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Question

If the line x23=y15=z+22 lies in the plane x+3yαz+β=0. Then (α,β) equals


A

(6, 7)

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B

(6, 17)

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C

(5, 5)

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D

(5, 15)

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Solution

The correct option is A

(6, 7)


The normal to the plane is perpendicular to the line.

Normal to the given plane is ^i+3^jα^k=0

So, (3^i5^j+2^k).(^i+3^jα^k)=0

3152α=0α=6

Also, as the line lies in the plane, the point (2, 1, 2) also lies in the plane.

1(2)+3(1)(6)(2)+β=0β=7

(α, β)=(6, 7)


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