The direction ration of a normal to the plane passing through (1, 0, 0), (0, 1, 0) and making an angle $\frac{\pi}{4}$ with the plane x + y = 3 are :
Answer & explanation
Correct answer: option 2
Let the plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
$\Rightarrow \frac{1}{a}=1, \frac{1}{b}=1$
$\Rightarrow a=b=1$
Also, $\sin \frac{\pi}{4}=\frac{\left|\frac{1}{a}+\frac{1}{b}\right|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}} \sqrt{1+1}}$
$\Rightarrow c= \pm \frac{1}{\sqrt{2}}$
Thus direction rations are $(1,1, \sqrt{2})$ or $(1,1,-\sqrt{2})$