Question

Find the equation of the plane through the point $(1,1,1)$ and perpendicular to the line $x-2y+z=2,4x+3y-z+1=0$

Answer 1

Given equation of lines $x-2y+z=2$ and $4x+3y-z+1=0$ which are lines (1) and (2) through which the point $(1,1,1)$ passes.

Now\quad the normal vector for line (1) is $\overrightarrow{n_1}=\hat{i}-2\hat{j}+\hat{k}$

and the normal vector for line (2) is $\overrightarrow{n_1}=4\hat{i}+3\hat{j}-\hat{k}$

Now we find $\overrightarrow{n}=\overrightarrow{n_1}\times \overrightarrow{n_2}=(\hat{i}-2\hat{j}+\hat{k})\times (4\hat{i}+3\hat{j}-\hat{k})$

$⟹\overrightarrow{n}=-\hat{i}+5\hat{j}+11\hat{k}$ is the required plane that passes through $(1,1,1)$

Now the position vector for the point $(1,1,1)$ is $\hat{i}+\hat{j}+\hat{k}$

As we know equation of plane to the given vector $\overrightarrow{n}$ is given by

$⟹(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$

$⟹\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}$

So, $(x\hat{i}+y\hat{j}+z\hat{k})(-\hat{i}+5\hat{j}+11\hat{k})=(\hat{i}+\hat{j}+\hat{k})(-\hat{i}+5\hat{j}+11\hat{k})$

$⟹-x+5y+11z=-1+5+11$

$⟹-x+5y+11z=15$

$⟹x-5y-11z+15=0$ which is the required equation of plane.