Question

Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes.
(i) 4x + 3y − 6z − 12 = 0
(ii) 2x + 3y − z = 6
(iii) 2x − y + z = 5

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{plane}\mathrm{is} \\ 4x+3y-6z-12=0 \\ \Rightarrow 4x+3y-6z=12 \\ \text{Dividng both sides by 12, we get} \\ \frac{4x}{12}+\frac{3y}{12}+\frac{(-6z)}{12}=\frac{12}{12} \\ \Rightarrow \frac{4x}{12}+\frac{3y}{12}-\frac{6z}{12}=\frac{12}{12} \\ \Rightarrow \frac{x}{3}+\frac{y}{4}+\frac{z}{-2}=1...\left(1\right) \\ \text{We know that the equation of the plane whose intercepts on the coordianate axes are}\mathit{\text{a}}\text{,}\mathit{\text{b}}\text{and}\mathit{\text{c}}is \\ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1...\left(2\right) \\ \text{Comparing (1) and (2), we get} \\ \mathrm{a}=3;\mathrm{b}=4;\mathrm{c}=-2 \end{gathered}$$

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

$$\begin{gathered} \left(iii\right)\mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{plane}\mathrm{is} \\ 2x-y+z=5 \\ \text{Dividng both sides by 5, we get} \\ \frac{2x}{5}+\frac{-y}{5}+\frac{z}{5}=\frac{5}{5} \\ \Rightarrow \frac{x}{\left({\displaystyle \frac{5}{2}}\right)}+\frac{y}{-5}+\frac{z}{5}=1...\left(1\right) \\ \text{We know that the equation of the plane whose intercepts on the coordianate axes are}\mathit{\text{a}}\text{,}\mathit{\text{b}}\text{and}\mathit{\text{c}}\text{is} \\ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1...\left(2\right) \\ \text{Comparing (1) and (2), we get} \\ a=\frac{5}{2};b=-5;c=5 \end{gathered}$$