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Q.Find the magnitude of the resultant of two vectors A⃗\vec{A} and B⃗\vec{B} in terms of their magnitudes and the angle θ\theta between them. Obtain the condition for maximum and minimum values of the resultant vector.

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2026Subjective· 5mImportance★★★★★
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Parallelogram law: R=A2+B2+2ABcos⁡θR = \sqrt{A^{2}+B^{2}+2AB\cos\theta}. cos⁡θ\cos\theta is largest (+1+1) at 0∘0^\circ giving Rmax=A+BR_{max}=A+B, and smallest (−1-1) at 180∘180^\circ giving Rmin=∣A−B∣R_{min}=|A-B|.

Let A⃗\vec A and B⃗\vec B be inclined at angle θ\theta (parallelogram law). Representing A⃗\vec A by OP and B⃗\vec B by OQ and completing the parallelogram, the resultant OS has magnitude found by dropping a perpendicular from S to the extended OP at N:

R2=(A+Bcos⁡θ)2+(Bsin⁡θ)2=A2+B2+2ABcos⁡θ.R^{2} = (A + B\cos\theta)^{2} + (B\sin\theta)^{2} = A^{2} + B^{2} + 2AB\cos\theta.

R=A2+B2+2ABcos⁡θ.\boxed{R = \sqrt{A^{2} + B^{2} + 2AB\cos\theta}.}

Maximum value: RR is largest when cos⁡θ\cos\theta is largest, i.e. cos⁡θ=+1\cos\theta = +1 (θ=0∘\theta = 0^\circ, vectors parallel):

Rmax=A2+B2+2AB=(A+B)2=A+B.R_{max} = \sqrt{A^{2} + B^{2} + 2AB} = \sqrt{(A + B)^{2}} = A + B.

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