Question

Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.

Answer 1

$$\begin{gathered} \text{The given equation of the plane is} \\ 2x+2y+2z=8 \\ \Rightarrow \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(2\hat{i}+2\hat{j}+2\hat{k}\right)=8 \\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}+2\hat{j}+2\hat{k}\right)=8,\text{which is the vector equation of the plane.} \\ \text{(Because the vector equation of the plane is}\overrightarrow{r}\text{.}\overrightarrow{n}\text{=}\overrightarrow{a}\text{.}\overrightarrow{n}\text{,} \\ \text{where the normal to the plane,}\overrightarrow{n}\text{=}2\hat{i}+2\hat{j}+2\hat{k}.) \\ \left|\overrightarrow{n}\right|=\sqrt{4+4+4}=2\sqrt{3} \\ \text{So, the unit vector perpendicular to}\overrightarrow{n}\text{=}\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}\text{=}\frac{2\hat{i}+2\hat{j}+2\hat{k}}{2\sqrt{3}}\text{=}\frac{1}{\sqrt{3}}\hat{i}\text{+}\frac{1}{\sqrt{3}}\hat{j}\text{+}\frac{1}{\sqrt{3}}\hat{k} \\ \text{So, the}\mathrm{direction\; cosines\; of\; the\; normal\; to\; the\; plane\; are}l=\frac{1}{\sqrt{3}},m=\frac{1}{\sqrt{3}},n=\frac{1}{\sqrt{3}} \\ \text{Let α, β and γ be the angles made by the given plane with the coordinate axes.} \\ \text{Then,} \\ l=\cos \alpha =\frac{1}{\sqrt{3}};m=\cos \beta =\frac{1}{\sqrt{3}};n=\cos \gamma =\frac{1}{\sqrt{3}} \\ \Rightarrow \cos \alpha =\cos \beta =\cos \gamma \\ \Rightarrow \alpha =\beta =\gamma \\ \text{So, the given plane is equally inclined to the coordinate axes.} \end{gathered}$$