Class 12: Scalar or Dot Product

$\displaystyle \textbf{Question 1: }~\text{Find }\overrightarrow{a}\cdot\overrightarrow{b},\text{ when:}$
$\displaystyle \text{(i) }\overrightarrow{a}=\widehat{i}-2\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=4\widehat{i}-4\widehat{j}+7\widehat{k}$
$\displaystyle \text{(ii) }\overrightarrow{a}=\widehat{j}+2\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}+\widehat{k}$
$\displaystyle \text{(iii) }\overrightarrow{a}=\widehat{j}-\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}+3\widehat{j}-2\widehat{k}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{We have}$
$\displaystyle \overrightarrow{a}=\widehat{i}-2\widehat{j}+\widehat{k}\ \text{and}\ \overrightarrow{b}=4\widehat{i}-\widehat{j}+7\widehat{k}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(\widehat{i}-2\widehat{j}+\widehat{k})\cdot(4\widehat{i}-\widehat{j}+7\widehat{k})$
$\displaystyle =(1)(4)+(-2)(-1)+(1)(7)$
$\displaystyle =4+2+7$
$\displaystyle =13$
$\displaystyle \text{(ii) }$
$\displaystyle \text{We have}$
$\displaystyle \overrightarrow{a}=\widehat{j}+2\widehat{k}=0\widehat{i}+\widehat{j}+2\widehat{k}\ \text{and}\ \overrightarrow{b}=2\widehat{i}+\widehat{k}=2\widehat{i}+0\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(0\widehat{i}+\widehat{j}+2\widehat{k})\cdot(2\widehat{i}+0\widehat{j}+\widehat{k})$
$\displaystyle =(0)(2)+(1)(0)+(2)(1)$
$\displaystyle =0+0+2$
$\displaystyle =2$
$\displaystyle \text{(iii) }$
$\displaystyle \text{We have}$
$\displaystyle \overrightarrow{a}=\widehat{j}-\widehat{k}=0\widehat{i}+\widehat{j}-\widehat{k}\ \text{and}\ \overrightarrow{b}=2\widehat{i}+3\widehat{j}-2\widehat{k}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(0\widehat{i}+\widehat{j}-\widehat{k})\cdot(2\widehat{i}+3\widehat{j}-2\widehat{k})$
$\displaystyle =(0)(2)+(1)(3)+(-1)(-2)$
$\displaystyle =3+2$
$\displaystyle =5$

$\displaystyle \textbf{Question 2: }~\text{For what value of }\lambda\text{ are the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ perpendicular to each other, where:}$
$\displaystyle \text{(i) }\overrightarrow{a}=\lambda\widehat{i}+2\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=4\widehat{i}-9\widehat{j}+2\widehat{k}$
$\displaystyle \text{(ii) }\overrightarrow{a}=\lambda\widehat{i}+2\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=5\widehat{i}-9\widehat{j}+2\widehat{k}$
$\displaystyle \text{(iii) }\overrightarrow{a}=2\widehat{i}+3\widehat{j}+4\widehat{k}\text{ and }\overrightarrow{b}=3\widehat{i}+2\widehat{j}-\lambda\widehat{k}$
$\displaystyle \text{(iv) }\overrightarrow{a}=\lambda\widehat{i}+3\widehat{j}+2\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}-\widehat{j}+3\widehat{k}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{If the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are perpendicular to each other, then}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow (\lambda \widehat{i}+2\widehat{j}+\widehat{k})\cdot(4\widehat{i}-9\widehat{j}+2\widehat{k})=0$
$\displaystyle \Rightarrow 4\lambda+(2)(-9)+(1)(2)=0$
$\displaystyle \Rightarrow 4\lambda-18+2=0$
$\displaystyle \Rightarrow 4\lambda-16=0$
$\displaystyle \Rightarrow 4\lambda=16$
$\displaystyle \Rightarrow \lambda=4$
$\displaystyle \text{(ii) }$
$\displaystyle \text{If the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are perpendicular to each other, then}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow (\lambda \widehat{i}+2\widehat{j}+\widehat{k})\cdot(5\widehat{i}-9\widehat{j}+2\widehat{k})=0$
$\displaystyle \Rightarrow ( \lambda)(5)+(2)(-9)+(1)(2)=0$
$\displaystyle \Rightarrow 5\lambda-18+2=0$
$\displaystyle \Rightarrow 5\lambda-16=0$
$\displaystyle \Rightarrow 5\lambda=16$
$\displaystyle \Rightarrow \lambda=\frac{16}{5}$
$\displaystyle \text{(iii) }$
$\displaystyle \text{If the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are perpendicular to each other, then}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow (2\widehat{i}+3\widehat{j}+4\widehat{k})\cdot(3\widehat{i}+2\widehat{j}-\lambda \widehat{k})=0$
$\displaystyle \Rightarrow (2)(3)+(3)(2)+(4)(-\lambda)=0$
$\displaystyle \Rightarrow 6+6-4\lambda=0$
$\displaystyle \Rightarrow 12-4\lambda=0$
$\displaystyle \Rightarrow 4\lambda=12$
$\displaystyle \Rightarrow \lambda=3$
$\displaystyle \text{(iv) }$
$\displaystyle \text{If the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are perpendicular to each other, then}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow (\lambda \widehat{i}+3\widehat{j}+2\widehat{k})\cdot(\widehat{i}-\widehat{j}+3\widehat{k})=0$
$\displaystyle \Rightarrow (\lambda)(1)+(3)(-1)+(2)(3)=0$
$\displaystyle \Rightarrow \lambda-3+6=0$
$\displaystyle \Rightarrow \lambda+3=0$
$\displaystyle \Rightarrow \lambda=-3$

$\displaystyle \textbf{Question 3: }~\text{If }\lvert\overrightarrow{a}\rvert=4,\ \lvert\overrightarrow{b}\rvert=3\text{ and }\overrightarrow{a}\cdot\overrightarrow{b}=6,\text{ find the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}$
$\displaystyle \text{Given that}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=6$
$\displaystyle \Rightarrow |\overrightarrow{a}|\ |\overrightarrow{b}|\cos\theta=6$
$\displaystyle \Rightarrow (4)(3)\cos\theta=6$
$\displaystyle \Rightarrow 12\cos\theta=6$
$\displaystyle \Rightarrow \cos\theta=\frac{6}{12}=\frac{1}{2}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$

$\displaystyle \textbf{Question 4: }~\text{If }\overrightarrow{a}=\widehat{i}-\widehat{j}\text{ and }\overrightarrow{b}=-\widehat{j}+2\widehat{k},\text{ find }(\overrightarrow{a}-2\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b}).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have}$
$\displaystyle \overrightarrow{a}=\widehat{i}-\widehat{j}\ \text{and}\ \overrightarrow{b}=-\widehat{j}+2\widehat{k}$
$\displaystyle \overrightarrow{a}-2\overrightarrow{b}=(\widehat{i}-\widehat{j})-2(-\widehat{j}+2\widehat{k})=\widehat{i}-\widehat{j}+2\widehat{j}-4\widehat{k}=\widehat{i}+\widehat{j}-4\widehat{k}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}=\widehat{i}-\widehat{j}-\widehat{j}+2\widehat{k}=\widehat{i}-2\widehat{j}+2\widehat{k}$
$\displaystyle (\overrightarrow{a}-2\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})$
$\displaystyle =(\widehat{i}+\widehat{j}-4\widehat{k})\cdot(\widehat{i}-2\widehat{j}+2\widehat{k})$
$\displaystyle =(1)(1)+(1)(-2)+(-4)(2)$
$\displaystyle =1-2-8$
$\displaystyle =-9$

$\displaystyle \textbf{Question 5: }~\text{Find the angle between the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b},\text{ where:}$
$\displaystyle \text{(i) }\overrightarrow{a}=\widehat{i}-\widehat{j},\ \overrightarrow{b}=\widehat{j}+\widehat{k}$
$\displaystyle \text{(ii) }\overrightarrow{a}=3\widehat{i}-2\widehat{j}-6\widehat{k},\ \overrightarrow{b}=4\widehat{i}-\widehat{j}+8\widehat{k}$
$\displaystyle \text{(iii) }\overrightarrow{a}=2\widehat{i}-\widehat{j}+2\widehat{k},\ \overrightarrow{b}=4\widehat{i}+4\widehat{j}-2\widehat{k}$
$\displaystyle \text{(iv) }\overrightarrow{a}=2\widehat{i}-3\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}+\widehat{j}-2\widehat{k}$
$\displaystyle \text{(v) }\overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k},\ \overrightarrow{b}=\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(1)^2+(-1)^2}=\sqrt{2}$
$\displaystyle |\overrightarrow{b}|=\sqrt{(1)^2+(1)^2}=\sqrt{2}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0-1+0=-1$
$\displaystyle \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\ |\overrightarrow{b}|}=\frac{-1}{\sqrt{2}\sqrt{2}}=-\frac{1}{2}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}$
$\displaystyle \text{(ii) }$
$\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(3)^2+(-2)^2+(-6)^2}=\sqrt{49}=7$
$\displaystyle |\overrightarrow{b}|=\sqrt{(4)^2+(-1)^2+(8)^2}=\sqrt{81}=9$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=12+2-48=-34$
$\displaystyle \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\ |\overrightarrow{b}|}=\frac{-34}{(7)(9)}=-\frac{34}{63}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(-\frac{34}{63}\right)$
$\displaystyle \text{(iii) }$
$\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(2)^2+(-1)^2+(2)^2}=\sqrt{9}=3$
$\displaystyle |\overrightarrow{b}|=\sqrt{(4)^2+(4)^2+(-2)^2}=\sqrt{36}=6$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=8-4-4=0$
$\displaystyle \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\ |\overrightarrow{b}|}=\frac{0}{(3)(6)}=0$
$\displaystyle \Rightarrow \theta=\cos^{-1}(0)=\frac{\pi}{2}$
$\displaystyle \text{(iv) }$
$\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(2)^2+(-3)^2+(1)^2}=\sqrt{14}$
$\displaystyle |\overrightarrow{b}|=\sqrt{(1)^2+(1)^2+(-2)^2}=\sqrt{6}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=2-3-2=-3$
$\displaystyle \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\ |\overrightarrow{b}|}=\frac{-3}{\sqrt{14}\sqrt{6}}=-\frac{3}{\sqrt{84}}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(-\frac{3}{\sqrt{84}}\right)$
$\displaystyle \text{(v) }$
$\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(1)^2+(2)^2+(-1)^2}=\sqrt{6}$
$\displaystyle |\overrightarrow{b}|=\sqrt{(1)^2+(-1)^2+(1)^2}=\sqrt{3}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=1-2-1=-2$
$\displaystyle \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\ |\overrightarrow{b}|}=\frac{-2}{\sqrt{6}\sqrt{3}}=\frac{-2}{\sqrt{18}}=-\frac{\sqrt{2}}{3}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(-\frac{\sqrt{2}}{3}\right)$

$\displaystyle \textbf{Question 6: }~\text{Find the angles which the vector }\overrightarrow{a}=\widehat{i}-\widehat{j}+\sqrt{2}\widehat{k}\text{ makes with the} \\ \text{coordinate axes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\theta_1\text{ be the angle between }\overrightarrow{a}\text{ and x-axis}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(1)^2+(-1)^2+(\sqrt{2})^2}=\sqrt{4}=2$
$\displaystyle \overrightarrow{b}=\widehat{i}\ \text{(Because }\widehat{i}\text{ is the unit vector along x-axis)}$
$\displaystyle |\overrightarrow{b}|=\sqrt{(1)^2}=\sqrt{1}=1$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=1+0+0=1$
$\displaystyle \cos\theta_1=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\ |\overrightarrow{b}|}=\frac{1}{(2)(1)}=\frac{1}{2}$
$\displaystyle \Rightarrow \theta_1=\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$
$\displaystyle \text{Let }\theta_2\text{ be the angle between }\overrightarrow{a}\text{ and y-axis}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(1)^2+(-1)^2+(\sqrt{2})^2}=\sqrt{4}=2$
$\displaystyle \overrightarrow{b}=\widehat{j}\ \text{(Because }\widehat{j}\text{ is the unit vector along y-axis)}$
$\displaystyle |\overrightarrow{b}|=\sqrt{(1)^2}=\sqrt{1}=1$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0-1+0=-1$
$\displaystyle \cos\theta_2=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\ |\overrightarrow{b}|}=\frac{-1}{(2)(1)}=-\frac{1}{2}$
$\displaystyle \Rightarrow \theta_2=\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}$
$\displaystyle \text{Let }\theta_3\text{ be the angle between }\overrightarrow{a}\text{ and z-axis}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(1)^2+(-1)^2+(\sqrt{2})^2}=\sqrt{4}=2$
$\displaystyle \overrightarrow{b}=\widehat{k}\ \text{(Because }\widehat{k}\text{ is the unit vector along z-axis)}$
$\displaystyle |\overrightarrow{b}|=\sqrt{(1)^2}=\sqrt{1}=1$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0+0+\sqrt{2}=\sqrt{2}$
$\displaystyle \cos\theta_3=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\ |\overrightarrow{b}|}=\frac{\sqrt{2}}{(2)(1)}=\frac{1}{\sqrt{2}}$
$\displaystyle \Rightarrow \theta_3=\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$

