Discussion Forum

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$  lie in the plane x + 3y - αz+ β = 0. Then (α, β) equals

Answer:

Explanation:

$\because the line \frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane x + 3y - αz+ β = 0.

$\therefore Pt(2,1,-2)$ lies on the plane i.e. 2 + 3 + 2α+ β = 0

or 2α + β + 5 = 0 ....(I)

Also normal to plane will be perpendicular to line

.'. 3x 1 - 5 x 3 + 2 x (-α) = 0 → α = -6 

From equation (I) then, β = 7

.'. (α, β) = (-6, 7)