Question

A vector $\vec{n}$ is inclined to $x$-axis at $45^{\circ }$, to $y$-axis at $60^{\circ }$ and at an acute angle to $z$-axis. If $\vec{n}$ is a normal to a plane passing through the point $(\sqrt{2},-1,1)$, then the equation of plane is

• A. $3\sqrt{2}x-4y-3z=7$
• B. $4\sqrt{2}x+7y+z=2$
• C. $\sqrt{2}x+y+z=2$
• D. $\sqrt{2}x-y-z=2$

Answer 1

Answer: (C)

$\vec{n}=$ normal vector $=<\cos 45^{\circ },\cos 60^{\circ },\cos \theta >$
where $\frac{1}{2}+\frac{1}{4}+{\cos }^2\theta =1$
$\Rightarrow \cos \theta =\frac{1}{2}$
$\Rightarrow \theta =60^{\circ }$ So, equation of plane is
$\frac{(x-\sqrt{2})}{\sqrt{2}}+\frac{(y+1)}{2}+\frac{z-1}{2}=0$
$\Rightarrow \frac{x}{\sqrt{2}}+\frac{y}{2}+\frac{z}{2}=1$
$\Rightarrow \sqrt{2}x+y+z=2$