$\displaystyle \textbf{Question 7: }~\text{(i) Dot product of a vector with }\widehat{i}+\widehat{j}-3\widehat{k},\ \widehat{i}+3\widehat{j}-2\widehat{k}\text{ and }2\widehat{i}+\widehat{j}+4\widehat{k}\text{ are }0,5\text{ and }8\text{ respectively. Find the vector. }[\text{CBSE 2003}]$
$\displaystyle \text{(ii) Dot products of a vector with }\widehat{i}-\widehat{j}+\widehat{k},\ 2\widehat{i}+\widehat{j}-3\widehat{k}\text{ and } \\ \widehat{i}+\widehat{j}+\widehat{k}\text{ are }4,0\text{ and }2\text{ respectively. Find the vector. }[\text{CBSE 2013}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Let }a\widehat{i}+b\widehat{j}+c\widehat{k}\text{ be the required vector}$
$\displaystyle \text{Given that}$
$\displaystyle (a\widehat{i}+b\widehat{j}+c\widehat{k})\cdot(\widehat{i}+\widehat{j}-3\widehat{k})=0$
$\displaystyle \Rightarrow a+b-3c=0\ \text{...(1)}$
$\displaystyle (a\widehat{i}+b\widehat{j}+c\widehat{k})\cdot(\widehat{i}+3\widehat{j}-2\widehat{k})=5$
$\displaystyle \Rightarrow a+3b-2c=5\ \text{...(2)}$
$\displaystyle (a\widehat{i}+b\widehat{j}+c\widehat{k})\cdot(2\widehat{i}+\widehat{j}+4\widehat{k})=8$
$\displaystyle \Rightarrow 2a+b+4c=8\ \text{...(3)}$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle a=1,\ b=2,\ c=1$
$\displaystyle \text{So, }a\widehat{i}+b\widehat{j}+c\widehat{k}=\widehat{i}+2\widehat{j}+\widehat{k}$
$\displaystyle \text{(ii) }$
$\displaystyle \text{Let }a\widehat{i}+b\widehat{j}+c\widehat{k}\text{ be the required vector}$
$\displaystyle \text{Given that}$
$\displaystyle (a\widehat{i}+b\widehat{j}+c\widehat{k})\cdot(\widehat{i}-\widehat{j}+\widehat{k})=4$
$\displaystyle \Rightarrow a-b+c=4\ \text{...(1)}$
$\displaystyle (a\widehat{i}+b\widehat{j}+c\widehat{k})\cdot(2\widehat{i}+\widehat{j}-3\widehat{k})=0$
$\displaystyle \Rightarrow 2a+b-3c=0\ \text{...(2)}$
$\displaystyle (a\widehat{i}+b\widehat{j}+c\widehat{k})\cdot(\widehat{i}+\widehat{j}+\widehat{k})=2$
$\displaystyle \Rightarrow a+b+c=2\ \text{...(3)}$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle a=2,\ b=-1,\ c=1$
$\displaystyle \text{So, }a\widehat{i}+b\widehat{j}+c\widehat{k}=2\widehat{i}-\widehat{j}+\widehat{k}$

$\displaystyle \textbf{Question 8: }~\text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are unit vectors inclined at an angle }\theta,\text{ then prove that:}$
$\displaystyle \text{(i) }\cos\frac{\theta}{2}=\frac{1}{2}\lvert\overrightarrow{a}+\overrightarrow{b}\rvert\qquad \text{(ii) }\tan\frac{\theta}{2}=\frac{\lvert\overrightarrow{a}-\overrightarrow{b}\rvert}{\lvert\overrightarrow{a}+\overrightarrow{b}\rvert}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Given that }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are unit vectors}$
$\displaystyle \text{So, }|\overrightarrow{a}|=1,\ |\overrightarrow{b}|=1$
$\displaystyle \text{We have}$
$\displaystyle |\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2\,\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle =1+1+2|\overrightarrow{a}||\overrightarrow{b}|\cos\theta$
$\displaystyle =2+2\cos\theta$
$\displaystyle \Rightarrow \cos\theta=\frac{|\overrightarrow{a}+\overrightarrow{b}|^2-2}{2}\ \text{...(1)}$
$\displaystyle |\overrightarrow{a}-\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2\,\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle =1+1-2|\overrightarrow{a}||\overrightarrow{b}|\cos\theta$
$\displaystyle =2-2\cos\theta$
$\displaystyle \Rightarrow \cos\theta=\frac{2-|\overrightarrow{a}-\overrightarrow{b}|^2}{2}\ \text{...(2)}$
$\displaystyle \text{Now,}$
$\displaystyle \cos\frac{\theta}{2}=\sqrt{\frac{1+\cos\theta}{2}}$
$\displaystyle =\sqrt{\frac{1+\frac{|\overrightarrow{a}+\overrightarrow{b}|^2-2}{2}}{2}}\ \text{[From (1)]}$
$\displaystyle =\sqrt{\frac{|\overrightarrow{a}+\overrightarrow{b}|^2}{4}}$
$\displaystyle =\frac{1}{2}\,|\overrightarrow{a}+\overrightarrow{b}|$
$\displaystyle \text{(ii) }$
$\displaystyle \text{Given that }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are unit vectors}$
$\displaystyle \text{So, }|\overrightarrow{a}|=1,\ |\overrightarrow{b}|=1$
$\displaystyle \text{We have}$
$\displaystyle |\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2\,\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle =1+1+2|\overrightarrow{a}||\overrightarrow{b}|\cos\theta$
$\displaystyle =2+2\cos\theta$
$\displaystyle \Rightarrow \cos\theta=\frac{|\overrightarrow{a}+\overrightarrow{b}|^2-2}{2}\ \text{...(1)}$
$\displaystyle |\overrightarrow{a}-\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2\,\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle =1+1-2|\overrightarrow{a}||\overrightarrow{b}|\cos\theta$
$\displaystyle =2-2\cos\theta$
$\displaystyle \Rightarrow \cos\theta=\frac{2-|\overrightarrow{a}-\overrightarrow{b}|^2}{2}\ \text{...(2)}$
$\displaystyle \sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}}$
$\displaystyle =\sqrt{\frac{1-\frac{2-|\overrightarrow{a}-\overrightarrow{b}|^2}{2}}{2}}\ \text{[From (2)]}$
$\displaystyle =\sqrt{\frac{|\overrightarrow{a}-\overrightarrow{b}|^2}{4}}$
$\displaystyle =\frac{1}{2}|\overrightarrow{a}-\overrightarrow{b}|$
$\displaystyle \text{Now,}$
$\displaystyle \tan\frac{\theta}{2}=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}=\frac{\frac{1}{2}|\overrightarrow{a}-\overrightarrow{b}|}{\frac{1}{2}|\overrightarrow{a}+\overrightarrow{b}|}=\frac{|\overrightarrow{a}-\overrightarrow{b}|}{|\overrightarrow{a}+\overrightarrow{b}|}$

$\displaystyle \textbf{Question 9: }~\text{If the sum of two unit vectors is a unit vector, prove that the} \\ \text{magnitude of their difference is }\sqrt{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a},\ \overrightarrow{b}\text{ and }\overrightarrow{c}\text{ be three unit vectors}$
$\displaystyle \text{Given that the sum of the unit vectors is a unit vector}$
$\displaystyle \therefore \overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}$
$\displaystyle \text{or }|\overrightarrow{c}|^2=|\overrightarrow{a}+\overrightarrow{b}|^2$
$\displaystyle \text{or }|\overrightarrow{c}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{a}||\overrightarrow{b}|\cos\theta$
$\displaystyle \text{or }1=1+1+2\cos\theta\ \text{[}\because |\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1\text{]}$
$\displaystyle \Rightarrow \cos\theta=-\frac{1}{2}\ \text{...(1)}$
$\displaystyle \text{Now, }|\overrightarrow{a}-\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2|\overrightarrow{a}||\overrightarrow{b}|\cos\theta$
$\displaystyle =1+1-2\left(-\frac{1}{2}\right)\ \text{[From (1)]}$
$\displaystyle =3$
$\displaystyle \Rightarrow |\overrightarrow{a}-\overrightarrow{b}|=\sqrt{3}$

$\displaystyle \textbf{Question 10: }~\text{If }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are three mutually perpendicular unit vectors,} \\ \text{then prove that }\lvert\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\rvert=\sqrt{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\overrightarrow{a},\ \overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are unit vectors}$
$\displaystyle \text{So, }|\overrightarrow{a}|=1,\ |\overrightarrow{b}|=1\ \text{and}\ |\overrightarrow{c}|=1$
$\displaystyle \text{Since they are mutually perpendicular,}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{b}\cdot\overrightarrow{c}=\overrightarrow{c}\cdot\overrightarrow{a}=0$
$\displaystyle \text{Now,}$
$\displaystyle |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2\overrightarrow{a}\cdot\overrightarrow{b}+2\overrightarrow{b}\cdot\overrightarrow{c}+2\overrightarrow{c}\cdot\overrightarrow{a}$
$\displaystyle =1+1+1+0+0+0$
$\displaystyle =3$
$\displaystyle \therefore |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt{3}$

$\displaystyle \textbf{Question 11: }~\text{If }\lvert\overrightarrow{a}+\overrightarrow{b}\rvert=60,\ \lvert\overrightarrow{a}-\overrightarrow{b}\rvert=40\text{ and }\lvert\overrightarrow{b}\rvert=46,\text{ find }\lvert\overrightarrow{a}\rvert.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that}$
$\displaystyle |\overrightarrow{a}+\overrightarrow{b}|^2+|\overrightarrow{a}-\overrightarrow{b}|^2=2\left(|\overrightarrow{a}|^2+|\overrightarrow{b}|^2\right)$
$\displaystyle \Rightarrow 60^2+40^2=2\left(|\overrightarrow{a}|^2+46^2\right)\ \text{...(Given)}$
$\displaystyle \Rightarrow 3600+1600=2|\overrightarrow{a}|^2+4232$
$\displaystyle \Rightarrow 968=2|\overrightarrow{a}|^2$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2=484$
$\displaystyle \Rightarrow |\overrightarrow{a}|=22$

$\displaystyle \textbf{Question 12: }~\text{Show that the vector }\widehat{i}+\widehat{j}+\widehat{k}\text{ is equally inclined with the coordinate axes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\theta_1\text{ be the angle between }\overrightarrow{a}\text{ and x-axis}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$
$\displaystyle \overrightarrow{b}=\widehat{i}\ \text{(Because }\widehat{i}\text{ is the unit vector along x-axis)}$
$\displaystyle |\overrightarrow{b}|=\sqrt{(1)^2}=\sqrt{1}=1$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=1+0+0=1$
$\displaystyle \cos\theta_1=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\ |\overrightarrow{b}|}=\frac{1}{(\sqrt{3})(1)}=\frac{1}{\sqrt{3}}$
$\displaystyle \Rightarrow \theta_1=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\ \text{...(1)}$
$\displaystyle \text{Let }\theta_2\text{ be the angle between }\overrightarrow{a}\text{ and y-axis}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$
$\displaystyle \overrightarrow{b}=\widehat{j}\ \text{(Because }\widehat{j}\text{ is the unit vector along y-axis)}$
$\displaystyle |\overrightarrow{b}|=\sqrt{(1)^2}=\sqrt{1}=1$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0+1+0=1$
$\displaystyle \cos\theta_2=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\ |\overrightarrow{b}|}=\frac{1}{(\sqrt{3})(1)}=\frac{1}{\sqrt{3}}$
$\displaystyle \Rightarrow \theta_2=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\ \text{...(2)}$
$\displaystyle \text{Let }\theta_3\text{ be the angle between }\overrightarrow{a}\text{ and z-axis}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$
$\displaystyle \overrightarrow{b}=\widehat{k}\ \text{(Because }\widehat{k}\text{ is the unit vector along z-axis)}$
$\displaystyle |\overrightarrow{b}|=\sqrt{(1)^2}=\sqrt{1}=1$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0+0+1=1$
$\displaystyle \cos\theta_3=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\ |\overrightarrow{b}|}=\frac{1}{(\sqrt{3})(1)}=\frac{1}{\sqrt{3}}$
$\displaystyle \Rightarrow \theta_3=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\ \text{...(3)}$
$\displaystyle \text{From (1), (2) and (3), the given vector is equally inclined to the coordinate axes}$

