Asked by Anonymous
Find the distance from the point P(5, 3, 2) to the plane x + 2y + z = 6.
Answers
Answered by
Reiny
for a given point (p,q,r) and plane Ax + By + Cz + D = 0
the distance between the point and the plane is
|Ap + Bq + Cr + D| / √(A^2 + B^2 + C^2)
so you have (5,3,2) as the point and x + 2y + z - 6 = 0 as the plane
go for it
the distance between the point and the plane is
|Ap + Bq + Cr + D| / √(A^2 + B^2 + C^2)
so you have (5,3,2) as the point and x + 2y + z - 6 = 0 as the plane
go for it