Question

22. If |Āx B |= √3A.B, then the value of A+B is
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Answer 1

|Ā x B |=√3A.B

A B sin theta = √3 A B cos theta

(AB will cancel AB)

tan theta =√3

then theta = 60°

A+B= √A^2 + B^2 + 2AB cos theta

= √A^2 + B^2 +2AB 1/2

= √A^2 + B^2

therefore,

√A^2 + B^2 is your answer

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Answer 2

Answer:

The answer is  √|A|^2 + |B|^2 + AB

Explanation:

Given , |Ā x B |=√3A.B

=> AB sin theta = √3 AB cos theta

=> tan theta =√3

So , theta = 60

|A| + |B| = √(A^2 + B^2 + 2AB cos theta)

=> |A| + |B| = √(A^2 +B^2 + AB)

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