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{matrix} x-1 & (y+1) & 3 \\ x & y-2 & 3+1 \\ 2 & -1 & 3 \\ \end{matrix} \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 $