Find the point at which the line intersects the plane. Is the intersection perpendicular?

Question

Find the point at which the line $$x = 1 - t \\ y = 3 + t \\ z = 7 + 2t \\$$ intersects the plane $$x + 2y + z = 20$$ Is the intersection perpendicular? I have found the point of intersection to be (-1, 5, 11) but I cannot figure out if the intersection is perpendicular. I believe I need to take the cross or dot product of something.

Answer 1

Point of intersection: $P(1-t,3+t,7+2t)$. This point is on the plane, thus: $1-t+2(3+t)+7+2t=20\to 3t+14=20 \to t = 2 \to P=(-1,5,11)$ The normal line to the plane is $N = (1,2,1)$, and the the line is perpendicular to the plane if it is parallel to $N$, and we have $l \parallel N \iff (-1,1,2) = k(1,2,1)$ for some real $k$, but this equation has no solution in $k$. Thus the line is not perpendicular to the plane.

Answer 2

Tangent vector to the line will be given by:

$$\frac{d}{dt}\vec{r}=\frac{d}{dt}<x(t),y(t),z(t)>=<-1,1,2>$$

A vector normal to the plane is $\vec{n}=<1,2,1>$

$$\frac{d}{dt}\vec{r}\times\vec{n}=<-3,3,-3>$$

More importantly the cross product above is nonzero and so the vector tangent to the line is not co linear to the vector which is normal to the plane and so the intersection is not perpendicular.