Question

Write the intercepts made by the plane 2x − 3y + 4z = 12 on the coordinate axes.

Answer 1

$$\begin{gathered} \text{The given equation of the plane is} \\ 2x-3y+4z=12 \\ \text{Dividng both sides by 12, we get} \\ \Rightarrow \frac{2x}{12}+\frac{-3y}{12}+\frac{4z}{12}=\frac{12}{12} \\ \Rightarrow \frac{x}{6}+\frac{y}{-4}+\frac{z}{3}=1...\left(1\right) \\ \text{We know that the equation of the plane whose intercepts on the coordianate axes are}\mathit{\text{a}}\text{,}\mathit{\text{b}}\text{and}\mathit{\text{c}}\text{is} \\ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1...\left(2\right) \\ \text{Comparing (1) and (2), we get} \\ a=6,b=-4\text{and}c=3 \end{gathered}$$