Q.State parallelogram law of vector addition. Using the law, derive an expression for the magnitude and direction of the resultant of two vectors inclined at an angle $\theta$. What will be the magnitude and direction if $\theta = \dfrac{\pi}{2}$? OR Deduce the following relations analytically for a uniform motion along a straight line, where the terms have their usual meanings. i) $v = u + at$ ii) $s = ut + \dfrac{1}{2}at^{2}$ iii) $v^{2} - u^{2} = 2as$

Two vectors from a common point add as the diagonal of the parallelogram they form; the general formula reduces to $R=\sqrt{P^2+Q^2}$ at a right angle.

Parallelogram law of vector addition: If two vectors $\vec{P}$ and $\vec{Q}$ acting simultaneously at a point are represented, in magnitude and direction, by the two adjacent sides of a parallelogram drawn from that point, then their resultant $\vec{R}$ is represented, in magnitude and direction, by the diagonal of the parallelogram passing through the same point.

Derivation: Let $\vec{P}$ and $\vec{Q}$ be represented by sides $OA$ and $OB$ of a parallelogram $OACB$, inclined at angle $\theta$ to each other, with the resultant $\vec{R} = \overrightarrow{OC}$ being the diagonal from $O$. Drop a perpendicular from $C$ to the extended line $OA$, meeting it at $N$. In the right triangle $ONC$: $CN = Q\sin\theta$ and $AN = Q\cos\theta$.

In right triangle $ONC$, by Pythagoras (with $ON = OA+AN = P+Q\cos\theta$):

$R^2 = ON^2+CN^2 = (P+Q\cos\theta)^2+(Q\sin\theta)^2 = P^2+2PQ\cos\theta+Q^2\cos^2\theta+Q^2\sin^2\theta$

$R^2 = P^2+Q^2+2PQ\cos\theta \implies R = \sqrt{P^2+Q^2+2PQ\cos\theta}$

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