Suppose a cable car moves with direction vector $\vec{d}$ = $$\begin{pmatrix} 2\\ -2\\ 1\\ \end{pmatrix} $$ The question asks, "at what angle to the horizontal does the cable car travel?". By horizontal, if you assume the x-axis, the angle can be found using $$\frac{|\vec{a}\cdot\vec{b}|}{|\vec{a}\vec{b}|}$$) Using this, the angle I find between $\vec{d}$ and $$\begin{pmatrix} 2\\ 0\\ 0\\ \end{pmatrix} $$ The angle I find is $48.2^\circ$. If I use the z-vector instead: $$\begin{pmatrix} 0\\ 0\\ 1\\ \end{pmatrix}$$ The angle I find is $70.5^\circ$
The answer is $\approx 19.5^\circ$. How would I get this value?
Additional information: The cable car starts at $(10,3,0)$ and it moves in the direction of $\vec{d}$ at a speed of $4.5 \;m\,s^{-1}$
You have to find the angle between $\vec{d}$ and the projection of $\vec{d}$ onto the horizontal plane, which is the vector $\vec{p}=(2,-2,0)$. Try to apply the same equation with those two vectors.