{Find the angle between two vectors}\overrightarrow{a}\text{and}\overrightarrow{b}\text{with magnitudes}\sqrt{3}\text{and}2,\text{respectively}\\ \text{having}\overrightarrow{a}\cdot \overrightarrow{b}=\sqrt{6}\end{array}$

Answer. $\begin{array}{c}|\overrightarrow{a}|=\sqrt{3},|\overrightarrow{b}|=2\text{and,}\overrightarrow{a}\cdot \overrightarrow{b}=\sqrt{6}\\ \text{Now, we know that}\overline{a}\cdot \overrightarrow{b}=|\overrightarrow{a}||\overline{b}|\cos \theta \\ \therefore \sqrt{6}=\sqrt{3}\times 2\times \cos \theta \\ \Rightarrow \cos \theta =\frac{\sqrt{6}}{\sqrt{3}\times 2}\\ \Rightarrow \cos \theta =\frac{1}{\sqrt{2}}\\ \Rightarrow \theta =\frac{\pi }{4}\end{array}$ Hence , the angle between the given vectors $\overrightarrow{a}\text{and}\overrightarrow{b}\text{is}\frac{\pi }{4}$

Q2. Find the angle between the given vector $\hat{i}-2\hat{j}+3\hat{k}\text{and}3\hat{i}-2\hat{j}+\hat{k}$

Answer. $\begin{array}{cc}\text{The qiven vectors are}& \hat{a}=\hat{i}-2\hat{j}+3\hat{k}\text{and}\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}\\ |\overrightarrow{a}|=\sqrt{1^2+(-2{)}^2+3^2}& =\sqrt{1+4+9}=\sqrt{14}\\ |\overrightarrow{b}|=\sqrt{3^2+(-2{)}^2+1^2}=& \sqrt{9+4+1}=\sqrt{14}\\ \text{Now,}\overrightarrow{a}\cdot \overrightarrow{b}& =(\hat{i}-2\hat{j}+3\hat{k})(3\hat{i}-2\hat{j}+\hat{k})\\ & =3+4+3\\ & =10\end{array}$

Q3. Find the projection of the vector $\hat{i}-\hat{j}\text{on the vector}\hat{i}+\hat{j}$

Answer. $\begin{array}{c}\text{Let}\overline{a}=\hat{i}-\hat{j}\text{and}\overrightarrow{b}=\hat{i}+\hat{j}\\ \text{Now, projection of vector}\overrightarrow{a}\text{on}\overrightarrow{b}\text{is given by}\\ \frac{1}{|\overrightarrow{b}|}\left(\overrightarrow{a}\overrightarrow{b}\right)=\frac{1}{\sqrt{1+1}}\{1.1+(-1\left)\right(1\left)\right\}=\frac{1}{\sqrt{2}}(1-1)=0\end{array}$ Hence the projection of vector $\overrightarrow{a}\text{on}\overrightarrow{b}\text{is}0.$

Q4. Find the projection of the vector $\hat{i}+3\hat{j}+7\hat{k}\text{on the vector}7\hat{i}-\hat{ȷ}+8\hat{k}$

Answer. $\begin{array}{c}\text{Let}\overline{a}=\hat{i}+3\hat{j}+7\hat{k}\text{and}\hat{b}=7\hat{i}-\hat{j}+8\hat{k}\\ \text{Now, projection of vector}\overrightarrow{a}\text{on}\overrightarrow{b}\text{is given by}\end{array}$ $\begin{array}{c}\text{Now, projection of vector}\overrightarrow{a}on\overrightarrow{b}\text{is given by}\\ \frac{1}{|\overrightarrow{b}|}(\overrightarrow{a}\cdot \overrightarrow{b})=\frac{1}{\sqrt{7^2+(-1{)}^2+8^2}}\left\{1\right(7)+3(-1)+7(8\left)\right\}=\frac{7-3+56}{\sqrt{49+1+64}}=\frac{60}{\sqrt{114}}\end{array}$

Q5. Show that each of the given three vectors is a unit vector: $\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})\cdot \frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}),\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})$. Also, show that they are mutually perpendicular to each other.

