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Question

If the line x-23=y-1-5=z+22 lies on the plane x+3y-αz+β=0, then α,β=


A

6,-17

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B

-6,7

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C

5,-15

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D

-5,15

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Solution

The correct option is B

-6,7


The explanation for the correct option

The given equation of the line: L1:x-23=y-1-5=z+22.

The given equation of the plane: P1:x+3y-αz+β=0.

The direction ratios of the line L1 can be given by a1,b1,c1=3,-5,2.

The direction ratios of the normal to the plane P1 can be given by a2,b2,c2=1,3,-α.

As the line L1 lies on the plane P1, thus a1a2+b1b2+c1c2=0.

31+-53+2-α=03-15-2α=0-2α=12α=-6

It is also true as the line L1 lies on the plane P1, thus 2,1,-2 is on the plane.

So, x+3y-αz+β=0

2+31--6-2+β=0α=-62+3-12+β=0β=7

Therefore, α,β=-6,7.

Hence, (B) is the correct option.


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