JEE Advanced 2018 Paper 2 · Mathematics · Three Dimensional Geometry · medium

A variable plane is at a constant distance 2 from the origin and meets the coordinate axes at A, B, C. The centroid of triangle ABC lies on the…

Numerical answer

A variable plane is at a constant distance

2

from the origin and meets the coordinate axes at

A

B

C

. The centroid of triangle

ABC

lies on the surface

\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}=k

. Find

k

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 2.25

Let the plane be

\frac xa+\frac yb+\frac zc=1

meeting the axes at

A(a,0,0)

B(0,b,0)

C(0,0,c)

. Its distance from the origin is

\frac{1}{\sqrt{1/a^{2}+1/b^{2}+1/c^{2}}}=2

, so

\frac1{a^{2}}+\frac1{b^{2}}+\frac1{c^{2}}=\frac14

. The centroid of triangle

ABC

(x,y,z)=\left(\frac a3,\frac b3,\frac c3\right)

, hence

a=3x

b=3y

c=3z

. Substituting,

\frac{1}{9}\left(\frac1{x^{2}}+\frac1{y^{2}}+\frac1{z^{2}}\right)=\frac14

, so

k=\frac94=2.25

. A numerical check with random admissible intercepts

(a,b,c)

gives

2.25

every time. The result does not depend on the particular plane, which is why a single number

k

exists; any family of planes at fixed distance

p

gives the locus

\sum x^{-2}=\frac{9}{p^{2}}

Original question written in the JEE Advanced 2018 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake