The problem I am trying to solve is below: What is the angle formed by the line $(1,2,0) + t(-1,2,1)$ and a normal vector of the plane $x+y-z = 4?$ Give your answer in degrees.
I am having a little bit of trouble solving this problem. I have watched videos and looked at websites, as well as books I have, however, all of them aren't very helpful. I have also tried drawing a diagram, but I don't think I did it correctly. Any advice/answers?
On
For the line $\ell=\mathbf{a}+\mathbf{u}t$, the line is passing through the point $\mathbf{a}$ and has direction vector $\mathbf{u}$.
For the plane given $x+y-z=4$, the normal vector is $\mathbf{n}=(1,1,-1)$. So the angle between the normal and the direction vector of the line can be computed using $\mathbf{u} \cdot \mathbf{n}=\|\mathbf{n}\| \, \|\mathbf{u}\| \cos \theta.$
So in your case, $\mathbf{u}=(-1,2,1)$ and $\mathbf{n}=(1,1,-1)$ gives $$\cos \theta =\frac{\mathbf{u} \cdot \mathbf{n}}{\|\mathbf{n}\| \, \|\mathbf{u}\|}=0.$$
HINT
\begin{align*} \langle x,y\rangle = \|x\|\|y\|\cos(\theta) \end{align*}