Math miscellaneousHard
Question
Let the line 
lies in the plane x + 3y - αz + β = 0. Then(α, β)equals
lies in the plane x + 3y - αz + β = 0. Then(α, β)equalsOptions
A.(6, - 17)
B.( - 6, 7)
C.(5, - 15)
D.(- 5, 15)
Solution
Dr′s of line = (3, - 5, 2)
Dr′s of normal to the plane = (1, 3, - α)
Line is perpendicular to normal ⇒ 3(1) - 5(3) + 2(-α) = 0 ⇒ 3 -15 - 2α = 0 ⇒ 2α = - 12 ⇒ α = - 6
Also (2, 1, - 2) lies on the plane
+ 3 + 6(-2) + β = 0 ⇒ β = 7
∴ (α, β) = (-6, 7)
Dr′s of normal to the plane = (1, 3, - α)
Line is perpendicular to normal ⇒ 3(1) - 5(3) + 2(-α) = 0 ⇒ 3 -15 - 2α = 0 ⇒ 2α = - 12 ⇒ α = - 6
Also (2, 1, - 2) lies on the plane
+ 3 + 6(-2) + β = 0 ⇒ β = 7
∴ (α, β) = (-6, 7)
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