Question

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

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

Answer 1

Answer: (B)

$\sqrt{2}x+y+z=2$

Let $\overrightarrow{n}$ be inclined at $\alpha ,\beta ,\gamma$ angles to $x,y,z$ axis.

$\begin{array}{c}\alpha ={45}^{\circ }\\ \beta =60^{\circ }\end{array}$

$\begin{array}{c}\cos \alpha =\frac{1}{\sqrt{2}}\\ \cos \beta =\frac{1}{2}\\ \cos \gamma =?\\ {\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1\\ \frac{1}{2}+\frac{1}{4}+{\cos }^2\gamma =1\\ {\cos }^2\gamma =\frac{1}{4}\\ \gamma ={60}^{\circ }\end{array}$

$\overrightarrow{n}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}$

Equation of plane :

$\begin{array}{c}\frac{1}{\sqrt{2}}\left(x-\sqrt{2}\right)+\frac{1}{2}\left(y+1\right)+\frac{1}{2}\left(z-1\right)=0\\ \sqrt{2}\left(x-\sqrt{2}\right)+y+1+z-1=0\\ \sqrt{2}x+y+z=2\end{array}$

Hence, this is the answer.