Question

Derive an expression for the magnitude and direction of the resultant vector using parallelogram law

Answer 1

Parallelogram law states that if two vectors are considered to be the adjacent sides of a Parallelogram, then the resultant of two vectors is given by the vector which is a diagonal passing through the point of contact of two vectors.
In the figure $\overrightarrow{P}$ and $\overrightarrow{Q}$ are two vectors.with magnitudes equal to length OA and OB respectively and making angle $\theta$ between them. Complete the parallelogram, OACB,
Join diagonal OC , that makes angle $\alpha$ with vector $\overrightarrow{P}$.
According to parallelogram law of vectors the resultant is represented by the diagonal passing through the point of contact of two vectors.
To find the magnitude of resultant , produce a perpendicular CD to meet OA produced to D.
From $△$ OCD,
$$O{C}^2=O{D}^2+C{D}^2$$
Now $\overrightarrow{C}D$=$\overrightarrow{A}C$ $\sin \theta$=$\overrightarrow{Q}\sin \theta$
$AD=\overrightarrow{A}C\cos \theta =\overrightarrow{Q}\cos \theta$
$\overrightarrow{O}D=\overrightarrow{O}A+\overrightarrow{A}D=\overrightarrow{P}+\overrightarrow{Q}\cos \theta$
Putting these values and representing resultant vector OC by $\overrightarrow{R}$, magnitude of the resultant is given by
$$R^2=(\overrightarrow{P}+\overrightarrow{Q}\cos \theta {)}^2+(\overrightarrow{Q}\sin \theta {)}^2={\overrightarrow{P}}^2+{\overrightarrow{Q}}^2+2\overrightarrow{P}\overrightarrow{Q}\cos \theta$$
In $△$ OCD,
$$\tan \alpha =\frac{CD}{OD}=\frac{\overrightarrow{Q}\sin \theta }{\overrightarrow{P}+\overrightarrow{Q}\cos \theta }$$
Resultant acts in the direction making an angle $\alpha ={\tan }^{-1}\left(\frac{\overrightarrow{Q}\sin \theta }{\overrightarrow{P}+\overrightarrow{Q}\cos \theta }\right)$ with direction of vector P .