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If vectors ` vec A , vec` B` and vec C`have magnitudes `5, 12 and 13 units and ` vec A + vec B =vec C`, find theangle between ` vec B and vec C`. |
Answer» As ` vec A + vec B = vec C` :. ` vec A = vec C - vec B` Then ` vec A . Vec A =( vec C - vec B) .(vec C - vec B)` `=vec C. vec C -2 vec C . Vec B + vev B . Vec B` or `A^(2) =C^(2) -2 CB cos theta + B^(2)` or `cos theta = (C^(32) + B^(2) -A%(2))/(2 C B)` `=(13^(2) + 12^(2) 12^(2) -5^(2))/(2xx 13 xx 12)= 9288)/(312) =(12) /(13)` `theta = cos ^(-1) (12 // 13)`. |
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