Three Dimensional Geometry — Class 12 Maths Solution

Given equation of the line is $\frac{{x - 3}}{2} = \frac{{y - 3}}{1} = \frac{z}{1} = \lambda$
…….(i)

So, DR's of the line are 2,1,1 and DC's of the given line are $\frac{2}{{\sqrt 6 }},\frac{1}{{\sqrt 6 }},\frac{1}{{\sqrt {16} }}$.

Also, the required lines make angle $\frac{\pi }{3}$ with the given line.

From Eq. (i), $x = (2\lambda + 3),y = (\lambda + 3)$ and $z = \lambda$

$\therefore$ $\cos \frac{\pi }{3}$

$= \frac{{(4\lambda + 6) + (\lambda + 3) + (\lambda )}}{{\sqrt 6 \sqrt {{{(2\lambda + 3)}^2} + {{(\lambda + 3)}^2} + {\lambda ^2}} }}$

$\Rightarrow$ $\frac{1}{2} = \frac{{6\lambda + 9}}{{\sqrt 6 \sqrt {\left( {4{\lambda ^2} + 9 + 12\lambda + {\lambda ^2} + 9 + 6\lambda + {\lambda ^2}} \right)} }}$

$\Rightarrow$ $\frac{{\sqrt 6 }}{2} = \frac{{6\lambda + 9}}{{\sqrt {6{\lambda ^2} + 18\lambda + 18} }}$

$\Rightarrow$ $6\sqrt {\left( {{\lambda ^2} + 3\lambda + 3} \right)} = 2(6\lambda + 9)$

$\Rightarrow$ $36\left( {{\lambda ^2} + 3\lambda + 3} \right) = 36\left( {4{\lambda ^2} + 9 + 12\lambda } \right)$
$\Rightarrow$ ${\lambda ^2} + 3\lambda + 3 = 4{\lambda ^2} + 9 + 12\lambda$

$\Rightarrow$ $3{\lambda ^2} + 9\lambda + 6 = 0$

$\Rightarrow$ ${\lambda ^2} + 3\lambda + 2 = 0$

$\Rightarrow$ $\lambda (\lambda + 2) + 1(\lambda + 2) = 0$

$\Rightarrow$ $(\lambda + 1)(\lambda + 2) = 0$

$\therefore$ $\lambda = - 1, - 2$

So, the DC's are 1,2,-1 and -1,1,-2 .
Also, both the required lines passes through origin.

So, the equations of required lines are $\frac{x}{1} = \frac{y}{2} = \frac{z}{{ - 1}}$ and $\frac{x}{{ - 1}} = \frac{y}{1} = \frac{z}{{ - 2}}$.