Question

Find the equation of the plane which is parallel to the lines $\overrightarrow{r}=\hat{i}+\hat{j}+\lambda \left(2\hat{i}+\hat{j}\right)+4k$ and $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and is passing through the point $(0,1,-1)$.

Answer 1

Line (vector form equation)

$=\hat{i}+\hat{j}+\alpha (2\hat{i}+\hat{j})+4\hat{k}$

$=\hat{i}+\hat{j}+4\hat{k}+\alpha (2\hat{i}+\hat{j}+0\hat{k})$

Equation of line in the cartesian form

$=\frac{x-1}{2}=\frac{y-1}{1}=\frac{z-4}{0}$

$\therefore$ DR of the line $<2,1,0>$

DR of another $<-3,2,1>$

$\therefore$ If the plane is parallel to these lines, the normal to the plane will be perpendicular to these lines

$\therefore$ Let the DR of the normal of the plane $<a,b,c>$

$\therefore 2a+b+0c=0$

$-3a+2b+c=0$

On solving :

$\frac{a}{1}=\frac{-b}{2}=\frac{c}{4+3}$

Plane with normal $<1,-2,7>$ and passing through $(0,1,-1)$ is :

$\frac{x-0}{1}=\frac{y-1}{-2}=\frac{z+1}{7}$