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Exercise · Q15

Q.Two vectors have magnitudes AA and BB. Show, using the expression for the magnitude of their resultant, that the resultant is maximum when the vectors are parallel and minimum when they are antiparallel, and state these maximum and minimum values.

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The magnitude of the resultant of two vectors of magnitude AA and BB, with angle θ\theta between them, is R=A2+B2+2ABcos⁡θR = \sqrt{A^2+B^2+2AB\cos\theta}. Since AA, BB are fixed, RR depends on θ\theta only through the term 2ABcos⁡θ2AB\cos\theta, and RR is largest exactly when cos⁡θ\cos\theta is largest, and smallest when cos⁡θ\cos\theta is smallest. Since cos⁡θ\cos\theta ranges only from −1-1 to +1+1 as θ\theta ranges from 0∘0^\circ to 180∘180^\circ: at θ=0∘\theta = 0^\circ (vectors parallel, cos⁡θ=1\cos\theta=1), Rmax⁡=A2+B2+2AB=(A+B)2=A+BR_{\max} = \sqrt{A^2+B^2+2AB} = \sqrt{(A+B)^2} = A+B; at θ=180∘\theta = 180^\circ (vectors antiparallel, cos⁡θ=−1\cos\theta=-1), $R_{\m …

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