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Question

# The lines x=ay+b,z=cy+d and x=a′y+b′,z=c′y+d′ will be perpendicular if and only if

A
aa+bb+cc=0
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B
(a+a)(b+b)+c+c=0
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C
aa+cc+1=0
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D
aa+bb+cc+1=0
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Solution

## The correct option is C aa′+cc′+1=0The two given lines are:-L1≡x=ay+b,z=cy+d y=x−ba,y=z−dc we can write L1 as L1≡x−ba=z−dc=ySimilarly, L2≡x=a′y+b′,z=c′y+d′ ⇒y=x−b′a′,y=z′−d′c′L2 can be written as:- L2≡x−b′a′=z−d′c′=yDirection ratios of line L1(a,c,1)andDirection ratios of line L2(a′,c′,1)We know that two lines are perpendicular iff, product of their direction ratios are 0.Therefore, a×a′+c×c′+1×1=0 ⇒aa′+cc′+1=0Hence, C is the correct answer.

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