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Question

If a+2b+3c=0, then a×b+b×c+c×a=ka×b,
Where k is equal to ?

A
0
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B
1
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C
2
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D
3
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Solution

The correct option is C 2
Given,
a+2b+3c=0a=(2b+3c)a×a=(2(b×a)+3(c×a))0=(2b×a+3(c×a))..........(i)
Again,
2b=(a+3c)2b×b=(a×b+3c×b)0=(a×b+3c×b)..........(ii)
equation (i) and (ii)
(2b×a+3c×a)+(a×b)+3c×b=03a×b3c×a3b×c=0
We know that,
a×b=b×a
a×b=c×a+b×c...............(iii)
Now,ATQ;
a×b+b×c+c×a=Ka×b
1=2a×b=ka×b {from equation (iii)}
k=2
Option C is correct answer.

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