Finding the angle between a line and plane exercise

Question

The aim is to find the angle between a line and plane and if they intersect then the point where they do that.

given: $$ \frac{x+1}{2}=\frac{y-3}{4}=\frac{z}{3}, 3x-3y+2z-5=0 $$

What I have found out:

s=(2; 4; 3) n=(3; -3; 2) $$|s*n|=|3*2+(-3)*4+2*3|=6-12+6=0 $$ $$|s|=\sqrt{29}$$ $$|n|=\sqrt{22}$$

$$ sin=\frac{0}{\sqrt{29}\sqrt{22}} $$

angle is 0 degrees.

Solution

The angle between them is 0, this means they are parallel.

So I take the point P that is on the line

$$ P(-1,3,0) $$ and replace it in the plane equation:

$$ 3*(-1)-3*3+2*0-5=-17$$

-17 is not equal to 0

they don't intersect.

Answer 1

What you calculated is not the angle between the line and the plane, but between (a direction vector of) the line and the normal (vector) of the plane, which is a vector perpendicular to the plane.

What I have found out:

s=(2; 4; 3) n=(3; -3; 2) $$\mathbf{s}\cdot\mathbf{n}=3*2+(-3)*4+2*3=6-12+6=0 $$

This dot product is zero, so the line is perpendicular to the normal, not to the plane. But that means the line and the plane aren't perpendicular, but parallel!

That leaves two options:

• they do not intersect, so they have no points in common;
• the line lies within the plane, so they have all points in common.

It is easy to see that $(-1,3,0)$ is a point on the line; does it lie in the plane?