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Q.

The angle between the vectors 2i+3j+k and -3i+6k is

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a

0°

b

45°

c

60°

d

90°

answer is D.

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Detailed Solution

cos⁡θ=A⋅BABA.B = (2 i + 3j - k) . (- 3i + 6 k)=2(-3)+3(0) +1(6) = 0∴cos⁡θ=014×45=0 or θ=90∘

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