3d vector algebra question

Question

All the x,y,z of the black dots are known variables.

I have a line, as shown in the diagram, that has a $16.5^\circ$ angle relative to the $XY$ plane.

How can I get the yellow $(x,y,z)$ as shown?

Answer 1

$P= (x,y,z)$ is on the line that connects $P_1 = (121,53, 0) $ with $P_2 = (118, 41,6) $. We want to locate the point $P$ such that the angle between $\vec{P_3 P} $ and the $z$-axis is $90^\circ - 16.5^\circ = 73.5^\circ$, where $P_3 = (109, 41, 0) $. Now point $P$ is given by

$ P = (1 - t) P_1 + t P_2 = (1 - t) (121,53, 0) + t (118, 41, 6) = (121,53, 0) + t (-3, -12, 6 ) $

Hence, the equation we want to solve is

$ \cos 73.5^\circ = \dfrac{ \mathbf{k} \cdot (P - P_3) } { \| P - P_3 \| } $

Now, $P - P_3 = (121,53,0) + t (-3, -12, 6) - (109, 41, 0) = (12, 12, 0) + t (-3, -12, 6) $

So the above equation reduces to

$ \cos 73.5^\circ = \dfrac{ 6 t }{\sqrt{ (12 - 3 t)^2 + (12 - 12 t)^2 + (6 t)^2 }} $

From which,

$\cos^2 73.5^\circ ( 288 - 360 t + 189 t^2 ) = 36 t^2 $

Solving, we get

$ t = 0.568783919 $

The other value of $t$ is extraneous.

With this, point $P$ is found by substituting the found $t$ into the above expression for $P$, and this gives us

$P = (121 - 3 t, 53- 12 t, 6 t) = (119.2936, 46.17459, 3.412704) $

Answer 2

Calling

$$ \cases{ p=(x,y,z)\\ p_1= (121,53,0)\\ p_2=(118,41,0)\\ p_3=(109,41,0) } $$

we have

$$ \cases{ \frac{6}{z}-\frac{53-41}{53-y}=0\\ \frac{6}{z}-\frac{121-118}{121-x}=0\\ \frac{z}{\|p-p_3\|}-\sin(16.5^{\circ})=0 } $$

solving for $(x,y,z)$ we got

$$ p = (119.294,46.175,3.413) $$