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Given a=x//(y-z), b=y//(z-x), and c=z//(...

Given `a=x//(y-z), b=y//(z-x), and c=z//(x-y),w h e r e .x , y and z` are not all zero, then the value of `a b+b c+c a` is a.`0` b. `1` c.`-1""` d. none of these

A

0

B

1

C

-1

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
`a=x//(y-z ) rArr x-ay+ ax=0`
`b=y //(z-x) rArr bx+ y-bz=0`
`c=z//(x-y) rArr -cx +cy +z=0`
Since x,y,z are not all zero , the above system has non-trivial solution .So
`Delta =|{:(1,,-a,,a),(b,,1,,-b),(-c,,c,,1):}|=0`
`" or " 1+ab +bc +ca =0`
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