$\displaystyle \textbf{Question 13: }~\text{Show that the vectors }\overrightarrow{a}=\frac{1}{7}(2\widehat{i}+3\widehat{j}+6\widehat{k}),\ \overrightarrow{b}=\frac{1}{7}(3\widehat{i}-6\widehat{j}+2\widehat{k}),\ \overrightarrow{c}=\frac{1}{7}(6\widehat{i}+2\widehat{j}-3\widehat{k})\text{ are mutually perpendicular unit vectors.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have}$
$\displaystyle |\overrightarrow{a}|=\frac{1}{7}\sqrt{2^2+3^2+6^2}=\frac{1}{7}\sqrt{49}=\frac{7}{7}=1$
$\displaystyle |\overrightarrow{b}|=\frac{1}{7}\sqrt{3^2+(-6)^2+2^2}=\frac{1}{7}\sqrt{49}=\frac{7}{7}=1$
$\displaystyle |\overrightarrow{c}|=\frac{1}{7}\sqrt{6^2+2^2+(-3)^2}=\frac{1}{7}\sqrt{49}=\frac{7}{7}=1$
$\displaystyle \text{And}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=\frac{1}{7}(2\widehat{i}+3\widehat{j}+6\widehat{k})\cdot\frac{1}{7}(3\widehat{i}-6\widehat{j}+2\widehat{k})$
$\displaystyle =\frac{1}{49}(6-18+12)$
$\displaystyle =0$
$\displaystyle \overrightarrow{b}\cdot\overrightarrow{c}=\frac{1}{7}(3\widehat{i}-6\widehat{j}+2\widehat{k})\cdot\frac{1}{7}(6\widehat{i}+2\widehat{j}-3\widehat{k})$
$\displaystyle =\frac{1}{49}(18-12-6)$
$\displaystyle =0$
$\displaystyle \overrightarrow{c}\cdot\overrightarrow{a}=\frac{1}{7}(6\widehat{i}+2\widehat{j}-3\widehat{k})\cdot\frac{1}{7}(2\widehat{i}+3\widehat{j}+6\widehat{k})$
$\displaystyle =\frac{1}{49}(12+6-18)$
$\displaystyle =0$
$\displaystyle \text{So, }|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1\ \text{and }\overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{b}\cdot\overrightarrow{c}=\overrightarrow{c}\cdot\overrightarrow{a}=0$
$\displaystyle \text{So, the given vectors are mutually perpendicular unit vectors}$

$\displaystyle \textbf{Question 14: }~\text{For any two vectors }\overrightarrow{a}\text{ and }\overrightarrow{b},\text{ show that }(\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=0\iff\lvert\overrightarrow{a}\rvert=\lvert\overrightarrow{b}\rvert.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have}$
$\displaystyle (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=0$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2-|\overrightarrow{b}|^2=0$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2=|\overrightarrow{b}|^2$
$\displaystyle \Rightarrow |\overrightarrow{a}|=|\overrightarrow{b}|$

$\displaystyle \textbf{Question 15: }~\text{If }\overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}+\widehat{j}-2\widehat{k}\text{ and }\overrightarrow{c}=\widehat{i}+3\widehat{j}-\widehat{k},\text{ find }\lambda\text{ such that }\overrightarrow{a}\perp(\lambda\overrightarrow{b}+\overrightarrow{c}).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given vectors are }\overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}+\widehat{j}-2\widehat{k}\ \text{and}\ \overrightarrow{c}=\widehat{i}+3\widehat{j}-\widehat{k}$
$\displaystyle \text{Now,}$
$\displaystyle \lambda\overrightarrow{b}+\overrightarrow{c}=\lambda(\widehat{i}+\widehat{j}-2\widehat{k})+(\widehat{i}+3\widehat{j}-\widehat{k})$
$\displaystyle =(\lambda+1)\widehat{i}+(\lambda+3)\widehat{j}-(2\lambda+1)\widehat{k}$
$\displaystyle \text{It is given that }\overrightarrow{a}\perp(\lambda\overrightarrow{b}+\overrightarrow{c})$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot(\lambda\overrightarrow{b}+\overrightarrow{c})=0$
$\displaystyle \Rightarrow (2\widehat{i}-\widehat{j}+\widehat{k})\cdot\big[(\lambda+1)\widehat{i}+(\lambda+3)\widehat{j}-(2\lambda+1)\widehat{k}\big]=0$
$\displaystyle \Rightarrow 2(\lambda+1)-(\lambda+3)-(2\lambda+1)=0$
$\displaystyle \Rightarrow 2\lambda+2-\lambda-3-2\lambda-1=0$
$\displaystyle \Rightarrow -\lambda-2=0$
$\displaystyle \Rightarrow \lambda=-2$

$\displaystyle \textbf{Question 16: }~\text{If }\overrightarrow{p}=5\widehat{i}+\widehat{j}-3\widehat{k}\text{ and }\overrightarrow{q}=\widehat{i}+3\widehat{j}-5\widehat{k},\text{ then find the value of }\lambda\text{ so that }\overrightarrow{p}+\overrightarrow{q}\text{ and }\overrightarrow{p}-\overrightarrow{q}\text{ are perpendicular vectors. } \ \ \ \ \ \ \ \ \ \ \ [\text{CBSE 2013}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that}$
$\displaystyle \overrightarrow{p}=5\widehat{i}+\lambda \widehat{j}-3\widehat{k}$
$\displaystyle \text{and }\overrightarrow{q}=\widehat{i}+3\widehat{j}-5\widehat{k}$
$\displaystyle \overrightarrow{p}+\overrightarrow{q}=(5\widehat{i}+\lambda \widehat{j}-3\widehat{k})+(\widehat{i}+3\widehat{j}-5\widehat{k})=6\widehat{i}+(\lambda+3)\widehat{j}-8\widehat{k}$
$\displaystyle \overrightarrow{p}-\overrightarrow{q}=(5\widehat{i}+\lambda \widehat{j}-3\widehat{k})-(\widehat{i}+3\widehat{j}-5\widehat{k})=4\widehat{i}+(\lambda-3)\widehat{j}+2\widehat{k}$
$\displaystyle \text{Given that }\overrightarrow{p}+\overrightarrow{q}\text{ is orthogonal to }\overrightarrow{p}-\overrightarrow{q}$
$\displaystyle \Rightarrow (\overrightarrow{p}+\overrightarrow{q})\cdot(\overrightarrow{p}-\overrightarrow{q})=0$
$\displaystyle \Rightarrow [6\widehat{i}+(\lambda+3)\widehat{j}-8\widehat{k}]\cdot[4\widehat{i}+(\lambda-3)\widehat{j}+2\widehat{k}]=0$
$\displaystyle \Rightarrow 24+(\lambda+3)(\lambda-3)-16=0$
$\displaystyle \Rightarrow 24+\lambda^2-9-16=0$
$\displaystyle \Rightarrow \lambda^2-1=0$
$\displaystyle \Rightarrow \lambda^2=1$
$\displaystyle \therefore \lambda=\pm 1$

$\displaystyle \textbf{Question 17: }~\text{If }\overrightarrow{\alpha}=3\widehat{i}+4\widehat{j}+5\widehat{k}\text{ and }\overrightarrow{\beta}=2\widehat{i}+\widehat{j}-4\widehat{k},\text{ then express }\overrightarrow{\beta}\text{ in the form }\overrightarrow{\beta}=\overrightarrow{\beta}_1+\overrightarrow{\beta}_2,\text{ where }\overrightarrow{\beta}_1\parallel\overrightarrow{\alpha}\text{ and }\overrightarrow{\beta}_2\perp\overrightarrow{\alpha}. \ \ \ \ \ \ \ \ \ \ \ [\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\overrightarrow{\alpha}=3\widehat{i}+4\widehat{j}+5\widehat{k}\ \text{and }\overrightarrow{\beta}=2\widehat{i}+\widehat{j}-4\widehat{k}$
$\displaystyle \text{Also,}$
$\displaystyle \overrightarrow{\beta}=\overrightarrow{\beta}_1+\overrightarrow{\beta}_2$
$\displaystyle \Rightarrow \overrightarrow{\beta}_2=\overrightarrow{\beta}-\overrightarrow{\beta}_1\ \text{...(1)}$
$\displaystyle \text{Since }\overrightarrow{\beta}_1\text{ is parallel to }\overrightarrow{\alpha},$
$\displaystyle \overrightarrow{\beta}_1=t\overrightarrow{\alpha}$
$\displaystyle \Rightarrow \overrightarrow{\beta}_1=t(3\widehat{i}+4\widehat{j}+5\widehat{k})=3t\widehat{i}+4t\widehat{j}+5t\widehat{k}\ \text{...(2)}$
$\displaystyle \text{Substituting the values of }\overrightarrow{\beta}_1\text{ and }\overrightarrow{\beta}\text{ in (1), we get}$
$\displaystyle \overrightarrow{\beta}_2=2\widehat{i}+\widehat{j}-4\widehat{k}-(3t\widehat{i}+4t\widehat{j}+5t\widehat{k})=(2-3t)\widehat{i}+(1-4t)\widehat{j}+(-4-5t)\widehat{k}\ \text{...(3)}$
$\displaystyle \text{Since }\overrightarrow{\beta}_2\text{ is perpendicular to }\overrightarrow{\alpha},$
$\displaystyle \overrightarrow{\beta}_2\cdot\overrightarrow{\alpha}=0$
$\displaystyle \Rightarrow [(2-3t)\widehat{i}+(1-4t)\widehat{j}+(-4-5t)\widehat{k}]\cdot(3\widehat{i}+4\widehat{j}+5\widehat{k})=0$
$\displaystyle \Rightarrow 3(2-3t)+4(1-4t)+5(-4-5t)=0$
$\displaystyle \Rightarrow 6-9t+4-16t-20-25t=0$
$\displaystyle \Rightarrow -50t-10=0$
$\displaystyle \Rightarrow -50t=10$
$\displaystyle \Rightarrow t=-\frac{1}{5}$
$\displaystyle \text{From (2) and (3), we get}$
$\displaystyle \overrightarrow{\beta}_1=-\frac{1}{5}(3\widehat{i}+4\widehat{j}+5\widehat{k})$
$\displaystyle \overrightarrow{\beta}_2=\frac{13}{5}\widehat{i}+\frac{9}{5}\widehat{j}-3\widehat{k}=\frac{1}{5}(13\widehat{i}+9\widehat{j}-15\widehat{k})$

$\displaystyle \textbf{Question 18: }~\text{If either }\overrightarrow{a}=\overrightarrow{0}\text{ or }\overrightarrow{b}=\overrightarrow{0},\text{ then }\overrightarrow{a}\cdot\overrightarrow{b}=0. \\ \text{But the converse need not be true. Justify your answer with an example. }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let us assume that either }|\overrightarrow{a}|=0\ \text{or}\ |\overrightarrow{b}|=0$
$\displaystyle \text{Then, }\overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}|\ |\overrightarrow{b}|\cos\theta=0\ \text{(}\theta\text{ is the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{)}$
$\displaystyle \text{Now, let us assume that }\overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow |\overrightarrow{a}|\ |\overrightarrow{b}|\cos\theta=0$
$\displaystyle \text{But here we cannot say that either }|\overrightarrow{a}|=0\ \text{or}\ |\overrightarrow{b}|=0\ \text{(because even }\cos\theta\text{ can be zero)}$
$\displaystyle \text{For example, let}$
$\displaystyle \overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k}\ \text{and}\ \overrightarrow{b}=-3\widehat{i}+2\widehat{k}$
$\displaystyle \text{Here, }|\overrightarrow{a}|=\sqrt{4+1+9}=\sqrt{14}\neq 0$
$\displaystyle |\overrightarrow{b}|=\sqrt{9+4}=\sqrt{13}\neq 0$
$\displaystyle \text{But }\overrightarrow{a}\cdot\overrightarrow{b}=(2\widehat{i}+\widehat{j}+3\widehat{k})\cdot(-3\widehat{i}+2\widehat{k})=-6+0+6=0$

