Question

Let $P=0$ be the equation of a plane passing through the line of intersection of the planes $2x-y=0$ and $3z-y=0$ and perpendicular to the plane $4x+5y-3z=8$. Then the points which lie on the plane $P=0$ is/are:

• A. $(0,9,17)$
• B. $(\frac{1}{2},1,\frac{1}{3})$
• C. $(1,3,-4)$
• D. $(\frac{1}{7},1,\frac{1}{9})$

Answer 1

Answer: (A), (B)

$(\frac{1}{2},1,\frac{1}{3})$

$P$ is the equation of the plane passing through the line of intersection of the planes $2x-y=0$ and $3z-y=0$. i.e.,
$P:p_1+\lambda p_2=0$
$\Rightarrow 2x-y+\lambda (-y+3z)=0$
$\Rightarrow 2x-(\lambda +1)y+3\lambda z=0$
$P$ is perpendicular to$4x+5y-3z=8$
$8-5(\lambda +1)-9\lambda =0$
$3-14\lambda =0$
$\lambda =\frac{3}{14}$
The equation of the plane $P$ is $28x-17y+9z=0.$
Option $A,B$ only satisfies the above equation.