Question

Find the vector equation of the plane passing through three points with position vectors $\hat{i}+\hat{j}-2\hat{k},2\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k}$. Also find the coordinates of the point of intersection of this plane and the line $\overrightarrow{r}=3\hat{i}-\hat{j}-\hat{k}+\lambda (2\hat{i}-2\hat{j}+\hat{k})$.

Answer 1

$(\overrightarrow{r}-\overrightarrow{a}).(\overrightarrow{b}\times \overrightarrow{c})=0$

$\overrightarrow{n}=\overrightarrow{b}\times \overrightarrow{c}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 2& -1& 1\\ 1& 2& 1\end{array}\right|=-3\overrightarrow{i}-\overrightarrow{j}+5\overrightarrow{k}$

Therefore, $[\overrightarrow{r}-(\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}\left)\right].(-3\overrightarrow{i}-\overrightarrow{j}+5\overrightarrow{k})=0$

Equation of plane is

$\overrightarrow{r}.(3\overrightarrow{i}-\overrightarrow{j}+5\overrightarrow{k})=-14$

Now, $\overrightarrow{r}=3\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}+\lambda (2\overrightarrow{i}-2\overrightarrow{j}+\overrightarrow{k})$

This line intersects the plane,

Therefore, $\left[\right(3+21\left)\right(3+2\lambda )\overrightarrow{i}+(-1-2\lambda \left)\overrightarrow{j}\right(-1+\lambda \left)\overrightarrow{k}\right].[3\overrightarrow{i}+\overrightarrow{j}-5\overrightarrow{k}]=14$

$-\lambda =1$

$\lambda =-1$

Therefore, Point of intersection is

$\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}$