Cross Product Proof: Proving Distance Formula for a Point Not on a Plane

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Homework Statement



Let P be a point not on the plane that passes through the points Q, R and S. show that the distance d from P to the plane is d = (|a.(bxc)|)/(|axb|)
where a = QR and b = QS and c = QP (those are lines between given two points)

Homework Equations



|axb| = (|a||b|sin[tex]\theta[/tex])n

|a||b| = (a.b)/cos[tex]\theta[/tex]

distance eq for a line =

The Attempt at a Solution



no idea.
i struggle with proofs.
help please!
 
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It's pretty straightforward to see what's going on geometrically. Start by thinking about what the vector [itex]\mathbf{a}\times\mathbf{b}[/itex] is in relation to the plane.
 
assuming P is arbitrary, then the vector a x b is the unit vector n parallel to plane of QRS.
say that distance d = vector PQ then a x b = vector n normal to plane. so then distance of P to the plane is the absolute value of the scalar projection of c onto n.

yeah?
 
a x b isn't a unit vector; also you say it's both parallel and normal to the plane. But I think you have the basic idea.
 
lanedance said:
how can a point pass through points? doesn't make sense...
"Let P be a point not on the plane that passes through the points Q, R and S."

A bit ambuguous but it is the plane that passses thorugh P, R, and S.
 
oh i don't mean parallel to plane. my bad. a x b gives vector n normal to plane QRS. then PQ = c and distance d is absolute value of scalar projection of c onto n.