Question

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

• A. $x+2y-z=3$
• B. $3x-2y-z=6$
• C. $3x+2y+z=6$
• D. $x-2y-z=5$

Answer 1

The correct option is A $x+2y-z=3$

Given: $L_1:\overrightarrow{r}=\hat{i}+\hat{j}+\hat{k}+\lambda (2\hat{i}+\hat{j}+4\hat{k})$ and $L_2:\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$
As the plane is parallel to lines $L_1,L_2$ having DR's $(2,1,4)$ and $(-3,2,1)$ respectively.
Let normal of the plane be $(a,b,c)$
Hence, normal of the plane will be
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 4\\ -3& 2& 1\end{array}\right|=-7\hat{i}-14\hat{j}+7\hat{k}\cdots \left(i\right)$
$\therefore a=-7,b=-14,c=7\cdots \left(i\right)$
As plane is passing through $(0,1,-1)$
$a\left(x\right)+b(y-1)+c(z+1)=0$
puttin from $\left(i\right)$
$-7\left(x\right)-14(y-1)+7(z+1)=0$
$\Rightarrow x+2y-2-z-1=0$
$\therefore$ Required equation of plane $x+2y-z=3$