- #1
DougD720
- 47
- 0
I know this has been posted before, and I've read the post concerning the same problem and I've googled this a million times, but i can't seem to get it. So here's the problem:
Given a Point and a Line, find the Line that passes through the point (3,1,-2) and is perpendicular to the line x=-1+t, y=-2+t, z=-1+t.
Okay, so i found the direction vector of the given line (which i'll call l) to be (1,1,1).
I know that a vector lying on l and a vector lying on the line i need to find (which i'll call l'), when you apply the dot product to those 2 vectors it must equal 0 in order for them to be perpendicular.
I know i have to use orthagonal projection, well i think i need to. So far i took the point given (which i called Q) and a point on the line given (which i called P) to find the vector QP, which (in my case) is (4,3,-1).
From here I'm lost with what to do, I've been sitting here racking my brain trying different things but i just can't seem to figure out how to find the line. I'm not asking for a solution but a *simple* set of steps or something like that to follow would be greatly appreciated.
Thanks!
Given a Point and a Line, find the Line that passes through the point (3,1,-2) and is perpendicular to the line x=-1+t, y=-2+t, z=-1+t.
Okay, so i found the direction vector of the given line (which i'll call l) to be (1,1,1).
I know that a vector lying on l and a vector lying on the line i need to find (which i'll call l'), when you apply the dot product to those 2 vectors it must equal 0 in order for them to be perpendicular.
I know i have to use orthagonal projection, well i think i need to. So far i took the point given (which i called Q) and a point on the line given (which i called P) to find the vector QP, which (in my case) is (4,3,-1).
From here I'm lost with what to do, I've been sitting here racking my brain trying different things but i just can't seem to figure out how to find the line. I'm not asking for a solution but a *simple* set of steps or something like that to follow would be greatly appreciated.
Thanks!