Q6. Find $|\overrightarrow{a}|\text{and}|\overrightarrow{b}|,\text{if}(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=8\text{and}|\overrightarrow{a}|=8|\overrightarrow{b}|$

Answer. $\begin{array}{c}(\overrightarrow{a}\cdot \overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=8\\ \Rightarrow \overrightarrow{a}\cdot \overrightarrow{a}-\overrightarrow{a}\overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{a}-\overrightarrow{b}\cdot \overrightarrow{b}=8\\ \Rightarrow |\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2=8\\ \Rightarrow (8|\overrightarrow{b}|{)}^2-|\overrightarrow{b}{|}^2=8\\ \Rightarrow 64|\overrightarrow{b}{|}^2-|\overrightarrow{b}{|}^2=8\\ \Rightarrow 63|\overrightarrow{b}{|}^2=8\end{array}$ $\begin{array}{c}\Rightarrow |\overrightarrow{b}{|}^2=\frac{8}{63}\\ \Rightarrow |\overrightarrow{b}|=\sqrt{\frac{8}{63}}\\ \Rightarrow |\overrightarrow{b}|=\frac{2\sqrt{2}}{3\sqrt{7}}\\ |\overrightarrow{a}|=8|\overrightarrow{b}|=\frac{8\times 2\sqrt{2}}{3\sqrt{7}}=\frac{16\sqrt{2}}{3\sqrt{7}}\end{array}$

Q7. Evaluate the product $(3\overrightarrow{a}-5\overrightarrow{b})\cdot (2\overrightarrow{a}+7\overrightarrow{b})$

Answer. $\begin{array}{c}(3\overrightarrow{a}-5\overrightarrow{b})\cdot (2\overrightarrow{a}+7\overrightarrow{b})\\ =3\overrightarrow{a}\cdot 2\overrightarrow{a}+3\overrightarrow{a}\cdot 7\overrightarrow{b}-5\overrightarrow{b}\cdot 2\overrightarrow{a}-5\overrightarrow{b}\cdot 7\overrightarrow{b}\\ =6\overrightarrow{a}\cdot \overrightarrow{a}+21\overrightarrow{a}\cdot \overrightarrow{b}-10\overrightarrow{a}\cdot \overrightarrow{b}-35\overrightarrow{b}\cdot \overrightarrow{b}\\ =6|\overrightarrow{a}{|}^2+|1\overrightarrow{a}\cdot \overrightarrow{b}-35|{\overrightarrow{b}|}^2\end{array}$

Q8. Find the magnitude of two vectors $\overrightarrow{a}\text{and}\overrightarrow{b}$ , having the same magnitude and such that the angle between them is ${60}^{\circ }$ and their scalar product is $\frac{1}{2}$.

Answer. $\begin{array}{c}\text{Let}\theta \text{be the angle between the vectors}\overrightarrow{a}\text{and}\overrightarrow{b}\text{.}\\ |\overrightarrow{a}|=|\overrightarrow{b}|,\overrightarrow{a}\cdot \overrightarrow{b}=\frac{1}{2},\text{and}\theta ={60}^{\circ }\text{.}\\ \text{We know that}\overline{a}\cdot \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos \theta \end{array}$ $\begin{array}{c}\text{we know that}\overline{a}\cdot \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos \theta \\ \therefore \frac{1}{2}=|\overrightarrow{a}||\overrightarrow{a}|\cos {60}^{\circ }\\ \Rightarrow \frac{1}{2}=|\overrightarrow{a}{|}^2\times \frac{1}{2}\\ \Rightarrow {\frac{1}{|a|}}^2=1\\ \Rightarrow |\overrightarrow{a}|=|\overrightarrow{b}|=1\end{array}$

Q9. $Find|\overrightarrow{x}|,\text{if for a unit vector}\overrightarrow{a},(\overrightarrow{x}-\overrightarrow{a})\cdot (\overrightarrow{x}+\overrightarrow{a})=12$

Answer.