JEE Advanced 2020 Paper 1 · Mathematics · Vector Algebra · medium

Points A(1,0,2), B(3,1,0) and C(2,3,1) are given. The perpendicular distance of the point D(2,4,5) from the plane ABC is d. Find d.

Numerical answer

Points

A(1,0,2)

B(3,1,0)

and

C(2,3,1)

are given. The perpendicular distance of the point

D(2,4,5)

from the plane

ABC

d

. Find

d

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 2.83

Form

\vec{AB}=(2,1,-2)

and

\vec{AC}=(1,3,-1)

. A normal to the plane is

\vec n=\vec{AB}\times\vec{AC}=(1\cdot(-1)-(-2)\cdot3,\ (-2)\cdot1-2\cdot(-1),\ 2\cdot3-1\cdot1)=(5,0,5)

, with

|\vec n|=5\sqrt2

. Now

\vec{AD}=(1,4,3)

, and

\vec{AD}\cdot\vec n=5+0+15=20

. The distance equals the projection of

\vec{AD}

on the unit normal:

d=\dfrac{|\vec{AD}\cdot\vec n|}{|\vec n|}=\dfrac{20}{5\sqrt2}=2\sqrt2\approx2.83

. Equivalently, the plane is

x+z=3

, and

\dfrac{|2+5-3|}{\sqrt2}=2\sqrt2

, which matches the vector computation.

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