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

Answer 1

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

Let equation plane in intercept form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Given areas of triangle formed by intercepts and the axes $xy,yz,zx$ are $12,9,6$ respectively.
$\Rightarrow \frac{1}{2}ab=12,\frac{1}{2}bc=9$ and $\frac{1}{2}ca=6$
$\Rightarrow ab=24,bc=18$ and $ca=12$
Solving these, we get $a=4,b=6,c=3$
Hence required plane is
$\frac{x}{4}+\frac{y}{6}+\frac{z}{3}=1$