Find the distance between the line $x=-2+t$, $y=1+2t$, $z=1-t$, and the plane $x+2y+z=5$
I answered with zero, because the line and the plane are not parallel but my professor told me I was wrong. Am I?
My exact answer was:
From the line equation:
$V=\langle 1,2,-1\rangle$ (direction vector)
From the plane equation:
$n=\langle 1,2,1 \rangle$ (normal vector)
$ n \cdot v= (1\cdot 1)+(2\cdot 2)+(1\cdot(-1))=1+4-1=4$ (dot product)
$|n|=\sqrt{1+4+1}=\sqrt{6}$
$|v|=\sqrt{1+4+1}=\sqrt{6}$
$n\cdot v=|n||v|\cos \vartheta$
$θ=\arccos (n\cdot v/(|n||v|))=\arccos (4/6)=48.189$ degrees
The line and the plane are not parallel and so, they must intersect at some point, therefore the distance between the line and the plane is zero.