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Question

The distance between the line r=2^i2^j+3^k+s(^i^j+4^k) and the plane r.(^i+5^j+^k)=5 is

A
1033
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B
109
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C
103
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D
310
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Solution

The correct option is A 1033
The vector along normal to the plane is ^i+5^j+^k=n say
The vector along the line is b=^i^j+4^k and n.b=(^i+5^j+^k).(^i^j+4^k)
=15+4=0
The line is parallel to the plane .The distance of the line from the plane is the distance of the point (2,2,3) from the plane r.(^i+5^j+^k)=5
i.e,x+5y+z=5
|210+35|12+52+1=1027=1033

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