$\displaystyle \textbf{Question 19: }~\text{Show that the vectors }\overrightarrow{a}=3\widehat{i}-2\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}-3\widehat{j}+5\widehat{k},\\ \overrightarrow{c}=2\widehat{i}+\widehat{j}-4\widehat{k}\text{ form a right-angled triangle. }[\text{CBSE 2005}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }ABC\text{ be the given triangle and}$
$\displaystyle \overrightarrow{AC}=\overrightarrow{b}=\widehat{i}-3\widehat{j}+5\widehat{k}$
$\displaystyle \overrightarrow{CB}=\overrightarrow{a}=3\widehat{i}-2\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{AB}=\overrightarrow{c}=2\widehat{i}+\widehat{j}-4\widehat{k}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=3+6+5=14$
$\displaystyle \overrightarrow{b}\cdot\overrightarrow{c}=2-3-20=-21$
$\displaystyle \overrightarrow{c}\cdot\overrightarrow{a}=6-2-4=0$
$\displaystyle \text{So, }\overrightarrow{AB}\text{ is perpendicular to }\overrightarrow{CB}$
$\displaystyle \text{Thus, }\triangle ABC\text{ is a right-angled triangle}$

$\displaystyle \textbf{Question 20: }~\text{If }\overrightarrow{a}=2\widehat{i}+2\widehat{j}+3\widehat{k},\ \overrightarrow{b}=-\widehat{i}+2\widehat{j}+\widehat{k}\text{ and }\overrightarrow{c}=3\widehat{i}+\widehat{j}\text{ are such that }\overrightarrow{a}+\lambda\overrightarrow{b}\perp\overrightarrow{c},\text{ then find the value of }\lambda.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have}$
$\displaystyle \overrightarrow{a}=2\widehat{i}+2\widehat{j}+3\widehat{k}$
$\displaystyle \overrightarrow{b}=-\widehat{i}+2\widehat{j}+\widehat{k}$
$\displaystyle \text{and}$
$\displaystyle \overrightarrow{c}=3\widehat{i}+\widehat{j}$
$\displaystyle \overrightarrow{a}+\lambda\overrightarrow{b}=2\widehat{i}+2\widehat{j}+3\widehat{k}+\lambda(-\widehat{i}+2\widehat{j}+\widehat{k})=(2-\lambda)\widehat{i}+(2+2\lambda)\widehat{j}+(3+\lambda)\widehat{k}$
$\displaystyle \text{Given that }\overrightarrow{a}+\lambda\overrightarrow{b}\text{ is perpendicular to }\overrightarrow{c}$
$\displaystyle \Rightarrow (\overrightarrow{a}+\lambda\overrightarrow{b})\cdot\overrightarrow{c}=0$
$\displaystyle \Rightarrow [(2-\lambda)\widehat{i}+(2+2\lambda)\widehat{j}+(3+\lambda)\widehat{k}]\cdot(3\widehat{i}+\widehat{j}+0\widehat{k})=0$
$\displaystyle \Rightarrow 3(2-\lambda)+1(2+2\lambda)+0=0$
$\displaystyle \Rightarrow 6-3\lambda+2+2\lambda=0$
$\displaystyle \Rightarrow 8-\lambda=0$
$\displaystyle \therefore \lambda=8$

$\displaystyle \textbf{Question 21: }~\text{Find the angles of a triangle whose vertices are }A(0,-1,-2),\\ B(3,1,4)\text{ and }C(5,7,1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that}$
$\displaystyle \overrightarrow{OA}=0\widehat{i}-\widehat{j}-2\widehat{k},\ \overrightarrow{OB}=3\widehat{i}+\widehat{j}+4\widehat{k},\ \overrightarrow{OC}=5\widehat{i}+7\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=3\widehat{i}+2\widehat{j}+6\widehat{k}\Rightarrow |\overrightarrow{AB}|=\sqrt{9+4+36}=7$
$\displaystyle \overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}=-3\widehat{i}-2\widehat{j}-6\widehat{k}\Rightarrow |\overrightarrow{BA}|=\sqrt{9+4+36}=7$
$\displaystyle \overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=2\widehat{i}+6\widehat{j}-3\widehat{k}\Rightarrow |\overrightarrow{BC}|=\sqrt{4+36+9}=7$
$\displaystyle \overrightarrow{CB}=\overrightarrow{OB}-\overrightarrow{OC}=-2\widehat{i}-6\widehat{j}+3\widehat{k}\Rightarrow |\overrightarrow{CB}|=\sqrt{4+36+9}=7$
$\displaystyle \overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}=-5\widehat{i}-8\widehat{j}-3\widehat{k}\Rightarrow |\overrightarrow{CA}|=\sqrt{25+64+9}=\sqrt{98}=7\sqrt{2}$
$\displaystyle \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=5\widehat{i}+8\widehat{j}+3\widehat{k}\Rightarrow |\overrightarrow{AC}|=\sqrt{25+64+9}=\sqrt{98}=7\sqrt{2}$
$\displaystyle \cos A=\frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|\overrightarrow{AB}|\ |\overrightarrow{AC}|}=\frac{15+16+18}{(7)(7\sqrt{2})}=\frac{49}{49\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\displaystyle \Rightarrow A=\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$
$\displaystyle \cos B=\frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{|\overrightarrow{BA}|\ |\overrightarrow{BC}|}=\frac{-6-12+18}{(7)(7)}=\frac{0}{49}=0$
$\displaystyle \Rightarrow B=\cos^{-1}(0)=\frac{\pi}{2}$
$\displaystyle \cos C=\frac{\overrightarrow{CB}\cdot\overrightarrow{CA}}{|\overrightarrow{CB}|\ |\overrightarrow{CA}|}=\frac{10+48-9}{(7)(7\sqrt{2})}=\frac{49}{49\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\displaystyle \Rightarrow C=\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$

$\displaystyle \textbf{Question 22: }~\text{Find the magnitude of two vectors }\overrightarrow{a}\text{ and }\overrightarrow{b},\text{ having the same magnitude} \\ \text{and such that the angle between them is }60^{\circ}\text{ and their scalar product is }\frac{1}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }30^\circ$
$\displaystyle \text{Also, }|\overrightarrow{a}|=|\overrightarrow{b}|,\ \overrightarrow{a}\cdot\overrightarrow{b}=\frac{1}{2}$
$\displaystyle \text{We know that}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}|\ |\overrightarrow{b}|\cos\theta$
$\displaystyle \Rightarrow \frac{1}{2}=|\overrightarrow{a}|\ |\overrightarrow{a}|\cos 60^\circ$
$\displaystyle \Rightarrow \frac{1}{2}=|\overrightarrow{a}|^2\left(\frac{1}{2}\right)$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2=1$
$\displaystyle \Rightarrow |\overrightarrow{a}|=1$
$\displaystyle \therefore |\overrightarrow{a}|=|\overrightarrow{b}|=1$

$\displaystyle \textbf{Question 23: }~\text{Show that the points whose position vectors are }\overrightarrow{a}=4\widehat{i}-3\widehat{j}+\widehat{k},\ \overrightarrow{b}=2\widehat{i}-4\widehat{j}+5\widehat{k},\ \overrightarrow{c}=\widehat{i}-\widehat{j}\text{ form a right triangle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that}$
$\displaystyle \overrightarrow{a}=\overrightarrow{OA}=4\widehat{i}-3\widehat{j}+\widehat{k},\ \overrightarrow{b}=\overrightarrow{OB}=2\widehat{i}-4\widehat{j}+5\widehat{k},\ \overrightarrow{c}=\overrightarrow{OC}=\widehat{i}-\widehat{j}+0\widehat{k}$
$\displaystyle \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-2\widehat{i}-\widehat{j}+4\widehat{k}$
$\displaystyle \overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=-\widehat{i}+3\widehat{j}-5\widehat{k}$
$\displaystyle \overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}=3\widehat{i}-2\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{AB}\cdot\overrightarrow{BC}=2-3-20=-21\neq 0$
$\displaystyle \overrightarrow{BC}\cdot\overrightarrow{CA}=-3-6-5=-14\neq 0$
$\displaystyle \overrightarrow{AB}\cdot\overrightarrow{CA}=-6+2+4=0$
$\displaystyle \text{So, }\overrightarrow{AB}\text{ is perpendicular to }\overrightarrow{CA}$
$\displaystyle \text{So, }\triangle ABC\text{ is a right-angled triangle}$

$\displaystyle \textbf{Question 24: }~\text{If the vertices }A,B,C\text{ of }\triangle ABC\text{ have position vectors } \\ (1,2,3),\ (-1,0,0),\ (0,1,2) \text{ respectively, what is the magnitude of }\angle ABC?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\overrightarrow{OA}=\widehat{i}+2\widehat{j}+3\widehat{k},\ \overrightarrow{OB}=-\widehat{i}+0\widehat{j}+0\widehat{k},\ \overrightarrow{OC}=0\widehat{i}+\widehat{j}+2\widehat{k}$
$\displaystyle \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-2\widehat{i}-2\widehat{j}-3\widehat{k}\Rightarrow |\overrightarrow{AB}|=\sqrt{4+4+9}=\sqrt{17}$
$\displaystyle \overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=\widehat{i}+\widehat{j}+2\widehat{k}\Rightarrow |\overrightarrow{BC}|=\sqrt{1+1+4}=\sqrt{6}$
$\displaystyle \overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}=\widehat{i}+\widehat{j}+\widehat{k}\Rightarrow |\overrightarrow{CA}|=\sqrt{1+1+1}=\sqrt{3}$
$\displaystyle \cos\angle ABC=\frac{|\overrightarrow{BA}\cdot\overrightarrow{BC}|}{|\overrightarrow{BA}|\ |\overrightarrow{BC}|}$
$\displaystyle =\frac{|(-2)(1)+(-2)(1)+(-3)(2)|}{(\sqrt{17})(\sqrt{6})}$
$\displaystyle =\frac{|-2-2-6|}{\sqrt{102}}$
$\displaystyle =\frac{10}{\sqrt{102}}$
$\displaystyle \Rightarrow \angle ABC=\cos^{-1}\left(\frac{10}{\sqrt{102}}\right)$

$\displaystyle \textbf{Question 25: }~\text{If }A,B,C\text{ have position vectors }(0,1,1),\ (3,1,5),\ (0,3,3) \\ \text{ respectively, show that }\triangle ABC\text{ is right angled at }C.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that}$
$\displaystyle \overrightarrow{OA}=0\widehat{i}+\widehat{j}+\widehat{k},\ \overrightarrow{OB}=3\widehat{i}+\widehat{j}+5\widehat{k},\ \overrightarrow{OC}=0\widehat{i}+3\widehat{j}+3\widehat{k}$
$\displaystyle \overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=-3\widehat{i}+2\widehat{j}-2\widehat{k}$
$\displaystyle \overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}=0\widehat{i}-2\widehat{j}-2\widehat{k}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{BC}\cdot\overrightarrow{CA}=0-4+4=0$
$\displaystyle \text{So, }\overrightarrow{BC}\text{ is perpendicular to }\overrightarrow{CA}$
$\displaystyle \text{So, }\triangle ABC\text{ is right-angled at }C$

