Question

Show that the plane whose vector equation is $\overrightarrow{r}$. $(\hat{i}+2\hat{j}-\hat{k})=3$ contains the line $\overrightarrow{r}=\hat{i}+\hat{j}+\lambda (2\hat{i}+\hat{j}+4\hat{k})$.

Answer 1

A line contains line if dot product of normal vector of plane and direction cosine of line is $0$ and point on line satisfies the plane.

We have:

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

Normal vector $=\overrightarrow{n}=(\hat{i}+2\hat{j}-\hat{k})$

Equation of line is

$\overrightarrow{r}=\hat{i}+\hat{j}+\lambda (2\hat{i}+\hat{j}+4\hat{k})$

Direction ratio $=\overrightarrow{m}=2\hat{i}+\hat{j}+4\hat{k}$

$\overrightarrow{m}.\overrightarrow{n}=\left(2\hat{i}+\hat{j}+4\hat{k}\right).\left(\hat{i}+2\hat{j}-\hat{k}\right)=2+2-4=0$

$\Rightarrow \overrightarrow{m}.\overrightarrow{n}=0$

As line passes through $(\hat{i}+\hat{j})$, hence it must satisfy equation of plane in ($i$)

$\begin{array}{c}L.H.S.=\overrightarrow{r}\left(\hat{i}+2\hat{j}-\hat{k}\right)\\ =\left(\hat{i}+\hat{j}\right)\left(\hat{i}+2\hat{j}-\hat{k}\right)\\ =1+2+0=3=R.H.S.\end{array}$

Hence point satisfies the plane

As both conditions are satisfied, hence plane contains the line.