Question

State and prove parallelogram law of vector addition and write its special cases.

Answer 1

According to parallelogram law of vector addition

1. If two vectors are acting simultaneously at a point, then it can be represented both in magnitude and direction by the adjacent sides drawn from a point.
2. If two vectors are considered to be the adjacent sides of a parallelogram and the resultant of these two vectors is given by the vector that is diagonal and passes through the point of contact of the two vectors.
3. The magnitude of the resultant vector is given by, $R=\sqrt{P^2+Q^2+2PQ\cos \theta }$

Proof of the parallelogram law of vector addition

Let $\overrightarrow{P}$ and $\overrightarrow{Q}$ be the two vectors represented by two lines $\overrightarrow{OA}$ and $\overrightarrow{OD}$ drawn from the same point.

Let us complete the parallelogram and name it $OABD$.

From the figure it is clear that $OA\parallel DB$ and $OD\parallel AB$ and $BC\perp OC$

Consider $∆ABC$

$$\begin{gathered} \cos \theta =\frac{AC}{AB} \\ \Rightarrow AC=AB\cos \theta \\ \Rightarrow AC=OD\cos \theta \left(\because AC=OD=Q\right) \\ \Rightarrow AC=Q\cos \theta ----------------------\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} \sin \theta =\frac{BC}{AB} \\ \Rightarrow BC=AB\sin \theta \\ \Rightarrow BC=OD\sin \theta \left(\because AC=OD=Q\right) \\ \Rightarrow BC=Q\sin \theta ------------------------\left(2\right) \end{gathered}$$

Now, to find the magnitude of resultant, consider $∆OBC$,

$$\begin{gathered} O{B}^2=O{C}^2+B{C}^2 \\ \Rightarrow O{B}^2={\left(OA+AC\right)}^2+B{C}^2 \\ \Rightarrow R^2={(P+Q\cos \theta )}^2+{\left(Q\sin \theta \right)}^2\left(From\left(1\right)and\left(2\right)\right) \\ \Rightarrow R^2={P^2+\left(Q\cos \theta \right)}^2+2PQ\cos \theta +{\left(Q\sin \theta \right)}^2 \\ \Rightarrow R^2=P^2+Q^2({\sin }^2\theta +{\cos }^2\theta )+2PQ\cos \theta \\ \Rightarrow R^2=P^2+Q^2+2PQ\cos \theta \left(({\sin }^2\theta +{\cos }^2\theta )=1\right) \\ \Rightarrow R=\sqrt{P^2+Q^2+2PQ\cos \theta } \end{gathered}$$

Hence proved.

Some special cases of the parallelogram law of vector

1. When the Two Vectors are Parallel (Same Direction)

If vectors $P$ and are parallel, then we have $\theta =0°$. Substituting this in the formula of Parallelogram Law of Vector Addition, we have

$\begin{array}{rcl}\left|R\right|& =& \sqrt{P^2+2PQ\cos 0+Q^2}\\ & =& \sqrt{P^2+2PQ+Q^2}\left[\because \cos 0=1\right]\\ & =& \sqrt{{\left(P+Q\right)}^2}\\ & =& P+Q\end{array}$

$\begin{array}{rcl}\beta & =& {\tan }^{-1}\left[\frac{Q\sin 0}{P+Q\cos 0}\right]\\ & =& {\tan }^{-1}\left[\frac{0}{P+Q}\right]\\ & =& 0\end{array}$

2. When the Two Vectors are Acting in Opposite Direction

If vectors P and Q are acting in opposite directions, then we have $\theta =180°$. Substituting this in the formula of Parallelogram Law of Vector Addition, we have

$\begin{array}{rcl}\left|R\right|& =& \sqrt{P^2+2PQ\cos 180+Q^2}\\ & =& \sqrt{P^2-2PQ+Q^2}\left[\because \cos 180=-1\right]\\ & =& \sqrt{{\left(P-Q\right)}^2}or\sqrt{{\left(Q-P\right)}^2}\\ & =& P-QorP-Q\end{array}$

$\begin{array}{rcl}\beta & =& {\tan }^{-1}\left[\frac{Q\sin 180}{P+Q\cos 180}\right]\\ & =& {\tan }^{-1}\left[\frac{0}{P+Q}\right]\\ & =& 0\end{array}$

3. When the Two Vectors are Perpendicular

If vectors P and Q are perpendicular to each other, then we have $\theta =90°$. Substituting this in the formula of Parallelogram Law of Vector Addition, we have

$\begin{array}{rcl}\left|R\right|& =& \sqrt{P^2+2PQ\cos 90+Q^2}\\ & =& \sqrt{P^2+0+Q^2}\left[\because \cos 90=0\right]\\ & =& \sqrt{{\left(P+Q\right)}^2}\end{array}$

$\begin{array}{rcl}\beta & =& {\tan }^{-1}\left[\frac{Q\sin 90}{P+Q\cos 90}\right]\\ & =& {\tan }^{-1}\left[\frac{Q}{P+0}\right]\\ & =& {\tan }^{-1}\left[\frac{Q}{P}\right]\end{array}$