Question

The equation of the plane containing the two lines

\[\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 0}{3} \text{ and }\frac{x}{- 2} = \frac{y - 2}{- 3} = \frac{z + 1}{- 1}\]

 

 

Options

•  8x + y − 5z − 7 = 0

•  8x + y + 5z − 7 = 0

• 8x − y − 5z − 7 = 0

•  None of these

   

Answer 1

 None of these

\[\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z- 0}{3}\] and 

\[\frac{x}{- 2} = \frac{y - 2}{- 3} = \frac{z + 1}{- 1}\]

Now, if these two lines lie on a plane, so the direction ratio of lines will be perpendicular to the plane's normal vector.