Question

A plane meets the co-ordinate axes in $A,B,C$ and $(\alpha,\beta,\gamma)$ is the centered of the triangle $ABC$. Then the equation of the plane is

• A. $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$
• B. $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 1$
• C. $\frac{3x}{\alpha} + \frac{3y}{\beta} + \frac{3z}{\gamma} = 1$
• D. $\alpha x + \beta y + \gamma z = 1$

Answer 1

Answer: (A)

$\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$

Answer 2

Let the co-ordinates of the points where the plane cuts the axes are (a, 0, 0), (0, b, 0), (0, 0, c). Since centroid is $( \alpha , \beta , \gamma )$ therefore $a = 3 \alpha$ $b = 3 \beta , c = 3 \gamma$

Equation of the plane will be $\frac { x } { a } + \frac { y } { b } + \frac { z } { c } = 1$