Question

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

• A. $8x-y+5z-8=0$
• B. $8x+y-5z-7=0$
• C. $x-8y+3z+6=0$
• D. $8x+y-5z+7=0$
• E. $x+y+z-6=0$

Answer 1

Answer: (B)

The equation of plane containing the line $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{3}$ and $\frac{x}{2}=\frac{y-2}{-1}=\frac{x+1}{3}$
The required equation is $\left|\begin{array}{ccc}x-1& (y+1)& 3\\ x& y-2& 3+1\\ 2& -1& 3\end{array}\right|=0$
$\Rightarrow$ $(x-1)(3y-6+z-1)-(y+1)$ $(3x-2z-2)+z(-x-2y+4)=0$ $-5x-3y-z+5+2y-3x+2z+2+4z=0$
$\Rightarrow$ $-8x-y+5z+7=0$
$\Rightarrow$ $8x+y-5z-7=0$