Vector Algebra — Class 12 Maths Solution

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

$\therefore$ $(3\widehat {\rm{i}} - 6\widehat {\rm{j}} + \widehat {\rm{k}}) \cdot (2\widehat {\rm{i}} - 4\widehat {\rm{j}} + \lambda \widehat {\rm{k}})$

$= |3\widehat {\rm{i}} - 6\widehat {\rm{j}} + \widehat {\rm{k}}| \cdot |2\widehat {\rm{i}} - 4\widehat {\rm{j}} + \lambda \widehat {\rm{k}}|$

$\Rightarrow$ $6 + 24 + \lambda = \sqrt {9 + 36 + 1} \sqrt {4 + 16 + {\lambda ^2}}$

$\Rightarrow$ $30 + \lambda = \sqrt {46} \sqrt {20 + {\lambda ^2}}$

$\Rightarrow$ $900 + {\lambda ^2} + 60\lambda = 46\left( {20 + {\lambda ^2}} \right)\quad$ [on squaring both sides]

$\Rightarrow$ ${\lambda ^2} + 60\lambda - 46{\lambda ^2} = 920 - 900$
$\Rightarrow$ $- 45{\lambda ^2} + 60\lambda - 20 = 0$

$\Rightarrow$ $- 45{\lambda ^2} + 30\lambda + 30\lambda - 20 = 0$

$\Rightarrow$ $- 15\lambda (3\lambda - 2) + 10(3\lambda - 2) = 0$

$\Rightarrow$ $(10 - 15\lambda )(3\lambda - 2) = 0$

$\therefore$ $\lambda = \frac{2}{3},\frac{2}{3}$

Alternate Method
Let $\overrightarrow {\rm{a}} = 3\widehat {\rm{i}} - 6\widehat {\rm{j}} + \widehat {\rm{k}}$

and $\overrightarrow {\rm{b}} = 2\widehat {\rm{i}} - 4\widehat {\rm{j}} + \lambda \widehat {\rm{k}}$

Since,
$\Rightarrow$ $\quad \frac{3}{2} = \frac{{ - 6}}{{ - 4}} = \frac{1}{\lambda } \Rightarrow \lambda = \frac{2}{3}$