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Question

If two sides of a triangle are represented by the vectors 3(^a×b) and b(^ab)^a where b is a non-zero vector and ^a is a unit vector in the direction of a, then the angles of triangle are

A
tan1(13); tan1(12); tan1(3+2123)
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B
tan1(3); tan1(13); cot1(0)
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C
tan1(3); tan1(2); tan1(3+2231)
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D
tan1(1); tan1(1); cot1(0)
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Solution

The correct option is B tan1(3); tan1(13); cot1(0)
Let V=3(^a×b)
and V2=b(^ab)^a=(^a×b)×^a
V1V2=3(^a×b)[(^a×b)×^a]
V1V2=0
So, the angle between V1 and V2 is A=90


Now, using sine law in ΔABC,
b(^ab)^asinθ=3^a×bcosθ
tanθ=13|b(^ab)^a||a×b|
tanθ=13|(^a×b)×^a||a×b|
tanθ=13|^a×b||a|sin90|a×b|
tanθ=13
θ=π6
The angles of ΔABC are π3,π6,π2 or tan1(3); tan1(13); cot1(0)

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