Question

Angle between the line $\frac{x+1}{1}=\frac{y}{2}=\frac{z-1}{1}$ and a normal to the plane $x-y+z=0$, is

• A. $0^o$
• B. ${30}^o$
• C. ${45}^o$
• D. ${90}^o$

Answer 1

The correct option is D ${90}^o$

$\frac{z+1}{1}=\frac{y}{2}=\frac{z-1}{1}$

$x-y+z=0$

$\sin \theta =\frac{\left(1\right)\left(1\right)+\left(2\right)(-1)+1\left(1\right)}{\sqrt{3}\sqrt{6}}$

$=0$

$\theta =0,\pi$.