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Q.

The angle between two vectors given by 6i^+6j^−3k^   and  7i^+4j^+4k^ is

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a

cos-1(13)

b

cos-1(513)

c

sin-1(23)

d

sin-1(53)

answer is D.

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Detailed Solution

cosθ = A→.B→AB = 42+24−1236+36+949+16+16 = 5481cosθ = 5481∴sinθ = 53  or  θ = sin-1(53)
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