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The two lines $x = a y + b, z = c y + d and x = a^{'} y + b^{'}, Z = c^{'} y + d^{'}$ will be perpendicular if and only if

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a

aa'+bb'+cc'+1=0

b

aa'+bb'+cc'=0

c

aa'+bb'+cc'-1=0

d

aa'+cc'+1=0

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detailed solution

Correct option is D

The given lines in symmetric forms are x−ba=y=z−dc,x−b'a'=y=z−d'c'Two lines x−x1a1=y−y1b1=z−z1c1 and x−x2a2=y−y2b2=z−z2c2 are to be perpendicular if and only if a1a2+b1b2+c1c2=0Hence, the condition that the given two lines are to be perpendicular is aa'+cc'+1=0

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