Distance from point to plane

In summary: You are saying that if you take the cross product of QR and P then the result is a vector that is normal to both QR and the position vector of P.
  • #1
Larrytsai
228
0

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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  • #2
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
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
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
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:
  • #6
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!
 

1. What is the formula for finding the distance from a point to a plane?

The formula for finding the distance from a point to a plane is:
d = |ax0 + by0 + cz0 + d| / √(a2 + b2 + c2)
Where (x0, y0, z0) is the coordinates of the point, and (a, b, c) is the normal vector of the plane.

2. What is the significance of the distance from a point to a plane in geometry?

The distance from a point to a plane is significant in geometry because it represents the shortest distance between the point and the plane. This distance can be used to determine the relationship between the point and the plane, such as whether the point lies on the plane, or whether it is above or below the plane.

3. Can the distance from a point to a plane be negative?

No, the distance from a point to a plane cannot be negative. The distance is always measured as a positive value, representing the perpendicular distance between the point and the plane.

4. How does the distance from a point to a plane relate to the equation of a plane?

The distance from a point to a plane is related to the equation of a plane through the use of the normal vector. The normal vector (a, b, c) of the plane is used in the formula for finding the distance, and it is also a component of the equation of a plane, which is ax + by + cz + d = 0.

5. Can the distance from a point to a plane be greater than the distance between two points?

No, the distance from a point to a plane cannot be greater than the distance between two points. The distance between two points is measured along a straight line, while the distance from a point to a plane is measured as the shortest distance between the point and the plane. Therefore, the distance from a point to a plane can never be longer than the distance between two points.

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