The Perpendicular Distance between a given Point and a given Plane
|A diagram of this is shown on the right.||P is the given point.
OP is the position vector of the given point.
The disc is a section of the plane.
ON is the normal to the plane which passes through the origin O.
n is the given vector to which the normal ON is parallel.
M is the point where ON crosses the plane.
F is the point where the perpendicular from P meets the plane.
FP is the required distance.
|A line is drawn from P, parallel to the plane and hence, at rightangles to ON, meeting ON at Q.||Since FP and ON are both perpendicular to the plane and PQ is parallel to the plane then, MQ = FP. Therefore, the length MQ is equal to the required distance.|
|By inspection, MQ = OQ - OM
||OM is the perpendicular distance of the plane from the origin. This distance is given in the problem as 2.
OM = 2
OQ is the projection of the vector OP on to ON which is parallel to the vector, n = [0, 4, 3]
OQ is then, the scalar (dot) product of OP and the unit vector = ;
OQ = [3, 1, 5].
We can now, finaly calculate MQ using;
MQ = OQ - OM
= - 2