How Do You Calculate the Distance from a Point to a Plane Using Cross Product?

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SUMMARY

The discussion focuses on calculating the distance from a point P(-4, -2, 3) to a plane defined by three points Q(1, -5, -2), R(-4, -7, 3), and S(6, -3, 0) using vector mathematics. The method involves finding a normal vector to the plane by taking the cross product of two vectors formed by the points in the plane, specifically QR and RS. The normal vector is calculated as n = QR × RS = <-2, 5, 0>. The distance from point P to the plane is determined using the formula dist = |(PQ · n) / |n||, where PQ is the vector from point P to any point on the plane.

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Homework Statement


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).

The Attempt at a Solution



I don't really know where to start.
What i was thinking was to use the cross product, but from which points?
 
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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.
 
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!
 
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?
 
is the vector the y-comp of QR, is it supposed to be positive 2 instead of -2? I am getting positive for some reason. Nvm i get why its -2 you subtracted the otherway around ^^
 
Last edited:
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?
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.

is the vector the y-comp of QR, is it supposed to be positive 2 instead of -2? I am getting positive for some reason. Nvm i get why its -2 you subtracted the otherway around ^^
Sorry but I'm not sure what you mean here. Nvm, I just realized what Nvm means!
 

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