If A→=i^+2k^−k^,B→=−i^+j^−2k^ then angle between A→ and B→ is
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a
π/2
b
0
c
π
d
π/3
answer is D.
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Detailed Solution
A→=i^+2j^−k^;B→=i^+j^−2k^ A=|A→|=12+22(−1)2=6 B=|B→|=(−1)2+12(−2)2=6 A→.B→=(i^+2j^−k^).(−i^+j^−2k^) =−1+2+2=3 If θ is angle between A→ and B→ , then A→.B→=ABcosθ 3=66cosθ cosθ=36=12 or θ=π3