Question

How do you prove the Parallelogram Law of vectors mathematically?
$${|a+b|}^2+{|a-b|}^2=2{\left|a\right|}^2+2{\left|b\right|}^2$$.

Answer 1

Hint: In this problem, we have to prove the Parallelogram Law of vectors mathematically. we should know that the Parallelogram law states that the sum of the squares of the length of the four sides of a parallelogram is equal to the sum of the squares of the length of the two diagonals. We can take the left-hand side of the given and expand it using the algebraic whole square formula, to prove for the right-hand side.

Complete step-by-step answer:
Here we have to prove the Parallelogram Law of vectors mathematically.
We know that Parallelogram law states that the sum of the squares of the length of the four sides of a parallelogram is equal to the sum of the squares of the length of the two diagonals.
The given parallelogram law of addition to be proved is,
$${|a+b|}^2+{|a-b|}^2=2{\left|a\right|}^2+2{\left|b\right|}^2$$
We can now take the left-hand side and expand it using the algebraic whole square formula
$$\begin{array}{rl}& {(a+b)}^2=a^2+b^2+2ab\\ & {(a-b)}^2=a^2+b^2-2ab\end{array}$$
We can take the Left-hand side,
LHS = $${|a+b|}^2+{|a-b|}^2$$
We can now expand it using the above formulas, we get
LHS = $$[{\left|a\right|}^2+{\left|b\right|}^2+2\left|a\right|\left|b\right|]+[{\left|a\right|}^2+{\left|b\right|}^2-2\left|a\right|\left|b\right|]$$
We can now simplify the above step, by cancelling the similar terms with opposite side, we get
 LHS = $${\left|a\right|}^2+{\left|b\right|}^2+{\left|a\right|}^2+{\left|b\right|}^2$$
We can now add the similar terms, we get
LHS = $$2{\left|a\right|}^2+2{\left|b\right|}^2$$
LHS = RHS
Hence proved.

Note: We should always remember the definition for the law of parallelogram, as Parallelogram law states that the sum of the squares of the length of the four sides of a parallelogram is equal to the sum of the squares of the length of the two diagonals. We should also remember the algebraic whole square formula, to prove these types of problems.