Question

The direction ratios of a normal to the plane through $(1,0,0),(0,1,0),$ which makes an angle of $\frac{\pi }{4}$ with the plane $x+y=3$ are

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

Answer 1

The correct option is B $1,1,\sqrt{2}$

Any plane through $(1,0,0)$ is

$A(x-1)+B(y-0)+C(z-0)=0$ ...$\left(1\right)$

It contains $(0,1,0)$ if $-A+B=0$ ....$\left(2\right)$

Also, $\left(1\right)$ makes an angle of $\frac{\pi }{4}$ with the plane $x+y=3$

Therefore, $\cos \frac{\pi }{4}=\frac{|A+B|}{\sqrt{A^2+B^2+C^2}\sqrt{1^2+1^2}}$

$\Rightarrow {(A+B)}^2=A^2+B^2+C^2\Rightarrow 2AB=C^2$ ....$\left(3\right)$

From (2) and (3), $C^2=2A^2\Rightarrow C=\pm \sqrt{2}A$

Hence, $A:B:C::A:A:\pm \sqrt{2}A$

$\therefore$ direction ratios are $1:1:\pm \sqrt{2}$