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If |A→×B→| = |A→.B→|, then angle between A→ and B→ will be

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a

300

b

450

c

600

d

900

answer is B.

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Detailed Solution

|A→ ×B→| = A→.B→ ⇒ ABsinθ = ABcosθ ⇒ tanθ = 1 ∴ θ = 450

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If |A→×B→| = |A→.B→|, then angle between A→ and B→ will be