Browse Questions

# If the line $\large\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z-2}{2}$ lie in the plane $x+3y-\alpha z+\beta=0, \;\;then\:\:(\alpha,\beta)=?$

$\begin {array} {1 1} (A)\;(-6, -17) & \quad (B)\;(-5, 5) \\ (C)\;(6, -17) & \quad (D)\;(5, -15) \end {array}$

Given equation of the line is $\large\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z-2}{2}$
Equation of the plane: $x+3y-\alpha z+\beta=0$
$\Rightarrow$ A known point on the line is $(2,1,2)$
D.R. of the line is $(3,-5,2)$ and D.R. of normal to the plane is $(1,3,-\alpha)$
Since the line lie on the plane every point on the line lie on the plane.
$\therefore \:(2,1,2)$ satis fy the plane eqn.
$\Rightarrow\:2+3-2\alpha+\beta=0\:\:or\:\:2\alpha-\beta=5$.........(i)
Also the normal to the plane is $\perp$ to every line on the plane.
$\therefore\:(3,-5,2).(1,3,-\alpha)=0$ $\Rightarrow\:3-15-2\alpha=0.\:or\:\:\alpha=-6............(ii)$
and $\beta=-17$
$\Rightarrow\:(\alpha,\beta)=(-6,-17)$