Question

The intercepts of the plane $2x-3y+4z=12$ on the coordinate axes are given by

• A. $3,-2,15$
• B. $6,-4,3$
• C. $6,-4,-3$
• D. $2,-3,4$

Answer 1

The correct option is B

$6,-4,3$

The explanation for the correct option

The given equation of the plane: $2x-3y+4z=12$.

Divide both sides of the equation by $12$.

$$\begin{gathered} \Rightarrow \frac{2x-3y+4z}{12}=\frac{12}{12} \\ \Rightarrow \frac{x}{6}+\frac{y}{-4}+\frac{z}{3}=1 \end{gathered}$$

The intercept form of the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, here $a,b$ and $c$ are $x,y$ and $z-$intercepts respectively.

Thus, the $x-$intercepts of the plane, $a=6$.

The $y-$intercepts of the plane, $b=-4$.

The $z-$intercepts of the plane, $c=3$.

Hence, the intercepts of the plane $2x-3y+4z=12$ on the co-ordinate axes are $6,-4,3$.