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Distance formula for between a point and a plane 
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#1
Mar311, 05:48 AM

P: 315

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=abcos(theta) definition of cross product is a x b= absin(theta) triple scalar product is a.(b x c)=b x cacos(theta) 3. The attempt at a solution putting point P above point S in plane, gives d = PS =PQsin(theta) = csin(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 ab gives d = csin(theta) = cb x c over cb = 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
Mar311, 05:19 PM

P: 315

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) = bsin(theta). theta is angle between QP=b and QR=a therefore by definition of cross product, sin(theta) = a x b over ab and so d = bsin(theta) = ba x b over ab = a x b over a but i dont know how to incorporate that into a plane with a dot product in the solution? 


#3
Mar511, 04:07 PM

P: 315

help?



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