JEE Advanced 2019 Paper 1 · Mathematics · Three Dimensional Geometry · easy

The plane x+2y+2z=12 meets the coordinate axes at A, B and C. Let H be the orthocentre and G the centroid of triangle ABC. Find the length HG.

Numerical answer

The plane

x+2y+2z=12

meets the coordinate axes at

A

B

and

C

. Let

H

be the orthocentre and

G

the centroid of triangle

ABC

. Find the length

HG

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 2.83

The intercepts give

A(12,0,0)

B(0,6,0)

C(0,0,6)

. Because

OA

OB

OC

are mutually perpendicular, the foot of the perpendicular from

O

to the plane

ABC

is the orthocentre of the triangle. The foot is

\frac{12}{1^2+2^2+2^2}(1,2,2)=\left(\frac43,\frac83,\frac83\right)

, so

H=\left(\frac43,\frac83,\frac83\right)

. One can verify

\vec{HA}\cdot\vec{BC}=0

. The centroid is

G=\left(4,2,2\right)

. Then

\vec{HG}=\left(\frac83,-\frac23,-\frac23\right)

and

HG=\frac13\sqrt{64+4+4}=\frac{\sqrt{72}}3=2\sqrt2\approx2.83

Original question written in the JEE Advanced 2019 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake