Finding a Perpendicular Vector in 3D Space

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To find a vector perpendicular to the plane formed by three points P, Q, and R in 3D space, first create two vectors from these points, such as Q-P and Q-R. The cross product of these two vectors, (Q-P) x (Q-R), yields a vector that is orthogonal to the plane defined by P, Q, and R. Additionally, the magnitude of this cross product relates to the area of the parallelogram formed by the vectors PQ and PR, with the area of triangle PQR being half of that value. Understanding these concepts is essential for solving problems in Calculus III.
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I'm given three points in 3D space (vectors), P, Q, and R. So I have to find a vector that is perpendicular to the plane formed by these points.

Anyone? Thanks.

I really just need some hint(s) for this.
 
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Draw two vectors from any of the points to other points. Only one plane defines those two vectors. The cross product of those two vectors is an orthogonal vector to that plane.
 
Umm, sorry, but could you elaborate? I don't quite get it...new to Calc III.
 
if p,q are two points, subtracting Q-P gives a vector pointing from P to Q.

Thus if P,Q,R are 3 points, spanning a plane, the two vectors Q-P and Q-R are both parellel to that plane.


then there is a construction called cross product for finding sa vector perpendiculr to two given vectors. hence perpendiculr to the plane they are parallel to.

thus (Q-P) x (Q-R) is perpendicualr to the plane spanned by P,Q,R.
 
Thanks a bunch!
 
One more thing, I'm also asked to find the area of the triangle PQR.
 
Think about the magnitude of the cross product. What is the relationship to the area of a parallelogram with two sides (PQ and PR) given by the vectors?

Then what is the relationship of the area of the parallelogram to the triangle formed by the two sides, e.g. PQ, PR and the third QR?
 
Thanks a lot.
 
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