Question

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

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

Answer 1

The correct option is B $\sqrt{2}x+y+z=2$

$\alpha ={45}^{\circ }\beta ={60}^{\circ }\gamma =?{\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1\frac{1}{2}+\frac{1}{4}+{\cos }^2\gamma =1{\cos }^2\gamma =\frac{1}{4}\cos \gamma =\frac{1}{2}$

$Direction\cos ineofnormal\overrightarrow{n}\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{1}{2}\right)\frac{1}{\sqrt{2}}\left(x-\sqrt{2}\right)+\frac{1}{2}\left(y+1\right)+\frac{1}{2}\left(z-1\right)=0\sqrt{2}\left(x-\sqrt{2}\right)+\left(y+1\right)+\left(z-1\right)=0\sqrt{2}x+y+z=2$

$Hence,theoptionBisthecorrectanswer.$