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JEE Main202229 Jun 2022Morning ShiftMathematicsThree Dimensional GeometryActual

Let d be the distance between the foot of perpendiculars of the points P 1 , 2 - 1 and Q 2 , - 1 , 3 on the plane - x + y + z = 1 . Then d 2 is equal to ______.

Correct answer

0

Step-by-step solution

Since - 1 + 2 - 1 - 1 < 0   &   - 2 - 1 + 3 - 1 < 0 so points P 1 , 2 , - 1 and Q 2 , - 1 , 3 lie on same side of the given plane. Perpendicular distance of point P from the plane is - 1 + 2 - 1 - 1 1 2 + 1 2 + 1 2 = 1 3 Perpendicular distance of point Q from the plane is = - 2 - 1 + 3 - 1 1 2 + 1 2 + 1 2 = 1 3 ⇒   P Q → is parallel to given plane. So, distance between P and Q = distance between their foot of perpendiculars. ⇒ P Q → = 1 - 2 2 + 2 + 1 2 + - 1 - 3 2

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