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Questions  

|A×B|2+|A.B|2  =

a
zero
b
A2B2
c
AB
d
AB

detailed solution

Correct option is B

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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