$\displaystyle \textbf{Question 26: }~\text{Find the projection of }\overrightarrow{b}+\overrightarrow{c}\text{ on }\overrightarrow{a},\text{ where }\overrightarrow{a}=2\widehat{i}-2\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}+2\widehat{j}-2\widehat{k}\text{ and }\overrightarrow{c}=2\widehat{i}-\widehat{j}+4\widehat{k}. [\text{CBSE 2007}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that}$
$\displaystyle \overrightarrow{a}=2\widehat{i}-2\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}-2\widehat{k}$
$\displaystyle \text{and }\overrightarrow{c}=2\widehat{i}-\widehat{j}+4\widehat{k}$
$\displaystyle \therefore \overrightarrow{b}+\overrightarrow{c}=\widehat{i}+2\widehat{j}-2\widehat{k}+2\widehat{i}-\widehat{j}+4\widehat{k}=3\widehat{i}+\widehat{j}+2\widehat{k}$
$\displaystyle \text{Projection of }\overrightarrow{b}+\overrightarrow{c}\text{ on }\overrightarrow{a}\text{ is}$
$\displaystyle \frac{(\overrightarrow{b}+\overrightarrow{c})\cdot\overrightarrow{a}}{|\overrightarrow{a}|}$
$\displaystyle =\frac{(3\widehat{i}+\widehat{j}+2\widehat{k})\cdot(2\widehat{i}-2\widehat{j}+\widehat{k})}{\sqrt{4+4+1}}$
$\displaystyle =\frac{6-2+2}{3}$
$\displaystyle =2$

$\displaystyle \textbf{Question 27: }~\text{If }\overrightarrow{a}=5\widehat{i}-\widehat{j}-3\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}+3\widehat{j}-5\widehat{k},\text{ then show that the vectors } \\ \overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{a}-\overrightarrow{b}\text{ are orthogonal. }[\text{CBSE 2004}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that}$
$\displaystyle \overrightarrow{a}=5\widehat{i}-\widehat{j}-3\widehat{k},\ \overrightarrow{b}=\widehat{i}+3\widehat{j}-5\widehat{k}$
$\displaystyle \therefore \overrightarrow{a}+\overrightarrow{b}=5\widehat{i}-\widehat{j}-3\widehat{k}+\widehat{i}+3\widehat{j}-5\widehat{k}=6\widehat{i}+2\widehat{j}-8\widehat{k}$
$\displaystyle \text{And }\overrightarrow{a}-\overrightarrow{b}=5\widehat{i}-\widehat{j}-3\widehat{k}-(\widehat{i}+3\widehat{j}-5\widehat{k})=4\widehat{i}-4\widehat{j}+2\widehat{k}$
$\displaystyle \text{Now,}$
$\displaystyle (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})$
$\displaystyle =(6\widehat{i}+2\widehat{j}-8\widehat{k})\cdot(4\widehat{i}-4\widehat{j}+2\widehat{k})$
$\displaystyle =24-8-16$
$\displaystyle =0$
$\displaystyle \text{So, }\overrightarrow{a}+\overrightarrow{b}\text{ is orthogonal to }\overrightarrow{a}-\overrightarrow{b}$

$\displaystyle \textbf{Question 28: }~\text{A unit vector }\overrightarrow{a}\text{ makes angles }\frac{\pi}{4}\text{ and }\frac{\pi}{3}\text{ with }\widehat{i}\text{ and }\widehat{j}\text{ respectively} \\ \text{and an acute angle }\theta\text{ with }\widehat{k}.\text{ Find the angle }\theta\text{ and components of }\overrightarrow{a}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k},\ \text{where }a_1,a_2\text{ and }a_3\text{ are components of }\overrightarrow{a}$
$\displaystyle \Rightarrow a_1^2+a_2^2+a_3^2=1\ \text{(Because }\overrightarrow{a}\text{ is a unit vector)}\ \text{...(1)}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{a}\cdot\widehat{i}=a_1$
$\displaystyle \Rightarrow |\overrightarrow{a}|\ |\widehat{i}|\cos\frac{\pi}{4}=a_1\ \text{(Because the angle between }\overrightarrow{a}\text{ and }\widehat{i}\text{ is }\frac{\pi}{4})$
$\displaystyle \Rightarrow (1)(1)\frac{1}{\sqrt{2}}=a_1\ \text{(Because }\overrightarrow{a}\text{ and }\widehat{i}\text{ are unit vectors)}$
$\displaystyle \Rightarrow a_1=\frac{1}{\sqrt{2}}$
$\displaystyle \text{Again,}$
$\displaystyle \overrightarrow{a}\cdot\widehat{j}=a_2$
$\displaystyle \Rightarrow |\overrightarrow{a}|\ |\widehat{j}|\cos\frac{\pi}{3}=a_2\ \text{(Because the angle between }\overrightarrow{a}\text{ and }\widehat{j}\text{ is }\frac{\pi}{3})$
$\displaystyle \Rightarrow (1)(1)\frac{1}{2}=a_2\ \text{(Because }\overrightarrow{a}\text{ and }\widehat{j}\text{ are unit vectors)}$
$\displaystyle \Rightarrow a_2=\frac{1}{2}$
$\displaystyle \text{Now from (1),}$
$\displaystyle \left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{2}\right)^2+a_3^2=1$
$\displaystyle \Rightarrow \frac{1}{2}+\frac{1}{4}+a_3^2=1$
$\displaystyle \Rightarrow \frac{3}{4}+a_3^2=1$
$\displaystyle \Rightarrow a_3^2=\frac{1}{4}$
$\displaystyle \Rightarrow a_3=\frac{1}{2}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{a}\cdot\widehat{k}=a_3$
$\displaystyle \Rightarrow |\overrightarrow{a}|\ |\widehat{k}|\cos\theta=\frac{1}{2}\ \text{(Because the angle between }\overrightarrow{a}\text{ and }\widehat{k}\text{ is }\theta)$
$\displaystyle \Rightarrow (1)(1)\cos\theta=\frac{1}{2}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$
$\displaystyle \text{And}$
$\displaystyle \overrightarrow{a}=\frac{1}{\sqrt{2}}\widehat{i}+\frac{1}{2}\widehat{j}+\frac{1}{2}\widehat{k}$

$\displaystyle \textbf{Question 29: }~\text{If two vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are such that }\lvert\overrightarrow{a}\rvert=2,\ \lvert\overrightarrow{b}\rvert=1\text{ and }\overrightarrow{a}\cdot\overrightarrow{b}=1,\text{ then find the value of }(3\overrightarrow{a}-5\overrightarrow{b})\cdot(2\overrightarrow{a}+7\overrightarrow{b}). [\text{ CBSE 2011}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that}$
$\displaystyle |\overrightarrow{a}|=2,\ |\overrightarrow{b}|=1\ \text{and}\ \overrightarrow{a}\cdot\overrightarrow{b}=1\ \text{...(1)}$
$\displaystyle \text{Now,}$
$\displaystyle (3\overrightarrow{a}-5\overrightarrow{b})\cdot(2\overrightarrow{a}+7\overrightarrow{b})$
$\displaystyle =6|\overrightarrow{a}|^2+21\,\overrightarrow{a}\cdot\overrightarrow{b}-10\,\overrightarrow{b}\cdot\overrightarrow{a}-35|\overrightarrow{b}|^2$
$\displaystyle =6|\overrightarrow{a}|^2+21\,\overrightarrow{a}\cdot\overrightarrow{b}-10\,\overrightarrow{a}\cdot\overrightarrow{b}-35|\overrightarrow{b}|^2\ \text{(We know that }\overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{b}\cdot\overrightarrow{a}\text{)}$
$\displaystyle =6|\overrightarrow{a}|^2+11\,\overrightarrow{a}\cdot\overrightarrow{b}-35|\overrightarrow{b}|^2$
$\displaystyle =6(2)^2+11(1)-35(1)^2\ \text{[From (1)]}$
$\displaystyle =24+11-35$
$\displaystyle =0$

$\displaystyle \textbf{Question 30: }~\text{If }\overrightarrow{x}\text{ is a unit vector, then find }\lvert\overrightarrow{x}\rvert\text{ in each of the following:}\\ \text{(i) }(\overrightarrow{x}-\overrightarrow{a})\cdot(\overrightarrow{x}+\overrightarrow{a})=8\qquad \text{(ii) }(\overrightarrow{x}-\overrightarrow{a})\cdot(\overrightarrow{x}+\overrightarrow{a})=12\quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Given that }\overrightarrow{a}\text{ is a unit vector}$
$\displaystyle \Rightarrow |\overrightarrow{a}|=1\ \text{...(1)}$
$\displaystyle (\overrightarrow{x}-\overrightarrow{a})\cdot(\overrightarrow{x}+\overrightarrow{a})=8$
$\displaystyle \Rightarrow |\overrightarrow{x}|^2-|\overrightarrow{a}|^2=8$
$\displaystyle \Rightarrow |\overrightarrow{x}|^2-1^2=8\ \text{[From (1)]}$
$\displaystyle \Rightarrow |\overrightarrow{x}|^2=9$
$\displaystyle \Rightarrow |\overrightarrow{x}|=3$
$\displaystyle \text{(ii) }$
$\displaystyle \text{Given that }\overrightarrow{a}\text{ is a unit vector}$
$\displaystyle (\overrightarrow{x}-\overrightarrow{a})\cdot(\overrightarrow{x}+\overrightarrow{a})=12$
$\displaystyle \Rightarrow |\overrightarrow{x}|^2-|\overrightarrow{a}|^2=12$
$\displaystyle \Rightarrow |\overrightarrow{x}|^2-1^2=12\ \text{[Because }\overrightarrow{a}\text{ is a unit vector]}$
$\displaystyle \Rightarrow |\overrightarrow{x}|^2=13$
$\displaystyle \Rightarrow |\overrightarrow{x}|=\sqrt{13}$

$\displaystyle \textbf{Question 31: }~\text{Find }\lvert\overrightarrow{a}\rvert\text{ and }\lvert\overrightarrow{b}\rvert,\text{ if}\\ \text{(i) }(\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=12\text{ and }\lvert\overrightarrow{a}\rvert=2\lvert\overrightarrow{b}\rvert\\ \text{(ii) }(\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=8\text{ and }\lvert\overrightarrow{a}\rvert=8\lvert\overrightarrow{b}\rvert\\ \text{(iii) }(\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=3\text{ and }\lvert\overrightarrow{a}\rvert=2\lvert\overrightarrow{b}\rvert\quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Given that}$
$\displaystyle |\overrightarrow{a}|=2|\overrightarrow{b}|\ \text{...(1)}$
$\displaystyle \text{And }(\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=12$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2-|\overrightarrow{b}|^2=12$
$\displaystyle \Rightarrow (2|\overrightarrow{b}|)^2-|\overrightarrow{b}|^2=12\ \text{[From (1)]}$
$\displaystyle \Rightarrow 4|\overrightarrow{b}|^2-|\overrightarrow{b}|^2=12$
$\displaystyle \Rightarrow 3|\overrightarrow{b}|^2=12$
$\displaystyle \Rightarrow |\overrightarrow{b}|^2=4$
$\displaystyle \Rightarrow |\overrightarrow{b}|=2$
$\displaystyle |\overrightarrow{a}|=2|\overrightarrow{b}|=2(2)=4$
$\displaystyle \therefore |\overrightarrow{a}|=4\ \text{and}\ |\overrightarrow{b}|=2$

$\displaystyle \text{(ii) }$
$\displaystyle \text{Given that}$
$\displaystyle |\overrightarrow{a}|=8|\overrightarrow{b}|\ \text{...(1)}$
$\displaystyle (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=8$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2-|\overrightarrow{b}|^2=8$
$\displaystyle \Rightarrow (8|\overrightarrow{b}|)^2-|\overrightarrow{b}|^2=8\ \text{[From (1)]}$
$\displaystyle \Rightarrow 64|\overrightarrow{b}|^2-|\overrightarrow{b}|^2=8$
$\displaystyle \Rightarrow 63|\overrightarrow{b}|^2=8$
$\displaystyle \Rightarrow |\overrightarrow{b}|^2=\frac{8}{63}$
$\displaystyle \Rightarrow |\overrightarrow{b}|=\sqrt{\frac{8}{63}}$
$\displaystyle |\overrightarrow{a}|=8|\overrightarrow{b}|=8\sqrt{\frac{8}{63}}=\frac{8\sqrt{8}}{\sqrt{63}}$
$\displaystyle \therefore |\overrightarrow{a}|=\frac{8\sqrt{8}}{\sqrt{63}}\ \text{and}\ |\overrightarrow{b}|=\sqrt{\frac{8}{63}}$

