Question

The equation of the plane through the point $(2,-1,-3)$ and parallel to the lines $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z}{-4}\text{and}\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is

• A. $8x+14y+13z+37=0$
• B. $8x-14y-13z-37=0$
• C. $8x-14y-13z+37=0$
• D. $8x-14y+13z+37=0$
• E. *$x+2y+2z+6=0$

Answer 1

The correct option is E *$x+2y+2z+6=0$

Given lines: l:$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z}{-4}$ and m:$\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$
Whose d.r.s are $(2,3,-4)$ and $(2,-3,2)$
Thus, the vectors vectors parallel to given lines are:
$b_1=2\hat{i}+3\hat{j}-4\hat{k}$ parallel to l and
$b_2=2\hat{i}-3\hat{j}+2\hat{k}$ parallel to m.
Then the vector normal to both the vectors $b_1$ and $b_2$ is
$\overrightarrow{n}$=$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& -4\\ 2& -3& 2\end{array}\right|$
$=\hat{i}(-8)-\hat{1}4j-13\hat{k}$
Thus, the d.r.s of normal is $(-8,-14,-13)$
Equation of line passing through $(x_1,y_1,z_1)=(2,-1,-3)$ and having d.r.s of normal $(a,b,c)=(-8,-14,-13)$ is $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
$\Rightarrow -8(x-2)-14(y+1)-13(z+3)=0$
$\Rightarrow -8x+16-14y-14-13z-39=0$
$-8x-14y-13z-37=0$
$\therefore 8x+14y+13z+37=0$ is the required plane equation.