Question

The equation of the plane containing the two lines $\frac{(x-1)}{2}=\frac{(y+1)}{-1}=\frac{z}{3}$and$\frac{x}{-1}=\frac{(y-2)}{3}=\frac{(z+1)}{-1}$ is

• A. $8x+y–5z–7=0$
• B. $8x+y+5z–7=0$
• C. $8x–y–5z–7=0$
• D. None of these

Answer 1

The correct option is A

$8x+y–5z–7=0$

Let $b_1$and $b_2$ be the direction vectors of the two given lines.

$\frac{(x-1)}{2}=\frac{(y+1)}{-1}=\frac{z}{3}$ and $\frac{x}{-1}=\frac{(y-2)}{3}=\frac{(z+1)}{-1}$

So,

$\overrightarrow{{\mathrm{b}}_1}=2\hat{i}-\hat{j}+3\hat{k}$

$\overrightarrow{{\mathrm{b}}_2}=-\hat{i}+3\hat{j}-\hat{k}$

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

$\overrightarrow{n}=-8\hat{i}-\hat{j}+5\hat{k}$

The point $(1,-1,0)$ lies on the plane.

Equation of plane passing through $(x_1,y_1,z_1)$ and perpendicular to a line with direction ratios $a,b,c$ is $a(x-x_1)+b(y-y_1)+c(z-z_1)=0\dots \left(i\right)$

Here $(a,b,c)=(-8,-1,5)$and$(x_1,y_1,z_1)=(1,-1,0)$

Substitute the values in (i), we get;

$$\begin{gathered} -8(x-1)-1(y+1)+5\left(z\right)=0 \\ \Rightarrow -8x+8–y–1+5z=0 \\ \Rightarrow -8x–y+5z+7=0 \\ \Rightarrow 8x+y–5z–7=0 \end{gathered}$$

Hence, option A is the answer.