$\displaystyle \text{(iii) }$
$\displaystyle \text{Given that}$
$\displaystyle |\overrightarrow{a}|=2|\overrightarrow{b}|\ \text{...(1)}$
$\displaystyle \text{And }(\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=3$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2-|\overrightarrow{b}|^2=3$
$\displaystyle \Rightarrow (2|\overrightarrow{b}|)^2-|\overrightarrow{b}|^2=3\ \text{[From (1)]}$
$\displaystyle \Rightarrow 4|\overrightarrow{b}|^2-|\overrightarrow{b}|^2=3$
$\displaystyle \Rightarrow 3|\overrightarrow{b}|^2=3$
$\displaystyle \Rightarrow |\overrightarrow{b}|^2=1$
$\displaystyle \Rightarrow |\overrightarrow{b}|=1$
$\displaystyle |\overrightarrow{a}|=2|\overrightarrow{b}|=2(1)=2$
$\displaystyle \therefore |\overrightarrow{a}|=2\ \text{and}\ |\overrightarrow{b}|=1$

$\displaystyle \textbf{Question 32: }~\text{Find }\lvert\overrightarrow{a}-\overrightarrow{b}\rvert,\text{ if}\\ \text{(i) }\lvert\overrightarrow{a}\rvert=2,\ \lvert\overrightarrow{b}\rvert=5\text{ and }\overrightarrow{a}\cdot\overrightarrow{b}=8\\ \text{(ii) }\lvert\overrightarrow{a}\rvert=3,\ \lvert\overrightarrow{b}\rvert=4\text{ and }\overrightarrow{a}\cdot\overrightarrow{b}=1\\ \text{(iii) }\lvert\overrightarrow{a}\rvert=2,\ \lvert\overrightarrow{b}\rvert=3\text{ and }\overrightarrow{a}\cdot\overrightarrow{b}=4\quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Given that}$
$\displaystyle |\overrightarrow{a}|=2,\ |\overrightarrow{b}|=5\ \text{and}\ \overrightarrow{a}\cdot\overrightarrow{b}=8\ \text{...(1)}$
$\displaystyle \text{We know that}$
$\displaystyle |\overrightarrow{a}-\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2\,\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle =2^2+5^2-2(8)\ \text{[Using (1)]}$
$\displaystyle =4+25-16$
$\displaystyle =13$
$\displaystyle \therefore |\overrightarrow{a}-\overrightarrow{b}|=\sqrt{13}$
$\displaystyle \text{(ii) }$
$\displaystyle \text{Given that}$
$\displaystyle |\overrightarrow{a}|=3,\ |\overrightarrow{b}|=4\ \text{and}\ \overrightarrow{a}\cdot\overrightarrow{b}=1\ \text{...(1)}$
$\displaystyle \text{We know that}$
$\displaystyle |\overrightarrow{a}-\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2\,\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle =3^2+4^2-2(1)\ \text{[Using (1)]}$
$\displaystyle =9+16-2$
$\displaystyle =23$
$\displaystyle \therefore |\overrightarrow{a}-\overrightarrow{b}|=\sqrt{23}$
$\displaystyle \text{(iii) }$
$\displaystyle \text{Given that}$
$\displaystyle |\overrightarrow{a}|=2,\ |\overrightarrow{b}|=3\ \text{and}\ \overrightarrow{a}\cdot\overrightarrow{b}=4\ \text{...(1)}$
$\displaystyle \text{We know that}$
$\displaystyle |\overrightarrow{a}-\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2\,\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle =2^2+3^2-2(4)\ \text{[Using (1)]}$
$\displaystyle =4+9-8$
$\displaystyle =5$
$\displaystyle \therefore |\overrightarrow{a}-\overrightarrow{b}|=\sqrt{5}$

$\displaystyle \textbf{Question 33: }~\text{Find the angle between two vectors }\overrightarrow{a}\text{ and }\overrightarrow{b},\text{ if}\\ \text{(i) }\lvert\overrightarrow{a}\rvert=\sqrt{3},\ \lvert\overrightarrow{b}\rvert=2\text{ and }\overrightarrow{a}\cdot\overrightarrow{b}=\sqrt{6}\\ \text{(ii) }\lvert\overrightarrow{a}\rvert=3,\ \lvert\overrightarrow{b}\rvert=3\text{ and }\overrightarrow{a}\cdot\overrightarrow{b}=1\quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}$
$\displaystyle \text{Given that}$
$\displaystyle |\overrightarrow{a}|=\sqrt{3},\ |\overrightarrow{b}|=2\ \text{and}\ \overrightarrow{a}\cdot\overrightarrow{b}=\sqrt{6}\ \text{...(1)}$
$\displaystyle \text{We know that}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}|\ |\overrightarrow{b}|\cos\theta$
$\displaystyle \Rightarrow \sqrt{6}=(\sqrt{3})(2)\cos\theta\ \text{[Using (1)]}$
$\displaystyle \Rightarrow \cos\theta=\frac{\sqrt{6}}{2\sqrt{3}}=\frac{1}{\sqrt{2}}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$
$\displaystyle \text{(ii) }$
$\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}$
$\displaystyle \text{Given that}$
$\displaystyle |\overrightarrow{a}|=3,\ |\overrightarrow{b}|=3\ \text{and}\ \overrightarrow{a}\cdot\overrightarrow{b}=1\ \text{...(1)}$
$\displaystyle \text{We know that}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}|\ |\overrightarrow{b}|\cos\theta$
$\displaystyle \Rightarrow 1=(3)(3)\cos\theta\ \text{[Using (1)]}$
$\displaystyle \Rightarrow \cos\theta=\frac{1}{(3)(3)}=\frac{1}{9}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{1}{9}\right)$

$\displaystyle \textbf{Question 34: }~\text{Express the vector }\overrightarrow{a}=5\widehat{i}-2\widehat{j}+5\widehat{k}\text{ as the sum of two vectors} \\ \text{such that one is parallel to the vector }\overrightarrow{b}=3\widehat{i}+\widehat{k}\text{ and the other is} \\ \text{perpendicular to }\overrightarrow{b}. [\text{CBSE 2005}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\overrightarrow{a}=5\widehat{i}-2\widehat{j}+5\widehat{k}\ \text{and}\ \overrightarrow{b}=3\widehat{i}+\widehat{k}$
$\displaystyle \text{Let }\overrightarrow{x}\text{ and }\overrightarrow{y}\text{ be such that}$
$\displaystyle \overrightarrow{a}=\overrightarrow{x}+\overrightarrow{y}$
$\displaystyle \Rightarrow \overrightarrow{y}=\overrightarrow{a}-\overrightarrow{x}\ \text{...(1)}$
$\displaystyle \text{Since }\overrightarrow{x}\text{ is parallel to }\overrightarrow{b},$
$\displaystyle \Rightarrow \overrightarrow{x}=t\overrightarrow{b}\ \text{(}t\text{ is a constant)}$
$\displaystyle \Rightarrow \overrightarrow{x}=t(3\widehat{i}+\widehat{k})=3t\widehat{i}+t\widehat{k}$
$\displaystyle \text{Substituting the values of }\overrightarrow{x}\text{ and }\overrightarrow{a}\text{ in (1), we get}$
$\displaystyle \overrightarrow{y}=5\widehat{i}-2\widehat{j}+5\widehat{k}-(3t\widehat{i}+t\widehat{k})=(5-3t)\widehat{i}-2\widehat{j}+(5-t)\widehat{k}\ \text{...(2)}$
$\displaystyle \text{Since }\overrightarrow{y}\text{ is perpendicular to }\overrightarrow{b},$
$\displaystyle \overrightarrow{y}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow [(5-3t)\widehat{i}-2\widehat{j}+(5-t)\widehat{k}]\cdot(3\widehat{i}+\widehat{k})=0$
$\displaystyle \Rightarrow 3(5-3t)+0+(5-t)=0$
$\displaystyle \Rightarrow 15-9t+5-t=0$
$\displaystyle \Rightarrow 20-10t=0$
$\displaystyle \Rightarrow t=2$
$\displaystyle \text{From (1) and (2), we get}$
$\displaystyle \overrightarrow{x}=6\widehat{i}+2\widehat{k}$
$\displaystyle \overrightarrow{y}=-\widehat{i}-2\widehat{j}+3\widehat{k}$

$\displaystyle \textbf{Question 35: }~\text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are two vectors of the same magnitude inclined} \\ \text{at an angle of }30^{\circ}\text{ such that }\overrightarrow{a}\cdot\overrightarrow{b}=3,\text{ find }\lvert\overrightarrow{a}\rvert,\ \lvert\overrightarrow{b}\rvert.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }30^\circ$
$\displaystyle \text{Also, }|\overrightarrow{a}|=|\overrightarrow{b}|\ \text{and}\ \overrightarrow{a}\cdot\overrightarrow{b}=3$
$\displaystyle \text{We know that}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}|\ |\overrightarrow{b}|\cos\theta$
$\displaystyle \Rightarrow 3=|\overrightarrow{a}|\ |\overrightarrow{a}|\cos 30^\circ$
$\displaystyle \Rightarrow 3=|\overrightarrow{a}|^2\left(\frac{\sqrt{3}}{2}\right)$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2=\frac{6}{\sqrt{3}}=2\sqrt{3}$
$\displaystyle \Rightarrow |\overrightarrow{a}|=\sqrt{2\sqrt{3}}=|\overrightarrow{b}|$
$\displaystyle \therefore |\overrightarrow{a}|=|\overrightarrow{b}|=\sqrt{2\sqrt{3}}$

$\displaystyle \textbf{Question 36: }~\text{Express }2\widehat{i}-\widehat{j}+3\widehat{k}\text{ as the sum of a vector parallel and a vector} \\ \text{perpendicular to }2\widehat{i}+4\widehat{j}-2\widehat{k}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=2\widehat{i}-\widehat{j}+3\widehat{k}\ \text{and}\ \overrightarrow{b}=2\widehat{i}+4\widehat{j}-2\widehat{k}$
$\displaystyle \text{and }\overrightarrow{x}\text{ and }\overrightarrow{y}\text{ be such that}$
$\displaystyle \overrightarrow{a}=\overrightarrow{x}+\overrightarrow{y}$
$\displaystyle \Rightarrow \overrightarrow{y}=\overrightarrow{a}-\overrightarrow{x}\ \text{...(1)}$
$\displaystyle \text{Since }\overrightarrow{x}\text{ is parallel to }\overrightarrow{b},$
$\displaystyle \overrightarrow{x}=t\overrightarrow{b}$
$\displaystyle \Rightarrow \overrightarrow{x}=t(2\widehat{i}+4\widehat{j}-2\widehat{k})=2t\widehat{i}+4t\widehat{j}-2t\widehat{k}\ \text{...(2)}$
$\displaystyle \text{Substituting the values of }\overrightarrow{x}\text{ and }\overrightarrow{a}\text{ in (1),}$
$\displaystyle \overrightarrow{y}=2\widehat{i}-\widehat{j}+3\widehat{k}-(2t\widehat{i}+4t\widehat{j}-2t\widehat{k})=(2-2t)\widehat{i}+(-1-4t)\widehat{j}+(3+2t)\widehat{k}\ \text{...(3)}$
$\displaystyle \text{Since }\overrightarrow{y}\text{ is perpendicular to }\overrightarrow{b},$
$\displaystyle \overrightarrow{y}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow [(2-2t)\widehat{i}+(-1-4t)\widehat{j}+(3+2t)\widehat{k}]\cdot(2\widehat{i}+4\widehat{j}-2\widehat{k})=0$
$\displaystyle \Rightarrow 2(2-2t)+4(-1-4t)-2(3+2t)=0$
$\displaystyle \Rightarrow 4-4t-4-16t-6-4t=0$
$\displaystyle \Rightarrow -24t-6=0$
$\displaystyle \Rightarrow -24t=6$
$\displaystyle \Rightarrow t=-\frac{1}{4}$
$\displaystyle \text{From (2) and (3),}$
$\displaystyle \overrightarrow{x}=2\left(-\frac{1}{4}\right)\widehat{i}+4\left(-\frac{1}{4}\right)\widehat{j}-2\left(-\frac{1}{4}\right)\widehat{k}=-\frac{1}{2}\widehat{i}-\widehat{j}+\frac{1}{2}\widehat{k}$
$\displaystyle \overrightarrow{y}=\left[2-2\left(-\frac{1}{4}\right)\right]\widehat{i}+\left[-1-4\left(-\frac{1}{4}\right)\right]\widehat{j}+\left[3+2\left(-\frac{1}{4}\right)\right]\widehat{k}=\frac{5}{2}\widehat{i}+\frac{5}{2}\widehat{k}=\frac{5}{2}(\widehat{i}+\widehat{k})$
$\displaystyle \text{So,}$
$\displaystyle \overrightarrow{a}=\overrightarrow{x}+\overrightarrow{y}=\left(-\frac{1}{2}\widehat{i}-\widehat{j}+\frac{1}{2}\widehat{k}\right)+\frac{5}{2}(\widehat{i}+\widehat{k})$

