Question

Find the coordinates of the point, where the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$ intersects the plane $x-y+z-5=0$. Also find the angle between the line the plane.

Answer 1

Let $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda$

Where, $x=3\lambda +2;y=4\lambda -1;z=2\lambda +2$

The point lies on the plane $x-y+z=5$
$(3\lambda +2)-(4\lambda -1)+(2\lambda +2)-5=0$

$3\lambda +2-4\lambda +1+2\lambda +2-5=0$

$\lambda =0$

Therefore, Point of intersection $=[2,-1,2]$
$\sin \theta =\frac{\left(3\right)\left(1\right)+4(-1)+2\left(1\right)}{\sqrt{9+16+4}\sqrt{1+1+1}}$

$\sin \theta =\frac{3-4+2}{\sqrt{29}\sqrt{3}}$

$\sin \theta =\frac{1}{\sqrt{87}}$

$\theta ={\sin }^{-1}\frac{1}{\sqrt{87}}$