Question

The direction ratio of normal of a plane passing through two points $(0,-1,0)$ & $(0,0,1)$ and makes an angle $\pi /4$ with the plane $y-z=5$ are

• A. $(2,-\sqrt{2},\sqrt{2})$
• B. $(\sqrt{2},2,1)$
• C. $(\sqrt{2},1,1)$
• D. $(\sqrt{2},-1,1)$

Answer 1

Answer: (D)

$(\sqrt{2},-1,1)$

Let equation of plane through $(0,-1,0)$ & $(0,0,1)$ is $ax+by+cz+d=0$

Substituting, $d-b=0$

& $d+c=0$

$\Rightarrow c+b=0⟶\left(1\right)$

$\overrightarrow{n}=(a,b,c)$ is unit vector normal to plane

$=\sqrt{a^2+b^2+c^2}=1$

$\because$ This plane make $\pi /4$ with $y-z=5$

$\therefore$ $\frac{1}{\sqrt{2}}=\frac{(a,b,c).(0,1,-1)}{1.\sqrt{2}}$

$\Rightarrow b-c=1⟶\left(2\right)$

Solving $\left(1\right)$ & $\left(2\right)$, $2b=1$

$\Rightarrow b=1/2$

$\therefore$ $c=-1/2$

and $a^2+\frac{1}{4}+\frac{1}{4}=1$

$\Rightarrow a^2=\frac{1}{2}\Rightarrow a=\pm \frac{1}{\sqrt{2}}$

$\therefore$ $dr$ of normal are $\pm \frac{1}{\sqrt{2}},\frac{1}{2},\frac{-1}{2}$

$=(\pm \sqrt{2},1,-1)$ [D]