Question

Find the vector equation of a plane which is at a distance of 8 units from the origin and which is normal to the vector $2\hat{i}+\hat{j}+2\hat{k}$

Answer 1

As equation of plane is $\overrightarrow{r}.\hat{n}=d$ in which $\overrightarrow{r}$ is position vector on place, $\hat{n}$ is unit vector of normal and d is the distance. So we will simply put these values to find the equation of the plane.

Complete step by step answer:
Moving ahead with the question in step-to-step manner;
Distance of plane from origin $=8$
Normal vector $=2\hat{i}+\hat{j}+2\hat{k}$
So unit vector along normal, we know that unit vector of any vector is that vector upon vector mode i.e. unit vector $=\dfrac{\overrightarrow{a}}{|a|}$ , so unit vector of normal vector is equal to;
$\hat{n}=\dfrac{\overrightarrow{n}}{|n|}$ . So by putting the value we will get;
$\begin{aligned} & \hat{n}=\dfrac{\overrightarrow{n}}{|n|} \\\ & \hat{n}=\dfrac{2\hat{i}+\hat{j}+2\hat{k}}{\sqrt{{{2}^{2}}+{{1}^{2}}+{{2}^{2}}}} \\\ & \hat{n}=\dfrac{2\hat{i}+\hat{j}+2\hat{k}}{\sqrt{9}} \\\ & \hat{n}=\dfrac{2\hat{i}+\hat{j}+2\hat{k}}{3} \\\ \end{aligned}$
And $\overrightarrow{r}$ is position vector of any point on the place i.e. $x\hat{i}+y\hat{j}+z\hat{k}$
So as we know, the equation of the plane when the normal vector and distance of the plane from the origin is given is $\overrightarrow{r}.\hat{n}=d$ . So to find the equation put the value, from where we will get;
$\begin{aligned} & \overrightarrow{r}.\hat{n}=d \\\ & \overrightarrow{r}.\left( \dfrac{2\hat{i}+\hat{j}+2\hat{k}}{3} \right)=8 \\\ \end{aligned}$
On further simplifying it, we will get;
$\overrightarrow{r}.\left( 2\hat{i}+\hat{j}+2\hat{k} \right)=24$
So the equation of the plane is $\overrightarrow{r}.\left( 2\hat{i}+\hat{j}+2\hat{k} \right)=24$ .
Hence the answer is $\overrightarrow{r}.\left( 2\hat{i}+\hat{j}+2\hat{k} \right)=24$ .

Note: There are many other forms of equation of plane rather than being $\overrightarrow{r}.\hat{n}=d$ , based on the given condition in questions. Moreover we can write $\overrightarrow{r}.\left( 2\hat{i}+\hat{j}+2\hat{k} \right)=24$ equation as $\left( x\hat{i}+y\hat{j}+z\hat{k} \right).\left( 2\hat{i}+\hat{j}+2\hat{k} \right)=24$ both are correct.