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

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Homework Help Overview

The problem involves 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). The discussion centers around the use of the cross product to find a normal vector to the plane.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss finding a normal vector to the plane using the cross product of vectors formed by the given points. There is uncertainty about which vectors to use and how to apply the cross product correctly. Some participants question the implications of using different vectors and the resulting normal vectors.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to finding the normal vector and questioning assumptions about vector components. Some guidance has been offered regarding the calculation of the normal vector, but no consensus has been reached on the overall method.

Contextual Notes

There are indications of confusion regarding vector components and the relationships between the points, as well as the implications of using different vectors in calculations. Participants are also navigating terminology and notation related to vector operations.

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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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