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Question

The vector from the point ^i+2^j+6^k to the straight line through the point (2,3,4) and parallel to the vector 6^i+3^j+4^k, is

A
29761i+11861j53461k
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B
29761i11861j53461k
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C
29761i+11861j+53461k
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D
none of these
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Solution

The correct option is A 29761i+11861j53461k
Equation of the straight line will be
x26=y33=z+44=t
Therefore
x=6t+2
y=3t+3
z=4t4
Therefore, let any point P on the line be
P=(6t+2,3t+3,4t4)
And A=(1,2,6)
Therefore
AP=(6t+3)i+(3t+1)j+(4t10)k
Now for perpendicular distance
AP.(6i+3j+4k)=0
Or
((6t+3)i+(3t+1)j+(4t10)k).(6i+3j+4k)=0
6(6t+3)+3(3t+1)+4(4t10)=0
36t+18+9t+3+16t40=0
61t19=0
t=1961
Hence
AP=29761i+11861j53461k
Hence
d=|AP|

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