Question

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

• A. $(1,1,\sqrt{2})$
• B. $(1,1,-\frac{1}{\sqrt{2}})$
• C. $(1,1,\frac{1}{\sqrt{2}})$
• D. $(1,1,0)$

Answer 1

The correct option is A $(1,1,\sqrt{2})$

The plane's equation in the intercept form is
$\frac{x}{1}+\frac{y}{1}+\frac{z}{c}=1$
So, direction ratios of the normal are $(1,1,\frac{1}{c})$ and that of the given plane are $(1,1,0)$
$\Rightarrow \cos \frac{\pi }{4}=\frac{1\cdot 1+1\cdot 1+\frac{1}{c}\cdot 0}{\sqrt{1+1}\cdot \sqrt{1/c^2+1+1}}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}\cdot \sqrt{1/c^2+1+1}}$
$\Rightarrow \frac{1}{c^2}+2=4\Rightarrow c=\pm \frac{1}{\sqrt{2}}$

Therefore, direction ratios of the normal to the plane is $(1,1,\pm \sqrt{2})$