Homework Help: Distance from point to plane

1. Sep 11, 2010

Larrytsai

1. The problem statement, all variables and given/known data
Find the distance from the point P(-4, -2, 3) to the plane through the three points
Q(1, -5, -2), R(-4, -7, 3), and S(6, -3, 0).

3. The attempt at a solution

I dont really know where to start.
What i was thinking was to use the cross product, but from which points?

2. Sep 11, 2010

Dick

You can find a normal to the plane by taking the cross product of two difference vectors in the plane, right? That would be a good start. You have several choices for a difference. Q-R would be one.

3. Sep 12, 2010

Lanthanum

You need to find a vector normal to the plane. First find two vectors that lie allong the plane. ie. QR and RS;
QR=<-5,-2,5> and RS=<10,4,-3>
A vector normal to the plane (n) will be normal to both of these. so;
n=<-5,-2,5>X<10,4,-3>
=<-14,35,0> (you can divide this by 7 cos it will still be in the same direction)
n=<-2,5,0>

Now the distance from P to the plane can be calculated using vector projections. Specifically,
dist = |projnPQ|
(You could use any other vector connecting P to the plane instead of PQ.)
So
dist = |[(PQ.n)/|n|]|
I'll leave the rest for you!

4. Sep 12, 2010

Larrytsai

just for understanding purposes, if I were to take QR and cross it with the point P i would be finding the normal to the QR-P plane right?

5. Sep 12, 2010

Larrytsai

is the vector the y-comp of QR, is it supposed to be positive 2 instead of -2? im getting positive for some reason. Nvm i get why its -2 you subtracted the otherway around ^^

Last edited: Sep 12, 2010
6. Sep 13, 2010

Lanthanum

What you get in that case is a vector which is normal to both QR and the position vector of P.
This is not the same as the normal to the plane containing QR and the point P.

Sorry but I'm not sure what you mean here. Nvm, I just realised what Nvm means!