Question

The value of $\lambda$ for which the vectors $3\hat{i}-6\hat{j}+\hat{k}and2\hat{i}-4\hat{j}+\lambda \hat{k}$ are parallel,is

(a) $\frac{2}{3}$ (b) $\frac{3}{2}$ (c) $\frac{5}{2}$ (d) $\frac{2}{5}$

Answer 1

(a) Since,two vectors are parallel i.e.,angle between them is zero.

$\therefore (3\hat{i}-6\hat{j}+\hat{k}).(2\hat{i}-4\hat{j}+\lambda \hat{k})=|3\hat{i}-6\hat{j}+\hat{k}|.|2\hat{i}-4\hat{j}+\lambda \hat{k}|[\because \overrightarrow{a}.\overrightarrow{b}=|a||b|cos{0}^0\Rightarrow \overrightarrow{a}.\overrightarrow{b}=|\overrightarrow{a}|\overrightarrow{b}|]\Rightarrow 6+24+\lambda =\sqrt{9+36+1}\sqrt{4+16+{\lambda }^2}\Rightarrow 30+\lambda =\sqrt{46}\sqrt{20+{\lambda }^2}\Rightarrow 900+{\lambda }^2+60\lambda =46(20+{\lambda }^2)\left[onsquaringbothsides\right]\Rightarrow {\lambda }^2+60\lambda -46{\lambda }^2=920-900\Rightarrow -45{\lambda }^2+60\lambda -20=0\Rightarrow -45{\lambda }^2+30\lambda +30\lambda -20=0\Rightarrow -15\lambda (3\lambda -2)+10(3\lambda -2)=0\Rightarrow (10-15\lambda )(3\lambda -2)=0\therefore \lambda =\frac{2}{3},\frac{2}{3}\text{Alternate~Method}Let\overrightarrow{a}=3\hat{i}-6\hat{j}+\hat{k}and\overrightarrow{b}=2\hat{i}-4\hat{j}+\lambda \hat{k}Since,\overrightarrow{a}||\overrightarrow{b}\Rightarrow \frac{3}{2}=\frac{-6}{-4}=\frac{1}{\lambda }\Rightarrow \lambda =\frac{2}{3}$