Question

Find the angle at which the normal vector to the plane $4x+8y+z=5$ is inclined to the coordinate axes.

Answer 1

$4x+8y+z=5$

The d r's of normal to plane $=<4,8,1>$

d c's $=<\frac{4}{\sqrt{4^2+8^2+1}},\frac{8}{\sqrt{4^2+8^2+1}},\frac{1}{\sqrt{4^2+8^2+1}}>$

$=<\frac{4}{9},\frac{8}{9},\frac{1}{9}>$

Angle with $X-axis={\cos }^{-1}\frac{4}{9}$

$Y-axis={\cos }^{-1}\frac{8}{9}$

$Z-axis={\cos }^{-1}\frac{1}{9}$