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Question

Let a=2^i+^j2^k and b=^i+^j. Let c be a vector such that |ca|=3|(a×b)×c|=3 and the angle between c and a×b be 30o. Then a.c is equal to

A
2
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B
5
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C
18
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D
258
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Solution

The correct option is A 2
Given,
α=2^ı+^ȷ2^k
b=^r+y
and r(¯a)=(aׯb)ׯc)=3.
Given, angle between |a×b| and c is π/b.
Given
|a×b|c12=3[|a×b|=|a|bsinθ]

|a×b||c|=6(i)
Now,
¯aׯ¯¯5=2^ı2j+k
|¯a×b|=4+4+1=3
Then, |c|=2 [from (i)]

Now, |ca|=3
Squaring both sides
|c|2+|a|22ca=9
4+92ca=9
y=2ca
ca=2

Option A

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