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If `vec(A)=vec(B)+vec(C )`, and the magnitudes of `vec(A)`,`vec(B)`,`vec(C )` are 5,4, and 3 units, then the angle between `vec(A)` and `vec(C )` isA. cos^(-1) ( 3/5)`B. cos^(-1) (4/5)`C. ` pi //2`D. sin ^(-1) (3/5)` |
Answer» Correct Answer - A Herem ` | vec A| = 5 , | vec B|= 4, |vec C | = 3` As ` vec A = vec B+ vec C. so vec B = vec A - vec C` As ` vec B` is now resultant of ` vec A` and ` vec C`. So it is given by ` `B^2 = A^2 + C^2 -2 AC cos theta` ` (4)^2= (5)^2 + (3)^2 - 2 xx 5 xx 3 cos theta` ` - 16 - 25 - 9 =- 30 cos theta` ` -18 =- 30 cos theta ` ` cos theta = (-18)/(-30) = 3/2 theta = cos^(-1) ( 3/5)`. |
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