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Q.

|A→ ×B→|2+|A→.B→|2 =

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a

zero

b

A2B2

c

AB

d

AB

answer is B.

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Detailed Solution

Let θ be angle between vectors A→ and B→ ∴ |A→ ×B→| = ABsinθand |A→ .B→| = ABcosθ |A→×B→|2 = |A.→B→ |2 = (ABsinθ)2+ABcosθ2= A2B2sin2θ+A2B2cos2θ = A2B2

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