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. have no common point

Answer 1

The correct option is B have a line in common

D.C's of line of intersection of first two planes are l,m,n $\frac{l}{1}=\frac{m}{6}=\frac{n}{-4}$
$6\left(\frac{1}{\sqrt{53}}\right)+5\left(\frac{6}{\sqrt{53}}\right)+9\left(\frac{-4}{\sqrt{53}}\right)=0$
Three planes have a line in common.
On solving the 3 equations are found to be linearly dependent and no unique solution is there So, planes do not meet at a point.
For planes to be perpendiulcar
$l_1+l_2+m_1{m}_2+n_1{n}_2=0$
$\therefore$ we can easily see no two planes are perpendicular.
Now, $e{q}^n$ of a plane passing through the intersect of first two planes is
$(4y+6x-5)+\lambda (2x+3y+5z-5)=0$
$\Rightarrow \left(2\lambda \right)x+(4+3\lambda )y+(6+5\lambda )z-5-5\lambda =0$ -(1)
If $3^{rd}$ plane passes through this line of intersection.
then, $\frac{2\lambda }{6}=\frac{4+3\lambda }{5}=\frac{6+5\lambda }{9}=\frac{5+5\lambda }{10}$
$\Rightarrow \lambda =3$
$\therefore$ Three plane pass through a common line.