Let d be the distance between the foot of perpendiculars of the points P(l, 2 - 1) and Q(2, -1, 3) on the plane -x + y + z= 1. Then d 2 is equal to_______.
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Points P(l, 2, -1) and Q (2, -1, 3) lie on same side of the plane.
Perpendicular distance of point P from plane is
Perpendicular distance of point Q from plane is
is parallel to given plane. So, distance between P and Q = distance between their foot of perpendiculars.
Alternate
-x + y+ z-1 = 0
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