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Question

The angle between the line x-12=y-21=z+3-2 and the plane x+y+4=0 is


A

0°

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B

30°

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C

45°

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D

90°

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Solution

The correct option is C

45°


Explanation for correct option:

Given the line x-12=y-21=z+3-2and the plane x+y+4=0

We know that the general equation of plane is , ax+by+cz+d=0.

Then the normal vector n to the plane can be defined as n=i^+j^ using the equation of the plane x+y+4=0.

And the vector b is the direction ratios along the line is defined as b=2i^+j^-2k^.

Now we know that ,

cos(90-θ)=|n.b||n||b|=2·1+1·11+14+1+4=2+11+14+1+4=329=32×3=12

We know that cos90-θ=sinθ, using the same we have,

cos(90-θ)=12sin(θ)=12θ=sin-112θ=45°

Therefore the angle between the line and the plane is 45°

Hence, the correct answer is option (C).


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