Question

How do you find the angle of resultant vector?

Answer 1

Step 1. Draw the diagram of showing angle of resultant vector:

Let $P$ and $Q$ be two vectors acting at the same instant at a point and represented both in magnitude and direction by two adjacent sides $OA$ and $OD$ of a parallelogram $OABD$ as shown in figure.

Let $\theta$ be the angle between $P$ and $Q$ and $R$ be the resultant vector. Then, as stated by the parallelogram law of vector addition, diagonal $OB$ represents the resultant of $P$ and $Q$.

So, we have

$R=P+Q$

Now, expand $A$ to $C$ and draw $BC$ perpendicular to $OC$.

From triangle $OCB$,

$O{B}^2=C^2+B{C}^2$

$\Rightarrow$$O{B}^2={(OA+AC)}^2+B{C}^2$ ......(i)

In triangle $ABC$, we have

$\cos \theta =\frac{AC}{AB}$

$\Rightarrow$$AC=AB\cos \theta$

$\Rightarrow$$AC=OD\cos \theta$

$=Q\cos \theta$ $[\because AB=OD=Q]$

Also,

$\cos \theta =\frac{BC}{AB}$

$\Rightarrow$$BC=AB\sin \theta$

$\Rightarrow$$BC=OD\sin \theta$

$=Q\sin \theta$ $[\because AB=OD=Q]$

Step 2. Find the Magnitude of resultant:

Substituting value of $AC$ and $BC$ in (i), we get

$R^2={(P+Q\cos \theta )}^2+{\left(Q\sin \theta \right)}^2$

$\Rightarrow$$R^2=P^2+2PQ\cos \theta +Q^2{\cos }^2\theta +Q^2{\sin }^2\theta$

$\Rightarrow$$R^2=P^2+2PQ\cos \theta +Q^2$

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

​Which is the magnitude of resultant.

Step 3. Find the Direction of resultant :

Let $\varphi$ be the angle made by resultant $R$ with $P$. Then,

From triangle $OBC$, we have

$\begin{array}{rcl}\tan \varphi & =& \frac{BC}{OC}\\ & =& \frac{BC}{OA+AC}\end{array}$

$\Rightarrow$$\tan \varphi =\frac{Q\sin \theta }{P+Q\cos \theta }$

$\Rightarrow$ $\varphi ={\tan }^{-1}\left(\frac{Q\sin \theta }{P+Q\cos \theta }\right)$

which is the direction of resultant.

Hence, we can find the angle of resultant vector as $\mathit{\Phi }\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{Q}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{P}\mathbf{+}\mathbf{Q}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}\mathbf{\right)}$