$\displaystyle \textbf{Question 37: }~\text{Decompose the vector }6\widehat{i}-3\widehat{j}-6\widehat{k}\text{ into vectors} \\ \text{which are parallel and perpendicular to the vector }\widehat{i}+\widehat{j}+\widehat{k}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=6\widehat{i}-3\widehat{j}-6\widehat{k}\ \text{and}\ \overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \text{and }\overrightarrow{x}\text{ and }\overrightarrow{y}\text{ be such that}$
$\displaystyle \overrightarrow{a}=\overrightarrow{x}+\overrightarrow{y}$
$\displaystyle \Rightarrow \overrightarrow{y}=\overrightarrow{a}-\overrightarrow{x}\ \text{...(1)}$
$\displaystyle \text{Since }\overrightarrow{x}\text{ is parallel to }\overrightarrow{b},$
$\displaystyle \overrightarrow{x}=t\overrightarrow{b}$
$\displaystyle \Rightarrow \overrightarrow{x}=t(\widehat{i}+\widehat{j}+\widehat{k})=t\widehat{i}+t\widehat{j}+t\widehat{k}\ \text{...(2)}$
$\displaystyle \text{Substituting the values of }\overrightarrow{x}\text{ and }\overrightarrow{a}\text{ in (1), we get}$
$\displaystyle \overrightarrow{y}=6\widehat{i}-3\widehat{j}-6\widehat{k}-(t\widehat{i}+t\widehat{j}+t\widehat{k})=(6-t)\widehat{i}+(-3-t)\widehat{j}+(-6-t)\widehat{k}\ \text{...(3)}$
$\displaystyle \text{Since }\overrightarrow{y}\text{ is perpendicular to }\overrightarrow{b},$
$\displaystyle \overrightarrow{y}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow [(6-t)\widehat{i}+(-3-t)\widehat{j}+(-6-t)\widehat{k}]\cdot(\widehat{i}+\widehat{j}+\widehat{k})=0$
$\displaystyle \Rightarrow 1(6-t)+1(-3-t)+1(-6-t)=0$
$\displaystyle \Rightarrow -3-3t=0$
$\displaystyle \Rightarrow t=-1$
$\displaystyle \text{From (2) and (3), we get}$
$\displaystyle \overrightarrow{x}=-\widehat{i}-\widehat{j}-\widehat{k}$
$\displaystyle \overrightarrow{y}=7\widehat{i}-2\widehat{j}-5\widehat{k}$
$\displaystyle \text{So,}$
$\displaystyle \overrightarrow{a}=\overrightarrow{x}+\overrightarrow{y}=(-\widehat{i}-\widehat{j}-\widehat{k})+(7\widehat{i}-2\widehat{j}-5\widehat{k})$

$\displaystyle \textbf{Question 38: }~\text{Let }\overrightarrow{a}=5\widehat{i}-\widehat{j}+7\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}-\widehat{j}+\lambda\widehat{k}.\text{ Find }\lambda\text{ such that }\overrightarrow{a}+\overrightarrow{b}\text{ is orthogonal to }\overrightarrow{a}-\overrightarrow{b}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that}$
$\displaystyle \overrightarrow{a}=5\widehat{i}-\widehat{j}+7\widehat{k},\ \overrightarrow{b}=\widehat{i}-\widehat{j}+\lambda\widehat{k}$
$\displaystyle \therefore \overrightarrow{a}+\overrightarrow{b}=5\widehat{i}-\widehat{j}+7\widehat{k}+\widehat{i}-\widehat{j}+\lambda\widehat{k}=6\widehat{i}-2\widehat{j}+(7+\lambda)\widehat{k}$
$\displaystyle \text{and }\overrightarrow{a}-\overrightarrow{b}=5\widehat{i}-\widehat{j}+7\widehat{k}-(\widehat{i}-\widehat{j}+\lambda\widehat{k})=4\widehat{i}+0\widehat{j}+(7-\lambda)\widehat{k}$
$\displaystyle \text{Given that }\overrightarrow{a}+\overrightarrow{b}\text{ is orthogonal to }\overrightarrow{a}-\overrightarrow{b}$
$\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=0$
$\displaystyle \Rightarrow [6\widehat{i}-2\widehat{j}+(7+\lambda)\widehat{k}]\cdot[4\widehat{i}+0\widehat{j}+(7-\lambda)\widehat{k}]=0$
$\displaystyle \Rightarrow 24+0+(49-\lambda^2)=0$
$\displaystyle \Rightarrow \lambda^2=73$
$\displaystyle \Rightarrow \lambda=\sqrt{73}$

$\displaystyle \textbf{Question 39: }~\text{If }\overrightarrow{a}\cdot\overrightarrow{a}=0\text{ and }\overrightarrow{a}\cdot\overrightarrow{b}=0,\text{ what can you conclude about} \\ \text{the vector }\overrightarrow{b}? [\text{CBSE 2004}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\overrightarrow{a}\cdot\overrightarrow{a}=0$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2=0$
$\displaystyle \Rightarrow |\overrightarrow{a}|=0\ \text{...(1)}$
$\displaystyle \text{Also given that}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow |\overrightarrow{a}|\ |\overrightarrow{b}|\cos\theta=0\ \text{(Where }\theta\text{ is the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b})$
$\displaystyle \Rightarrow 0\cdot|\overrightarrow{b}|\cos\theta=0\ \text{[From (1)]}$
$\displaystyle \Rightarrow 0=0$
$\displaystyle \text{So, it means that for any vector }\overrightarrow{b},\ \text{the given equation }\overrightarrow{a}\cdot\overrightarrow{b}=0\text{ is satisfied}$

$\displaystyle \textbf{Question 40: }~\text{If }\overrightarrow{c}\text{ is perpendicular to both }\overrightarrow{a}\text{ and }\overrightarrow{b},\text{ then prove that it is} \\ \text{perpendicular to both }\overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{a}-\overrightarrow{b}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\overrightarrow{c}\text{ is perpendicular to both }\overrightarrow{a}\text{ and }\overrightarrow{b}$
$\displaystyle \Rightarrow \overrightarrow{c}\cdot\overrightarrow{a}=0\ \text{and}\ \overrightarrow{c}\cdot\overrightarrow{b}=0\ \text{...(1)}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{c}\cdot(\overrightarrow{a}+\overrightarrow{b})=\overrightarrow{c}\cdot\overrightarrow{a}+\overrightarrow{c}\cdot\overrightarrow{b}=0+0=0\ \text{[From (1)]}$
$\displaystyle \text{So, }\overrightarrow{c}\text{ is perpendicular to }\overrightarrow{a}+\overrightarrow{b}$
$\displaystyle \text{Again,}$
$\displaystyle \overrightarrow{c}\cdot(\overrightarrow{a}-\overrightarrow{b})=\overrightarrow{c}\cdot\overrightarrow{a}-\overrightarrow{c}\cdot\overrightarrow{b}=0-0=0\ \text{[From (1)]}$
$\displaystyle \text{So, }\overrightarrow{c}\text{ is perpendicular to }\overrightarrow{a}-\overrightarrow{b}$

$\displaystyle \textbf{Question 41: }~\text{If }\lvert\overrightarrow{a}\rvert=a\text{ and }\lvert\overrightarrow{b}\rvert=b,\text{ prove that }\left(\frac{\overrightarrow{a}}{a^{2}}-\frac{\overrightarrow{b}}{b^{2}}\right)^{2}=\left(\frac{\overrightarrow{a}-\overrightarrow{b}}{ab}\right)^{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \left(\frac{\overrightarrow{a}}{a^{2}}-\frac{\overrightarrow{b}}{b^{2}}\right)^{2}$
$\displaystyle =\left|\frac{\overrightarrow{a}}{a^{2}}\right|^{2}+\left|\frac{\overrightarrow{b}}{b^{2}}\right|^{2}-\frac{2\,\overrightarrow{a}\cdot\overrightarrow{b}}{a^{2}b^{2}}$
$\displaystyle =\frac{|\overrightarrow{a}|^{2}}{a^{4}}+\frac{|\overrightarrow{b}|^{2}}{b^{4}}-\frac{2\,\overrightarrow{a}\cdot\overrightarrow{b}}{a^{2}b^{2}}$
$\displaystyle =\frac{a^{2}}{a^{4}}+\frac{b^{2}}{b^{4}}-\frac{2\,\overrightarrow{a}\cdot\overrightarrow{b}}{a^{2}b^{2}}\ \text{(From the given information)}$
$\displaystyle =\frac{1}{a^{2}}+\frac{1}{b^{2}}-\frac{2\,\overrightarrow{a}\cdot\overrightarrow{b}}{a^{2}b^{2}}$
$\displaystyle =\frac{b^{2}+a^{2}-2\,\overrightarrow{a}\cdot\overrightarrow{b}}{a^{2}b^{2}}$
$\displaystyle =\frac{a^{2}+b^{2}-2\,\overrightarrow{a}\cdot\overrightarrow{b}}{a^{2}b^{2}}$
$\displaystyle =\frac{|\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}-2\,\overrightarrow{a}\cdot\overrightarrow{b}}{a^{2}b^{2}}\ \text{(From the given information)}$
$\displaystyle =\frac{(\overrightarrow{a}-\overrightarrow{b})^{2}}{a^{2}b^{2}}$
$\displaystyle =\left(\frac{\overrightarrow{a}-\overrightarrow{b}}{ab}\right)^{2}$

$\displaystyle \textbf{Question 42: }~\text{If }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are three non-coplanar vectors such that }\overrightarrow{d}\cdot\overrightarrow{a}=\overrightarrow{d}\cdot\overrightarrow{b}=\overrightarrow{d}\cdot\overrightarrow{c}=0,\text{ then show that }\overrightarrow{d}\text{ is the null vector.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\overrightarrow{d}\cdot\overrightarrow{a}=0$
$\displaystyle \text{So, either }\overrightarrow{d}=\overrightarrow{0}\text{ or }\overrightarrow{d}\perp\overrightarrow{a}$
$\displaystyle \text{Similarly, }\overrightarrow{d}\cdot\overrightarrow{b}=0$
$\displaystyle \text{So, either }\overrightarrow{d}=\overrightarrow{0}\text{ or }\overrightarrow{d}\perp\overrightarrow{b}$
$\displaystyle \text{Also, }\overrightarrow{d}\cdot\overrightarrow{c}=0$
$\displaystyle \text{So, either }\overrightarrow{d}=\overrightarrow{0}\text{ or }\overrightarrow{d}\perp\overrightarrow{c}$
$\displaystyle \text{But }\overrightarrow{d}\text{ cannot be perpendicular to }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ simultaneously since }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are non-coplanar}$
$\displaystyle \text{So, }\overrightarrow{d}=\overrightarrow{0}$
$\displaystyle \text{Hence, }\overrightarrow{d}\text{ is a null vector}$

$\displaystyle \textbf{Question 43: }~\text{If a vector }\overrightarrow{a}\text{ is perpendicular to two non-collinear vectors }\overrightarrow{b}\text{ and }\overrightarrow{c},\text{ then }\overrightarrow{a} \\ \text{ is perpendicular to every vector in the plane of }\overrightarrow{b}\text{ and }\overrightarrow{c}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\overrightarrow{a}\text{ is perpendicular to }\overrightarrow{b}\text{ and }\overrightarrow{c}$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}=0\ \text{and}\ \overrightarrow{a}\cdot\overrightarrow{c}=0\ \text{...(1)}$
$\displaystyle \text{Now, let }\overrightarrow{r}\text{ be any vector in the plane of }\overrightarrow{b}\text{ and }\overrightarrow{c}$
$\displaystyle \text{Then, }\overrightarrow{r}\text{ is the linear combination of }\overrightarrow{b}\text{ and }\overrightarrow{c}$
$\displaystyle \overrightarrow{r}=x\overrightarrow{b}+y\overrightarrow{c},\ \text{for some }x\text{ and }y$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{r}$
$\displaystyle =\overrightarrow{a}\cdot(x\overrightarrow{b}+y\overrightarrow{c})$
$\displaystyle =x(\overrightarrow{a}\cdot\overrightarrow{b})+y(\overrightarrow{a}\cdot\overrightarrow{c})$
$\displaystyle =x(0)+y(0)\ \text{[From (1)]}$
$\displaystyle =0$
$\displaystyle \text{Thus, }\overrightarrow{a}\text{ is perpendicular to }\overrightarrow{r}$
$\displaystyle \text{That is, }\overrightarrow{a}\text{ is perpendicular to every vector in the plane of }\overrightarrow{b}\text{ and }\overrightarrow{c}$

