Question

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

Answer 1

Hint: The parallelogram law of addition of vectors states that if two vectors are considered to be the two adjacent sides of a parallelogram with their tails meeting at the common point, then the diagonal of the parallelogram originating from the common point will be the resultant vector. By drawing a proper diagram and using geometry, we can solve this question.

Complete step-by-step answer:

The parallelogram law of vector addition states that if two vectors are considered to be the two adjacent sides of a parallelogram with their tails meeting at the common point, then the diagonal of the parallelogram originating from the common point will be the resultant vector.

Hence, let us draw a diagram.

In the figure we have two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ with magnitudes as $OP$ and $OR$ respectively. $\overrightarrow{A}$ and $\overrightarrow{B}$ make an angle $\theta $ between them. To get the resultant by the parallelogram law of vector addition, we draw the complete parallelogram $OPQR$.

Therefore, according to the parallelogram law of vector addition, $\overrightarrow{C}$ is the resultant vector of $\overrightarrow{A}$ and $\overrightarrow{B}$. The resultant has a magnitude $OQ$ and makes an angle of $\varphi $ with $\overrightarrow{B}$.

Now, we have to find the magnitude of the resultant, that is, we have to find $OQ$. To do that, we will employ geometry. Thus, we draw a perpendicular $QS$ to meet $S$ extended from $OR$.

Hence, in triangle $OQS$ using Pythagoras theorem, we get,

$O{{Q}^{2}}=Q{{S}^{2}}+O{{S}^{2}}$  -(1)

Now, $SQ=RQ\sin \theta =OP\sin \theta $ -(2)

As we know, in a parallelogram, opposite sides are equal in length

$\left( \because RQ=OP \right)$ 

Also, $OS=OR+RS=OR+RQ\cos \theta =OR+OP\cos \theta $ -(3)

$\left( \because RQ=OP \right)$ 

Hence, using (2) and (3) in (1), we get,

$O{{Q}^{2}}={{\left( OP\sin \theta \right)}^{2}}+{{\left( OR+OP\cos \theta \right)}^{2}}$

$\therefore O{{Q}^{2}}=O{{P}^{2}}{{\sin }^{2}}\theta +O{{R}^{2}}+2OR.OP\cos \theta +O{{P}^{2}}{{\cos }^{2}}\theta $

$\therefore O{{Q}^{2}}=O{{P}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+O{{R}^{2}}+2OR.OP\cos \theta $

$\therefore O{{Q}^{2}}=O{{P}^{2}}\left( 1 \right)+O{{R}^{2}}+2OR.OP\cos \theta $ -$\left( \because {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \right)$

$\therefore O{{Q}^{2}}=O{{P}^{2}}+O{{R}^{2}}+2OR.OP\cos \theta $

Square rooting both sides, we get,

$\sqrt{O{{Q}^{2}}}=\sqrt{O{{P}^{2}}+O{{R}^{2}}+2OR.OP\cos \theta }$

$\therefore OQ=\sqrt{O{{P}^{2}}+O{{R}^{2}}+2OR.OP\cos \theta }$ -(4)

Now, as defined earlier,

$OQ=\left| \overrightarrow{C} \right|,OP=\left| \overrightarrow{A} \right|,OR=\left| \overrightarrow{B} \right|$

Therefore, putting these values in (4), we get,

$\left| \overrightarrow{C} \right|=\sqrt{{{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|\cos \theta }$

Hence, we have got the value for the magnitude of the resultant.

Now, we will find the angle made by the resultant, that is, find out its direction.

Now, in triangle OQS,

$\tan \varphi =\dfrac{QS}{OS}=\dfrac{QS}{OR+RS}$

Now, using (2) and (3), we get,

$\tan \varphi =\dfrac{OP\sin \theta }{OR+OP\cos \theta }$

Now, as defined earlier,

$OQ=\left| \overrightarrow{C} \right|,OP=\left| \overrightarrow{A} \right|,OR=\left| \overrightarrow{B} \right|$

$\therefore \tan \varphi =\dfrac{\left| A \right|\sin \theta }{\left| B \right|+\left| A \right|\cos \theta }$

$\therefore \varphi ={{\tan }^{-1}}\left( \dfrac{\left| A \right|\sin \theta }{\left| B \right|+\left| A \right|\cos \theta } \right)$

Hence, we have also found out the angle made by the resultant with one of the vectors and thus found out its direction relative to that vector.

Note: Students should be aware of the proper originating points and the way of drawing the vectors in the parallelogram law of addition of vectors. Often students get confused between the triangle law of addition and the parallelogram law of addition of vectors. However, the way of drawing the vectors in both is different. In triangle law, the individual component vectors are joined head to tail while in parallelogram law the individual component vectors are joined tail to tail. Students must properly know the way of drawing and adding in both these methods properly as these form the basis of all further studies on vectors.