Question

The equation of the plane which makes with the coordinate axes, a triangle with centroid (α, β, γ) is given by 

A. αx + βy + γz = 1 

B. αx + βy + γz = 3

C. \(\frac{x}{α}\) + \(\frac{y} β\) + \(\frac{z}γ\) = 1

D. \(\frac{x}{α}\) + \(\frac{y} β\) + \(\frac{z}γ\) = 3

Answer 1

Given: Centroid of triangle is (α, β, γ) 

To find: Equation of plane. 

Formula Used: Equation of plane \(\frac{x}a\) + \(\frac{y}b\) + \(\frac{z}c\) = 1 where (x, y, z) is a point on the plane and a, b, c are intercepts on x-axis, y-axis and z-axis respectively.

Centroid of a triangle

Explanation: 

Let the equation of plane be

  \(\frac{x}a\) + \(\frac{y}b\) + \(\frac{z}c\) = 1.....(1)

Therefore, A = 3α, B = 3β, C = 3γ where (a, b, c) is the centroid of the triangle with vertices (A, 0, 0), (0, B, 0) and (0, 0, C) 

Substituting in (1),

⇒ \(\frac{x}{3a}\) + \(\frac{y}{3b}\) + \(\frac{z}{3c}\) = 1

Here a = α, b = β and c = γ

⇒ \(\frac{x}{α}\) + \(\frac{y} β\) + \(\frac{z}γ\) = 3

Therefore equation of required plane is \(\frac{x}{α}\) + \(\frac{y} β\) + \(\frac{z}γ\) = 3