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Find the angle between two vectors ` vec aa n d vec b`with magnitudes `sqrt(3)`and `2`respectively and such that ` vec adot vec b=sqrt(6.)` |
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Answer» Let `theta` be the angle between `vec(a)` and `vec(b)`. Then, `vec(a).vec(B)=sqrt(6) implies |vec(a)||vec(b)| cos theta = sqrt(6) implies (sqrt(3)) (2) cos theta =sqrt(6)` `implies cos theta=sqrt(6)/(2sqrt(3))=1/sqrt(2) implies theta = cos^(-1) (1/sqrt(2))=pi/4` Hence, the required angle is `pi/4`. |
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