Question

Find the equation of plane passing through the point $(1,1,1,)$ & containing the line $\overrightarrow{r}=-3\hat{i}+\hat{j}+5\hat{k}+\lambda (\widehat{3i}\widehat{-j}-5\hat{k})$. Also show containing the line $\overrightarrow{r}=-\hat{i}+2\hat{j}+5\hat{k}+m(\hat{i}-\widehat{2j}-5\hat{k})$.

Answer 1

$\begin{array}{c}\pi =a\left(x-1\right)+b\left(y-1\right)+c\left(z-1\right)=0\\ \overrightarrow{r}=-3\hat{i}+\hat{j}+5\hat{k}+\lambda \left(3\hat{i}-\hat{j}-5\hat{k}\right)\\ L_1:\frac{x+3}{3}=\frac{y-1}{-1}=\frac{z-5}{-5}\end{array}$

Plane also contains $(-3,1,5)$

$-4a+4c=0$

$c=a$

As plane contains line $L_1:$

$3a-b-5c=0$

$-2a-b=0$

$b=-2a$

$a(x-1)-2a(y-1)+a(z-1)=0$

$x-1-2(y-1)+z-1=0$

Equation of plane

$\begin{array}{c}\pi =x-2y+z=0\\ \overrightarrow{r_2}=-\hat{i}+2\hat{j}+5\hat{k}+m\left(\hat{i}-2\hat{j}-5\hat{k}\right)\\ L_2:\frac{x+1}{1}=\frac{y-2}{-2}=\frac{z-5}{-5}\\ \left(r-1,-2r+2,-5r+5\right)\\ r-1-2\left(-2r+2\right)-5r+5=0\end{array}$

Hence,

Plane also contains line $L_2$.