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Question

The straight line
r=(^i+^j+2^k)+t(2^i+5^j+3^k) is parallel to the plane r.(2^i+^j3^k)=5. Then, the distance between the straight line and the plane is

A
914
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B
814
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C
714
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D
614
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E
514
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Solution

The correct option is A 814
Line r=(^i+^j+2^k)+t(2^i+5^j+3^k)
Plane r.(2^i+^j3^k)=5
Since n=2^i+^j3^k is normal to plane
|n|2=14
So for some k (^i+^j+2^k)+kn will be on the plane
Distance between the plane is |kn|
Solve for k by applying on plane above point
((^i+^j+2^k)+kn).n=5
(^i+^j+2^k).n+k|n|2=5
2+16+14k=5k=814
Distance between the Line and plane is |kn|=81414=814

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