 # Suppose a cable car moves with direction vector vec(d) = ((2),(-2),(1)) 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 (|vec(a) * vec(b)|)/(|vec(a) vec(b)|). Using this, the angle I find between vec(d) and ((2),(0),(0)) The angle I find is 48.2^(circ) moiraudjpdn 2022-08-12 Answered
Suppose a cable car moves with direction vector $\stackrel{\to }{d}$
$\left(\begin{array}{c}2\\ -2\\ 1\end{array}\right)$
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
$\frac{|\stackrel{\to }{a}\cdot \stackrel{\to }{b}|}{|\stackrel{\to }{a}\stackrel{\to }{b}|}$
) Using this, the angle I find between $\stackrel{\to }{d}$and
$\left(\begin{array}{c}2\\ 0\\ 0\end{array}\right)$
The angle I find is ${48.2}^{\circ }$ . If I use the z-vector instead:
$\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$
The angle I find is ${70.5}^{\circ }$
The answer is $\approx {19.5}^{\circ }$. How would I get this value?
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You have to find the angle between $\stackrel{\to }{d}$ and the projection of $\stackrel{\to }{d}$ onto the horizontal plane, which is the vector $\stackrel{\to }{p}=\left(2,-2,0\right)$. Try to apply the same equation with those two vectors.
###### Not exactly what you’re looking for? Landon Wolf
Horizontal' here means the x,y plane.
If you sketch a picture, it could be clear that the angle of a vector and a plane and the angle with a normal vector of the plane adds up to ${90}^{\circ }$
So, since you already found the angle with the z -axis to be ${70.5}^{\circ }$ , the angle to the horizontal will be ${90}^{\circ }-{70.5}^{\circ }={19.5}^{\circ }$