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

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cos−1(13)

cos−1(53)

sin−1(23)

sin−1(53)

answer is D.

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Detailed Solution

If the angle between the vectors is ‘θ ’ then cosθ=A→.B→|A→|.|B→| ⇒cosθ=42+24−12(36+36+9)(49+16+16) cosθ=54(9)(9)=23 ∴sinθ=53 ⇒θ=sin−1(53)

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