A plane meets the co-ordinate axes in A,B,C such that the centroid of the triangle ABC is the point (p,q,r). The equation of the plane is

Engineering

Mathematics

Plane and Its Different Forms

Section Formulae and Centres of a Triangle

Distance from a Plane

Question

A plane meets the co-ordinate axes in A,B,C such that the centroid of the triangle ABC is the point (p,q,r). The equation of the plane is

$\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 2$

$\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$

none of these

$\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 1$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

The equation of plane is

$\frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1$ ……. (1)

Using centroid formula in XYZ plane is $(x = \frac{x_{1} + x_{2} + x_{3}}{3}, y = \frac{y_{1} + y_{2} + y_{3}}{3}, z = \frac{z_{1} + z_{2} + z_{3}}{3})$

Then,

$(\frac{A}{3}, \frac{B}{3}, \frac{C}{3}) = (p, q, r)$

Therefore,

A = 3p, B = 3q, C = 3r

Put the value of A,B and C in equation of plane (1) and we get

$\frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1$

$\frac{x}{3 p} + \frac{y}{3 q} + \frac{z}{3 r} = 1$

Hence,

$\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 3$

Option (D) is correct answer.