Question

The equation of the plane containing the straight line $\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z}{5}$ and perpendicular to the plane $x-y+z+2=0$ is

• A. $3x-2y+4z-2=0$
• B. $2x-4y-3z+6=0$
• C. $2x+3y+z+4=0$
• D. $x+2y+z+4=0$

Answer 1

The correct option is C $2x+3y+z+4=0$

$\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-0}{5}$ is perpendicular to the plane $x-y+z+2=0$
Required plane is parallel to the vector $\overrightarrow{v_1}=2\hat{i}-3\hat{j}+5\hat{k}$ and $\overrightarrow{v_2}=\hat{i}-\hat{j}+\hat{k}$
$\therefore$ Normal vector to the required plane is, $\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -3& 5\\ 1& -1& 1\end{array}\right|$

$=\hat{i}(-3+5)-\hat{j}(2-5)+\hat{k}(-2+3)$
$=2\hat{i}+3\hat{j}+\hat{k}$
Also, the plane contains the point $(1,-2,0)$
$\therefore$ Equation of plane is
$2(x-1)+3(y+2)+1(z-0)=0$
$\Rightarrow 2x+3y+z+4=0$