Question

Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane.

Answer 1

$$\begin{gathered} \text{The given equation of the plane is} \\ 2x+y-2z=3 \\ \text{Dividng both sides by 3, we get} \\ \frac{2x}{3}+\frac{y}{3}+\frac{-2z}{3}=\frac{3}{3} \\ \Rightarrow \frac{x}{\left({\displaystyle \frac{3}{2}}\right)}+\frac{y}{3}+\frac{z}{\left({\displaystyle \frac{-3}{2}}\right)}=1...\left(1\right) \\ \text{We know that the equation of the plane whose intercepts on the coordianate axes are}\mathit{\text{a}}\text{,}\mathit{\text{b}}\text{and}\mathit{\text{c}}\text{is} \\ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1...\left(2\right) \\ \text{Comparing (1) and (2), we get} \\ a=\frac{3}{2};b=3;c=\frac{-3}{2} \\ \mathbf{\text{Finding the direction cosines of the normal}} \\ \text{The}\text{given equation of the plane is} \\ 2x+y-2z=8 \\ \Rightarrow \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(2\hat{i}+\hat{j}-2\hat{k}\right)=8 \\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}+\hat{j}-2\hat{k}\right)=8,\text{which is the vector equation of the plane.} \\ \text{(Because the vector equation of the plane is}\overrightarrow{r}\text{.}\overrightarrow{n}\text{=}\overrightarrow{a}\text{.}\overrightarrow{n}\text{,} \\ \text{where the normal to the plane,}\overrightarrow{n}\text{=}2\hat{i}+\hat{j}-2\hat{k}.) \\ \left|\overrightarrow{n}\right|=\sqrt{4+1+4}=3 \\ \text{So, the unit vector perpendicular to}\overrightarrow{n}\text{=}\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}\text{=}\frac{2\hat{i}+\hat{j}-2\hat{k}}{3}\text{=}\frac{2}{3}\hat{i}\text{+}\frac{1}{3}\hat{j}\text{-}\frac{2}{3}\hat{k} \\ \text{So, the}\mathrm{direction\; cosines\; of\; the\; normal\; to\; the\; plane\; are,}\frac{2}{3},\frac{1}{3},\frac{-2}{3}. \end{gathered}$$