Question

Consider the 2 lines with equations (x+8)/1=(y+4)/3= (z-2)/1 and (x,y,z)=(3,3,3)+t (4,-1,-1), tER.

Find the point of intersection of the lines.

So I that these 2 lines are perpendicular because if you do the dot product of their direction vectors you get zero.

L1:
x=-8+s
y=-4+3s
z=2+s

L2:
x=3+4t
y=3-t
z=3-t

So I tried setting L1=L2 for each variable,x,y and z.

Then doing elimination yo solve for s and t and sub them back into my original parametric equations but I don't get the answer.

What do I need to do?

The answer key says the answer is (-5,5,5)

Answer 1

I set the x and y equations equal to each other to get:
4t - s = -11
t + 3s = 7
solving these, I got t = -2 and s = 3
which verified in the z value of 0 for both lines.
So sub s=3 and t = -2 and it the corresponding lines and you get
(x,y,z) = (-5,5,5)