Question

The equation of plane containing intersecting lines $\frac{x+3}{3}=\frac{y}{1}=\frac{z-2}{2}$ and $\frac{x-3}{4}=\frac{y-2}{2}=\frac{z-6}{3}$ is _________.

• A. $x+y+z+5=0$
• B. $2x-y+z+9=0$
• C. $x+y-2z+7=0$
• D. $x+2y-2z+9=0$

Answer 1

Answer: (C)

$x+y-2z+7=0$

Let $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$ be the direction vectors of the two lines respectively.

$\overrightarrow{d_1}=3\hat{i}+\hat{j}+2\hat{k}$

$\overrightarrow{d_2}=4\hat{i}+2\hat{j}+3\hat{k}$

Let $\overrightarrow{n}$ be a vector perpendicular to both lines.

$\overrightarrow{n}=\overrightarrow{d_1}\times \overrightarrow{d_2}$

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& 2\\ 4& 2& 3\end{array}\right|$

$=-\hat{i}-\hat{j}+2\hat{k}$

The second line passes through the point $(3,2,6)$.

So the plane also contains the point $(3,2,6)$.

Equation of plane passing through $(x_1,y_1,z_1)$ and having direction ratios $(a,b,c)$ for its normal is given by

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$

$⟹(-1)(x-3)+(-1)(y-2)+\left(2\right)(z-6)=0$

$⟹x+y-2z+7=0$

Hence, the answer is option (C).