Question

State and prove parallelogram law of vector addition and determine the magnitude and direction of the resultant vector.

Answer 1

Hint: Recall the statement of the parallelogram. According to the parallelogram law of vector addition if two vectors act along two adjacent sides of a parallelogram(having magnitude equal to the length of the sides) both pointing away from the common vertex, then the resultant is represented by the diagonal of the parallelogram passing through the same common vertex and in the same sense as the two vectors. Using triangle law of vector addition proves the parallelogram law of vector addition. Also, find the magnitude of the resultant vector using cosine law.

Complete step by step solution:

We will use bold notation to describe vectors, i.e. we use $\mathbf{AB}$ to denote the vector $\overrightarrow{AB}$
Parallelogram law of vector addition: Consider the vector v represented by AB side of the parallelogram and the vector u represented by side AD of the parallelogram ABCD. Then by the parallelogram law of vector addition AB+AD=AC i.e. u+v=w
Proof:
We know that if we move a vector parallel to itself without changing the magnitude, then the new vector is equal to the original vector.
Since BC||AD and BC = AD, hence, we have BC=AD , i.e. a=u.
In triangle ACB, we have from triangle law of vector addition AB+BC=AC.
Substituting the value of AB and BC, we get
a+v=w
Hence, we have u+v=w, which is the parallelogram law of vector addition.
The angle between vectors a and v is $\pi -B$
So let $\theta =\pi -B\Rightarrow B=\pi -\theta $
Now by cosine rule in triangle ACB, we have
$\cos B=\dfrac{{{a}^{2}}+{{v}^{2}}-{{w}^{2}}}{2av}\Rightarrow \cos \left( \pi -\theta \right)=\dfrac{{{a}^{2}}+{{v}^{2}}-{{w}^{2}}}{2av}$
We know that $\cos \left( \pi -x \right)=-\cos x$
Hence, we have
$-2av\cos \theta ={{a}^{2}}+{{v}^{2}}-{{w}^{2}}\Rightarrow {{w}^{2}}={{a}^{2}}+{{v}^{2}}+2av\cos \theta $
Taking square root on both sides, we get
$w=\sqrt{{{a}^{2}}+{{v}^{2}}+2av\cos \theta }$
But $a=u$
Hence, we have
$w=\sqrt{{{u}^{2}}+{{v}^{2}}+2uv\cos \theta }$
Hence, the magnitude of the resultant vector w is $\sqrt{{{u}^{2}}+{{v}^{2}}+2uv\cos \theta }$
Also, let the angle made by the resultant vector v be $\phi $

Hence, we have $AE=v+a\cos \theta $ and $CE=a\sin \theta $
Hence, in triangle AEC, we have
$\tan \phi =\dfrac{a\sin \theta }{v+a\cos \theta }\Rightarrow \phi ={{\tan }^{-1}}\left( \dfrac{u\sin \theta }{v+u\cos \theta } \right)$, which is the required direction of the resultant vector.

Note: 

1. Angle between two vectors is the angle between two directions of the vectors and not the angle between the line segments. This is why we chose the angle between a and v to be $\pi -B$ and not B. Many students make a mistake in the correct selection of the angle between two vectors.
2. Do not confuse the parallelogram law of vector addition with the parallelogram law in Euclidean geometry which states that the sum of the squares of diagonals of a parallelogram is equal to the sum of the squares of its sides, i.e. $A{{C}^{2}}+B{{D}^{2}}=2\left( A{{D}^{2}}+A{{B}^{2}} \right)$