Question

Find the equation of the plane passing through the point $a=2i+3j-k$ and perpendicular to the vector $3i-2j-2k$ and the distance of the plane from the origin?

Answer 1

Equation of plane is given by

$\overrightarrow{r}.\left(2\hat{i}+3\hat{j}-\hat{k}\right)=2$

Distance of plane from origin $=2$ units.

Now, to find the equation of the plane passing through the point $a=2\hat{i}+3\hat{j}-\hat{k}$ and perpendicular to a vectors $3\hat{i}-2\hat{j}-2\hat{k}$ and the distance of this plane from the origin.

We know that equation of a plane in vector form is given by $\overrightarrow{r}.\overrightarrow{n}=d$

here plane is passing through

$\begin{array}{c}a=2\hat{i}+3\hat{j}-\hat{k}\\ \overrightarrow{n}=3\hat{i}-2\hat{j}-2\hat{k}\end{array}$

So, the plane passes through vector $a$ and perpendicular to vector $n$ is given by equation.

$\begin{array}{c}\left(\overrightarrow{r}-\overrightarrow{a}\right).\overrightarrow{n}=0\\ \Rightarrow \overrightarrow{r}\overrightarrow{n}=\overrightarrow{a}\overrightarrow{n}\\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}+3\hat{j}-\hat{k}\right)=\left(2\hat{i}+3\hat{j}-\hat{k}\right).\left(3\hat{i}-2\hat{j}-2\hat{k}\right)\\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}+3\hat{j}-\hat{k}\right)=6-6+2\\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}+3\hat{j}-\hat{k}\right)=2\end{array}$

is the required equation of plane.

Distance of plane from origin $=2$ units.