The parallelogram law is a part of elementary geometry. It states that “the sum of the squares of the lengths of the two diagonals of a parallelogram is same as the sum of the squares of the lengths of its four sides.”


Proof of Parallelogram Law

ABCD is 
a parallelogram. We know that AB = CD and BC = AD. By parallelogram law we have

2(AB)2 + 2(BC)2 = (AC)2 + (BD)2

i.e., a2 + b2 + c2 + d2 = d12 + d22

In ∆ ABC we use cosine law, and get

d12 = a2 + d2 − 2ad cos $\beta$  …(i)

In ∆ ABD by use of cosine law we have

d22 =  c2 + d2 − 2cd cos $\alpha$ …(ii)

Adding (i) and (ii) we get,

d12 + d22 = a2 + d2 + c2 + d2 − 2ad cos $\beta$ − 2cd cos $\alpha$

⇒ d12 + d22 = a2 + d2 + c2 + b2 − 2ad cos $\beta$ + 2ad cos $\beta$

[since, cos $\alpha$  = − cos $\beta$ and a = c, b = d]

⇒ d12 + d22 = a2 + d2 + c2 + b2

Hence, proved.
In vector algebra, the parallelogram law is very useful. Then we commonly name it as parallelogram of vector addition. 
Parallelogram Law of Vector Addition

Let P and q be two vectors. Then the sum of the vectors P + q is obtained by first placing them head to tail and then by drawing the vector from the tail that is free to the head that is free.

Then we make use of parallelogram law and say that if the vectors form the sides of the parallelogram then the vector sum and difference represents the diagonals of the parallelogram and we have

|P + q|2 + |P − q|2 = 2 |P|2 + 2 |q|2

In general, the parallelogram law of vectors states that “if the two given vectors are acting at a common point and can be represented in magnitude with direction as the two adjacent sides of a parallelogram, then the resultant vector is represented in magnitude with direction by the diagonal which is passing through the tail that is common of two vectors.”

P and Q are two vectors representing sides of a parallelogram. We complete the parallelogram by joining all sides and make diagonal from common tail of the vectors and say it R vector.

We draw CD perpendicular to OA extended.

We need to prove that R is the sum of vectors P And Q


In right ∆ CDA

AC2 = AD2 + CD2          …(i)

Also, $\frac{AD}{AC}$ = cos $\theta$

⇒ AD = AC cos $\theta$  …(ii)

And CD = AC sin $\theta$ …(iii)

In right ∆ ODC

OC2 = OD2 + CD2

⇒ OC2 = (OA + AD)2 + AC2 − AD2                       [using (i)]

⇒ OC2 = OA2 + AD2 + 2.OA.AD + AC2 − AD2

⇒ OC2 = OA2 + AC2 + 2.OA.AD 

⇒ OC2 = OA2 + AC2 + 2.OA.AC. cos $\theta$            [using (ii)]

⇒ OC2 = OA2 + OB2 + 2.OA.OB. cos $\theta$           [since OB = AC]

⇒ R2 = P2 + Q2 + 2.P.Q.cos $\theta$

This is the magnitude of the resultant vector of P and Q.

For the direction consider ∆ OCD

This law is very useful is determining resultant vectors when two vectors can be represented easily as sides of the parallelogram.
In norm space, the parallelogram law statement is nothing but an equation that relates to norms.

2 ||x||2 + 2 ||y||2 = ||x + y||2 + ||x − y||2

In inner product space the norm is determined with the help of the inner product.

||x|| = <x, x>

Due to this definition, the parallelogram law serves as an algebraic identity in an inner product space.
||x + y||2 = <x + y, x + y> = <x, x> + <y, y> + <x, x> + <y, y> …(i)

||x − y||2 = <x − y, x − y> = <x, x> − <y, y> − <x, x> + <y, y> …(ii)

Adding (i) and (ii) we get

 ||x + y||2 + ||x − y||2 = 2 <x, x> + 2 <y, y> = 2 ||x||2 + 2 ||y||2
Let us see an example to get more understanding on use of parallelogram law.

Example:  

Find the resultant vector in the figure given.



Solution:

Let A = 4 and B = 5.

Let R be the resultant vector

Then by parallelogram law of vector addition we have R = A + B

Therefore, resultant vector R = 9 cm.

The direction of this vector cannot be determined as we do not know the angle between the given vectors.