$\displaystyle \textbf{Question 44: }~\text{If }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0},\text{ show that the angle }\theta\text{ between the vectors }\\ \overrightarrow{b}\text{ and }\overrightarrow{c}\text{ is given by}\\ \cos\theta=\frac{\lvert\overrightarrow{a}\rvert^{2}-\lvert\overrightarrow{b}\rvert^{2}-\lvert\overrightarrow{c}\rvert^{2}}{2\lvert\overrightarrow{b}\rvert\lvert\overrightarrow{c}\rvert}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given,}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\displaystyle \Rightarrow \overrightarrow{b}+\overrightarrow{c}=-\overrightarrow{a}$
$\displaystyle \Rightarrow |\overrightarrow{b}+\overrightarrow{c}|^{2}=|-\overrightarrow{a}|^{2}$
$\displaystyle \Rightarrow |\overrightarrow{b}|^{2}+|\overrightarrow{c}|^{2}+2\,\overrightarrow{b}\cdot\overrightarrow{c}=|\overrightarrow{a}|^{2}$
$\displaystyle \Rightarrow 2\,\overrightarrow{b}\cdot\overrightarrow{c}=|\overrightarrow{a}|^{2}-|\overrightarrow{b}|^{2}-|\overrightarrow{c}|^{2}$
$\displaystyle \Rightarrow 2|\overrightarrow{b}|\ |\overrightarrow{c}|\cos\theta=|\overrightarrow{a}|^{2}-|\overrightarrow{b}|^{2}-|\overrightarrow{c}|^{2}$
$\displaystyle \therefore \cos\theta=\frac{|\overrightarrow{a}|^{2}-|\overrightarrow{b}|^{2}-|\overrightarrow{c}|^{2}}{2|\overrightarrow{b}|\ |\overrightarrow{c}|}$

$\displaystyle \textbf{Question 45: }~\text{Let }\overrightarrow{u},\overrightarrow{v}\text{ and }\overrightarrow{w}\text{ be vectors such that }\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\overrightarrow{0}.\\ \text{ If }\lvert\overrightarrow{u}\rvert=3,\ \lvert\overrightarrow{v}\rvert=4\text{ and }\lvert\overrightarrow{w}\rvert=5,\text{ then find }\overrightarrow{u}\cdot\overrightarrow{v}+\overrightarrow{v}\cdot\overrightarrow{w}+\overrightarrow{w}\cdot\overrightarrow{u}. [\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that}$
$\displaystyle \overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=0$
$\displaystyle \Rightarrow |\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}|=0$
$\displaystyle \Rightarrow |\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}|^{2}=0$
$\displaystyle \Rightarrow |\overrightarrow{u}|^{2}+|\overrightarrow{v}|^{2}+|\overrightarrow{w}|^{2}+2(\overrightarrow{u}\cdot\overrightarrow{v}+\overrightarrow{v}\cdot\overrightarrow{w}+\overrightarrow{w}\cdot\overrightarrow{u})=0$
$\displaystyle \Rightarrow 3^{2}+4^{2}+5^{2}+2(\overrightarrow{u}\cdot\overrightarrow{v}+\overrightarrow{v}\cdot\overrightarrow{w}+\overrightarrow{w}\cdot\overrightarrow{u})=0\ \text{(Given : }|\overrightarrow{u}|=3,|\overrightarrow{v}|=4\text{ and }|\overrightarrow{w}|=5)$
$\displaystyle \Rightarrow 9+16+25+2(\overrightarrow{u}\cdot\overrightarrow{v}+\overrightarrow{v}\cdot\overrightarrow{w}+\overrightarrow{w}\cdot\overrightarrow{u})=0$
$\displaystyle \Rightarrow 50+2(\overrightarrow{u}\cdot\overrightarrow{v}+\overrightarrow{v}\cdot\overrightarrow{w}+\overrightarrow{w}\cdot\overrightarrow{u})=0$
$\displaystyle \Rightarrow 2(\overrightarrow{u}\cdot\overrightarrow{v}+\overrightarrow{v}\cdot\overrightarrow{w}+\overrightarrow{w}\cdot\overrightarrow{u})=-50$
$\displaystyle \therefore \overrightarrow{u}\cdot\overrightarrow{v}+\overrightarrow{v}\cdot\overrightarrow{w}+\overrightarrow{w}\cdot\overrightarrow{u}=\frac{-50}{2}=-25$

$\displaystyle \textbf{Question 46: }~\text{Let }\overrightarrow{a}=x^{2}\widehat{i}+2\widehat{j}-2\widehat{k},\ \overrightarrow{b}=\widehat{i}-\widehat{j}+\widehat{k}\text{ and }\overrightarrow{c}=x^{2}\widehat{i}+5\widehat{j}-4\widehat{k} \\ \text{ be three vectors.Find the values of }x\text{ for which the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b} \\ \text{ is acute and the angle between }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ is obtuse.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have}$
$\displaystyle \overrightarrow{a}=x^{2}\widehat{i}+2\widehat{j}-2\widehat{k},\ \overrightarrow{b}=\widehat{i}-\widehat{j}+\widehat{k}\ \text{and}\ \overrightarrow{c}=x^{2}\widehat{i}+5\widehat{j}-4\widehat{k}$
$\displaystyle \text{Let }\theta_{1}\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ and }\theta_{2}\text{ be the angle between }\overrightarrow{b}\text{ and }\overrightarrow{c}$
$\displaystyle \text{Given that }\theta_{1}\text{ is acute and }\theta_{2}\text{ is obtuse}$
$\displaystyle \Rightarrow \cos\theta_{1}>0\ \text{and}\ \cos\theta_{2}<0$
$\displaystyle \Rightarrow \frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\ |\overrightarrow{b}|}>0\ \text{and}\ \frac{\overrightarrow{b}\cdot\overrightarrow{c}}{|\overrightarrow{b}|\ |\overrightarrow{c}|}<0$
$\displaystyle \Rightarrow \frac{x^{2}-4}{\sqrt{x^{4}+4+4}\,\sqrt{1+1+1}}>0\ \text{and}\ \frac{x^{2}-9}{\sqrt{1+1+1}\,\sqrt{x^{4}+25+16}}<0$
$\displaystyle \Rightarrow x^{2}-4>0\ \text{and}\ x^{2}-9<0$
$\displaystyle \Rightarrow x\in(-\infty,-2)\cup(2,\infty)\ \text{and}\ x\in(-3,3)$
$\displaystyle \Rightarrow x\in(-3,-2)\cup(2,3)$

$\displaystyle \textbf{Question 47: }~\text{Find the values of }x\text{ and }y\text{ if the vectors }\overrightarrow{a}=3\widehat{i}+x\widehat{j}-\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}+\widehat{j}+y\widehat{k}\text{ are mutually perpendicular vectors of equal magnitude.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have}$
$\displaystyle \overrightarrow{a}=3\widehat{i}+x\widehat{j}-\widehat{k}\ \text{and}\ \overrightarrow{b}=2\widehat{i}+\widehat{j}+y\widehat{k}$
$\displaystyle \text{It is given that the vectors are perpendicular}$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow 3(2)+x(1)+(-1)(y)=0$
$\displaystyle \Rightarrow 6+x-y=0$
$\displaystyle \Rightarrow x-y=-6\ \text{...(1)}$
$\displaystyle \text{Also, it is given that}$
$\displaystyle |\overrightarrow{a}|=|\overrightarrow{b}|$
$\displaystyle \Rightarrow \sqrt{9+x^{2}+1}=\sqrt{4+1+y^{2}}$
$\displaystyle \Rightarrow \sqrt{10+x^{2}}=\sqrt{5+y^{2}}$
$\displaystyle \Rightarrow 10+x^{2}=5+y^{2}$
$\displaystyle \Rightarrow x^{2}-y^{2}=-5$
$\displaystyle \Rightarrow (x+y)(x-y)=-5$
$\displaystyle \Rightarrow -6(x+y)=-5\ \text{[Using (1)]}$
$\displaystyle \Rightarrow x+y=\frac{5}{6}\ \text{...(2)}$
$\displaystyle \text{(1) + (2) gives}$
$\displaystyle 2x=-\frac{31}{6}$
$\displaystyle \Rightarrow x=-\frac{31}{12}$
$\displaystyle \text{From (1),}$
$\displaystyle -\frac{31}{12}-y=-6$
$\displaystyle \Rightarrow y=-\frac{31}{12}+6=\frac{41}{12}$
$\displaystyle \therefore x=-\frac{31}{12}\ \text{and}\ y=\frac{41}{12}$

$\displaystyle \textbf{Question 48: }~\text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are two non-collinear unit vectors such that } \\ \lvert\overrightarrow{a}+\overrightarrow{b}\rvert=\sqrt{3},\text{ find }(2\overrightarrow{a}-5\overrightarrow{b})\cdot(3\overrightarrow{a}+\overrightarrow{b}).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have}$
$\displaystyle |\overrightarrow{a}+\overrightarrow{b}|=\sqrt{3}$
$\displaystyle \text{Squaring both sides, we get}$
$\displaystyle |\overrightarrow{a}+\overrightarrow{b}|^{2}=3$
$\displaystyle \Rightarrow |\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}=3$
$\displaystyle \Rightarrow 1+1+2\overrightarrow{a}\cdot\overrightarrow{b}=3\ \text{(Because }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are unit vectors)}$
$\displaystyle \Rightarrow 2+2\overrightarrow{a}\cdot\overrightarrow{b}=3$
$\displaystyle \Rightarrow 2\overrightarrow{a}\cdot\overrightarrow{b}=1$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}=\frac{1}{2}\ \text{...(1)}$
$\displaystyle \text{Now,}$
$\displaystyle (2\overrightarrow{a}-5\overrightarrow{b})\cdot(3\overrightarrow{a}+\overrightarrow{b})$
$\displaystyle =6|\overrightarrow{a}|^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}-15\overrightarrow{b}\cdot\overrightarrow{a}-5|\overrightarrow{b}|^{2}$
$\displaystyle =6|\overrightarrow{a}|^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}-15\overrightarrow{a}\cdot\overrightarrow{b}-5|\overrightarrow{b}|^{2}\ \text{(}\overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{b}\cdot\overrightarrow{a}\text{)}$
$\displaystyle =6|\overrightarrow{a}|^{2}-13\overrightarrow{a}\cdot\overrightarrow{b}-5|\overrightarrow{b}|^{2}$
$\displaystyle =6(1)-13\left(\frac{1}{2}\right)-5(1)\ \text{[From (1)]}$
$\displaystyle =1-\frac{13}{2}$
$\displaystyle =-\frac{11}{2}$

$\displaystyle \textbf{Question 49: }~\text{If }\overrightarrow{a},\overrightarrow{b}\text{ are two vectors such that }\left|\overrightarrow{a}+\overrightarrow{b}\right|=\left|\overrightarrow{b}\right|, \\ \text{ then prove that }\overrightarrow{a}+2\overrightarrow{b}\text{ is perpendicular to }\overrightarrow{a}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that}$
$\displaystyle |\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{b}|$
$\displaystyle \text{Squaring both sides, we get}$
$\displaystyle |\overrightarrow{a}+\overrightarrow{b}|^{2}=|\overrightarrow{b}|^{2}$
$\displaystyle \Rightarrow |\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{b}|^{2}$
$\displaystyle \Rightarrow |\overrightarrow{a}|^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}=0\ \text{...(1)}$
$\displaystyle \text{Now,}$
$\displaystyle (\overrightarrow{a}+2\overrightarrow{b})\cdot\overrightarrow{a}$
$\displaystyle =\overrightarrow{a}\cdot\overrightarrow{a}+2\overrightarrow{b}\cdot\overrightarrow{a}$
$\displaystyle =|\overrightarrow{a}|^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle =0\ \text{[Using (1)]}$
$\displaystyle \text{So, }\overrightarrow{a}+2\overrightarrow{b}\text{ is perpendicular to }\overrightarrow{a}$