Question

The three planes 4y + 6z = 5; 2x + 3y + 5z = 5; 6x + 5y + 9z = 10.

• A. meet in a point
• B. have a line in common
• C. form a triangular prism
• D. mutuvally perpendicular

Answer 1

The correct option is B have a line in common

Let DC’s of line of intersection of planes 4y + 6z = 5 and 2x + 3y + 5z = 5 are l,m,n then
$0+4m+6n=0$
$2l+3m+5n=0$
$\Rightarrow \frac{l}{1}=\frac{m}{6}=\frac{n}{-4}=\frac{1}{\sqrt{53}}$
$\Rightarrow l=\frac{1}{\sqrt{53}},m=\frac{6}{\sqrt{53}},n=-\frac{4}{\sqrt{53}}$
Also, $6\left(\frac{1}{\sqrt{53}}\right)+5\left(\frac{6}{\sqrt{53}}\right)+9\left(\frac{-4}{\sqrt{53}}\right)=0$
i.e. Line of intersection is perpendicular to normal of third plane.
Hence, three planes have a line in common.