Question

The areas of triangles formed by a plane with the positive $X,Y;Y,Z;Z,X$ axes respectively are $12,9,6$ square units, then the equation of the plane is

• A. $\frac{x}{6}+\frac{y}{4}+\frac{z}{3}=1$
• B. $\frac{x}{4}+\frac{y}{6}+\frac{z}{3}=1$
• C. $\frac{x}{3}+\frac{y}{6}+\frac{z}{4}=1$
• D. $\frac{x}{4}+\frac{y}{3}+\frac{z}{6}=1$

Answer 1

The correct option is B $\frac{x}{4}+\frac{y}{6}+\frac{z}{3}=1$

Let the plane cut the $X,Y,Z$ axes at points $A(a,0,0),B(0,b,0)$ and $C(0,0,c)$ respectively.
Given areas of triangle formed by intercepts and the axes $XY,YZ,ZX$ are $12,9,6$ respectively.
$\Rightarrow \frac{ab}{2}=12,\frac{bc}{2}=9,\frac{ac}{2}=6\Rightarrow ab=24,bc=18,ac=12$
Solving these, we get $a=4,b=6,c=3$
So, Equation of plane in intercept form is :
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
$\Rightarrow \frac{x}{4}+\frac{y}{6}+\frac{z}{3}=1$