Find Intersection of Perpendicular Lines: Solve L1 & L2

[Stang70Fastba]

Homework Statement

Suppose that a straight line L1 passes through the point (2, 1, 2), and is perpendicular to another line which is given by L2(t) = (1 + t, 2 + t, 3 - t); -inf < t < inf

Suppose that L1 and L2 intersects at a point (a, b, c): Find (a, b, c):

Homework Equations

Dot product (which equals 0 for perpendicular lines)

The Attempt at a Solution

I've tried quite a few solutions but I get the feeling I'm making this WAY more complicated than is necessary. The trouble is my teacher explained some method of getting a plane out of the given line and then using the plane's normal vector to find L2 and then determining the intersection, but I can't for the life of me figure out what he did. At this point I'm not even sure where to start. I attempted to do some projection of vectors from a point on the given line to the point that L1 passes through. However I get stuck after that point at figuring out what to do. I would appreciate any help! Thanks!

 

[HallsofIvy]

You are given that L2 is (1 + t, 2 + t, 3 - t). Any plane perpendicular to that has normal vector <1, 1, -1>, the "direction vector" of L2. Do you see that (a, b, c) is the point where L2 intersects the plane perpendicular to L2 and containing (2, 1, 2)?

That plane is (x- 2)+ (y- 1)- (z- 2)= 0. Set x= 1+ t, y= 2+ t, z= 3- t in that and solve for t.

 

[Stang70Fastba]

I understand what you say about the plane. So what you gave me was the equation of the plane, and what I'm doing is plugging in the L2 equation and solving for it = 0? In that case, I get t = 1/3, which gives me a point (4/3, 7/3, 8,3). Is this correct?

I appreciate the help. I'm always slow with this stuff and then one day it clicks. Could you quickly explain how you got the equation for the plane?