proof the parallelogram law of vector addition

Answer 1

The parallelogram law states that the sum of the squares of the lengths of the four sides of a parallelogram is equal to the sum of the squares of the lengths of the two diagonals.

Let's prove this by considering a parallelogram ABCD, where AB and AD are the adjacent sides, and the vectors representing these sides are given by **u** (AB) and **v** (AD). Also, let AC and BD be the diagonals of the parallelogram, represented by vectors **w** and **x** respectively.

Now, we can write the vector relationships for the diagonals as follows:

**w** = **u** + **v** (AC = AB + BC = AB + AD)
**x** = **u** - **v** (BD = BC + CD = AD + AB)

Now we calculate the squares of the lengths of the diagonals:

||**w**||^2 = (**u** + **v**) • (**u** + **v**) = **u** • **u** + **u** • **v** + **v** • **u** + **v** • **v** = ||**u**||^2 + 2(**u** • **v**) + ||**v**||^2 (Using dot product properties)

||**x**||^2 = (**u** - **v**) • (**u** - **v**) = **u** • **u** - **u** • **v** - **v** • **u** + **v** • **v** = ||**u**||^2 - 2(**u** • **v**) + ||**v**||^2 (Using dot product properties)

Now adding the squares of the diagonals:

||**w**||^2 + ||**x**||^2 = (||**u**||^2 + 2(**u** • **v**) + ||**v**||^2) + (||**u**||^2 - 2(**u** • **v**) + ||**v**||^2) = 2(||**u**||^2 + ||**v**||^2)

Now let's calculate the sum of the squares of the four sides of the parallelogram:

||**u**||^2 + ||**v**||^2 + ||**u**||^2 + ||**v**||^2 = 2(||**u**||^2 + ||**v**||^2)

Comparing both the above equations, we get:

||**w**||^2 + ||**x**||^2 = ||**u**||^2 + ||**v**||^2 + ||**u**||^2 + ||**v**||^2

This proves the parallelogram law of vector addition.