Question

Find the vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2\hat{i}-3\hat{j}+4\hat{k}.$ Also, find its Cartesian form.

Answer 1

$$\begin{gathered} \text{Given, normal vector,}\overrightarrow{n}\text{=2}\hat{i}-3\hat{j}+4\hat{k} \\ \text{Now,}\hat{n}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{4+9+16}}=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}=\frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k} \\ \text{The equation of the plane in normal form is} \\ \overrightarrow{r}.\hat{n}=d\text{(where}\mathit{\text{d}}\text{is the distance of the plane from the origin)} \\ \text{Substituting,}\hat{n}\text{=}\frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k}\text{and}\mathit{\text{d}}\text{=}\frac{6}{\sqrt{29}}\text{here, we get} \\ \overrightarrow{r}.\left(\frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k}\right)=\frac{6}{\sqrt{29}}...\left(1\right) \\ \mathbf{Cartesian}\mathbf{}\mathbf{form} \\ \text{For Cartesian form, substituting}\overrightarrow{r}\text{=}x\hat{i}+y\hat{j}+z\hat{k}\text{in (1), we get} \\ \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(\frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k}\right)=\frac{6}{\sqrt{29}} \\ \Rightarrow \frac{2x-3y+4z}{\sqrt{29}}=\frac{6}{\sqrt{29}} \\ \Rightarrow 2x-3y+4z=6 \end{gathered}$$