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Distance formula for between a point and a plane

  1. Mar 3, 2011 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant 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)

    3. 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 wouldnt be in the plane. im not sure where to go from here??

    thanks
     
  2. jcsd
  3. Mar 3, 2011 #2
    i can do it with 2 points on a line if thats 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 dont know how to incorporate that into a plane with a dot product in the solution?
     
  4. Mar 5, 2011 #3
    help?
     
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