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Question

Suppose the direction cosines of two lines are given by al+bm+cn=0 and fmn+gln+hlm=0 where f,g,h,a,b,c are arbitrary constants and l,m,n are direction cosines of the line. On the basis of the above information the given lines will be perpendicular if:

A
fagb+hc=0
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B
fa+gbhc=0
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C
fa+gb+hc=0
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D
fa+gb+hc=0
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Solution

The correct option is D fa+gb+hc=0
al+bm+cn=0.....(1)
fmn+gln+hlm=0....(2)
n=(al+bmc)
Eliminating n in the above equation,
(fm+gl)[(al+bmc)]+hlm=0
agl2+(bg+afch)lm+bfm2=0
l1l2m1m2=bfag
Similarly, m1m2n1n2=cgbh
l1l2fa=m1m2gb=n1n2hc
But for lines to be perpendicular, l1l2+m1m2+n1n2=0fa+gb+hc=0

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