Question

Consider a plane $x+y-z=1$ and point $A(1,2,-3)$. A line $L$ has the equation $x=1+3r,y=2-r$ and $z=3+4r$.
The equation of the plane containing line $L$ and point $A$ has the equation-

• A. $x-3y+5=0$
• B. $x+3y-7=0$
• C. $x-3y+8=2$
• D. $x-3y+59=2$

Answer 1

The correct option is B $x+3y-7=0$

Given:

Point $A(1,2,-3)$

Line: $x=1+3r,y=2-r,z=3+4r$

Point $B(1,2,3)$

Putting $r=1$

$x=4,y=1,z=7$

Point $R(4,1,7)$

$\begin{array}{c}\overrightarrow{AR}=3\hat{i}-\hat{j}+10\hat{k}\\ \overrightarrow{BR}=3\hat{i}-\hat{j}+4\hat{k}\\ \overrightarrow{n}=\overrightarrow{AR}\times \overrightarrow{BR}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 10\\ 3& -1& 4\end{array}\right|=6\hat{i}+18\hat{j}\end{array}$

Equation of plane

$\begin{array}{c}\left(\overrightarrow{r}-\overrightarrow{a}\right).\overrightarrow{n}=0\\ 6\left(x-1\right)+18\left(y-2\right)=0\\ x+3y-7=0\end{array}$