Question

(3) On page 136 of the workbook, we developed a formula for the shortest distance from a point to a plane. To briefly recap, suppose P = (21,41, 21) is a point with corresponding position vector p, and II is a plane with normal n = (a, b, c) given by ax + by + cz = d. Then the shortest distance from P to the plane is given by p-q|l, where Q is the point (with corresponding position vector q) which both lies on the plane II, and the line r =p+In. To convince you this was true, I appealed to your sense of geometric intuition and said, “The quickest way to get from P to the plane is by moving along a straight line in the direction of n." In this question, I'd like you to prove this rigorously. Let Q be a point on the plane (with position vector q') other than Q. Prove that p - q'|| > ||p-all. (Hint: Write p-q' = p -q+q-q'. What would Pythagoras do?)

Answer 1

Let us assume is the shortest distance from a point to the plane. Let us define be a point on the plane.

Let us define the position vector from to by
p-= (p-q) + (9-0)
.

plane

Now From the above graph, we understand that it is a right angle triangle. In which
p-
is the hypotenuce and and
b-b
are the other two sides. So from the pythagoras theorem, we can say that
z llib – b|| + |lb – d|| = || Þ – d||
. Since
97
,
115 – d|| < || 5 – d|| = 31b – d|| < z|| Þ – d||
. i.e is an unique position vector such that is the shortest distance from a point to the plane.