Distance formula for between a point and a plane

[ProPatto16]

Homework Statement

Let P be a point not on a plane that passes through points Q, R, S. show that the distance, d, from P to the plane is:

d = |a.(b x c)| over |a x b| where a=QR, b=QS and c=QP

Homework Equations

definition of dot product is a.b=|a||b|cos(theta)
definition of cross product is a x b= |a||b|sin(theta)
triple scalar product is |a.(b x c)|=|b x c||a|cos(theta)

The Attempt at a Solution

putting point P above point S in plane, gives d = |PS| =|PQ|sin(theta) = |c|sin(theta)
theta is angle between |QP| and |QS| which is c and b so by definition of the cross product sin(theta)=|a x b| over |a||b| gives d = csin(theta) = |c||b x c| over |c||b| = |b x c| over |b|

to incorporate dot product i need a cos theta and the only relavant one i can figure is angle between QR and QS which is a and b. which wouldn't be in the plane. I am not sure where to go from here??

thanks

 

[ProPatto16]

i can do it with 2 points on a line if that's any help i will show it.

let P be a point not on the line that passes through Q and R. show that the distance d from the point P to the line L is

d = |a x b| over |a|

have line QR with point P drawn above and an arbitrary point S on the line where the perpendicular of P meets. let QR be vector a and QP be vector b.

the distance d is PS. which is also QPsin(theta) = |b|sin(theta). theta is angle between QP=b and QR=a therefore by definition of cross product, sin(theta) = |a x b| over |a||b|
and so d = |b|sin(theta) = |b||a x b| over |a||b| = |a x b| over |a|


but i don't know how to incorporate that into a plane with a dot product in the solution?

 

[ProPatto16]

help?