Consider the plane defined by equation $3x+4y-z=2$ and a line defined by the following vector equation (in parametric form)
$x(t)=2-2t$, $y(t)=-1+3t$, $z(t)=-t$
(a) Find the point where the line intersects the plane.
(b) Find the normal vector to the plane.
(c) Find the angle at which the line intersects the plane (Hint: Use dot product)
(a) I have found the point of intersection at $(2,-1,0)$ by substituting the parametric vector equation into the equation of the plane.
(b) Normal is $(3,4,-1)$
(c) I'm a little stumped here. Do I use this formula $a.b=|a||b|\cos\theta$ to solve for the angle?