Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3 y-\alpha z+\beta=0$. Then $(\alpha, \beta)$ equals

JEE Main · 2009 · 12 Apr 2009 · Offline · Three Dimensional Geometry

Let the line

\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}

lies in the plane

x + 3 y - α z + β = 0

. Then

(α, β)

equals

(A) $(6, - 17)$
(B) $(- 6, 7)$ ✓ Correct
(C) $(5, - 15)$
(D) $(- 5, 15)$

Answer: (B)

Solution

For the line to lie in the plane its direction must be perpendicular to the plane's normal and one of its points must satisfy the plane. The direction

(3, - 5, 2)

and normal

(1, 3, - α)

give

3 (1) - 5 (3) + 2 (- α) = 0

, so

3 - 15 - 2 α = 0

and

α = - 6

. Putting the point

(2, 1, - 2)

into

x + 3 y - α z + β = 0

gives

2 + 3 - 12 + β = 0

, so

β = 7

, and hence

(α, β) = (- 6, 7)

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option

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