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

`\text{ Given, normal vector } , \vec{n} =\text{  2 } \hat{i} - \text{  3 }\hat{j}  +\text{   4 \hat{k}  `

\[\text{ Now } , \hat{n}  = \frac{\vec{n}}{\left| \vec{n} \right|} = \frac{2 \hat{i}  - \text{  3  }\hat{j}  +\text{   4 }\hat{k} }{\sqrt{4 + 9 + 16}} = \frac{2 \hat{i}  - \text{  3 }\hat{j} + 4 \hat{k} }{\sqrt{29}} = \frac{2}{\sqrt{29}} \hat{i}  - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k}  \]

\[ \text{ The equation of the plane in normal form is } \]

\[ \vec{r} . \hat{n}  =\text{   d }(\text{ where d is the distance of the plane from the origin } )\]

\[\text{ Substituting, } \hat{n} =\frac{2}{\sqrt{29}} \hat{i}  - \frac{3}{\sqrt{29}} \hat{j}  + \frac{4}{\sqrt{29}} \hat{k}  \text{ and } d=\frac{6}{\sqrt{29}} \text{ here, we get } \]

\[ \vec{r} . \left( \frac{2}{\sqrt{29}} \hat{i} - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k} \right) = \frac{6}{\sqrt{29}} . . . (1)\]

\[ \text{ Cartesian form } \]

\[\text{ For Cartesian form, substituting }  \vec{r} =x \hat{i}  + y \hat{j} + z \hat{k}  \text{ in (1), we get } \]

\[\left( x \hat{i} + y \hat{j} + z \hat{k} \right) . \left( \frac{2}{\sqrt{29}} \hat{i}  - \frac{3}{\sqrt{29}} \hat{j}  + \frac{4}{\sqrt{29}} \hat{k}  \right) = \frac{6}{\sqrt{29}}\]

\[ \Rightarrow \frac{2x - 3y + 4z}{\sqrt{29}} = \frac{6}{\sqrt{29}}\]

\[ \Rightarrow 2x - 3y + 4z = 6\]