Find a point on a line that creates a perpendicular in 3D space [closed]
Solution 1:
Since point $D$ will be on $AC$, then there exists a scalar $t \in \mathbb{R} $ such that
$ D = A + t (C - A) \hspace{20pt}(1)$
The vector $(C-A)$ is the direction vector of the ray $AC$. Now we want want $BD = D - B $ to be perpendicular to $(C - A)$. Then using dot product we must have
$ [(A - B) + t (C - A)] \cdot (C - A) = 0 \hspace{20pt} (2) $
From which, $ t = \dfrac{ (B - A) \cdot (C - A) }{ (C - A) \cdot (C - A) } \hspace{20pt}(3) $
Using $t$ from $(3)$ into $(1)$ gives the point $D$.