I have A(1,2,3) , B(-1,0,1), C(1,-1,1) which are points in $\mathbb{R}^3$. I'm trying to find another point H such that AH${\parallel}$AC and BH${\perp}$AC.
I set H = ($h_1$, $h_2$, $h_3$), and took the dot product of BH and AC set to $0$. This gave me an equation in terms of $h_2$ and $h_3$.
I then took the cross product of AH and AC set to $<0,0,0>$. I was going to have two equations and two unknowns and then solve for them. However, the cross product gave me two values for $h_1$, making it impossible to solve for the other equations.
Without giving me the answer, would someone just be able to explain why this didn't work? My prof ambiguously mentioned to extend the line, but I have no idea where to go with that. Does he mean potentially set up an equation with more unknown values and then subtract two vectors because I know AC?