Question

If a line which passes through the point $A(0,1,2)$ and makes angle $\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{2}$ with $x,y,\&z$ axes respectively. The line meets the plane $x+y+z=0$ at point $B$. The length $\sqrt{2}AB$ is equal to

• A. $3$
• B. $-3$
• C. $4$
• D. $3\sqrt{2}$

Answer 1

The correct option is A $3$

The D.C. of the line are $\left(\begin{array}{c}\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\end{array}\right)$
any point on the line at a distance $\lambda$ from $A(0,1,2)$
is $\left(\begin{array}{c}0+\frac{\lambda }{\sqrt{2}},1+\frac{\lambda }{\sqrt{2}},2+\lambda .0\end{array}\right)$
which lies on $x+y+z=0$
$\therefore \frac{\lambda }{\sqrt{2}}+\left(\begin{array}{c}1+\frac{\lambda }{\sqrt{2}}\end{array}\right)+2=0$
$\Rightarrow \frac{2\lambda }{\sqrt{2}}=-3\Rightarrow \lambda =\frac{-3}{\sqrt{2}}$
$\therefore B=\left(\begin{array}{c}-\frac{3}{2},-\frac{1}{2},2\end{array}\right)$
$A=(0,1,2)\&B\equiv \left(\begin{array}{c}-\frac{3}{2},-\frac{1}{2},2\end{array}\right)$
$AB=\sqrt{\frac{9}{4}+\frac{9}{4}}=\frac{3\sqrt{2}}{2}=\frac{3}{\sqrt{2}}$