Prove $|x + y|^2 + |x - y|^2 = 2|x|^2 + 2|y|^2$ and interpret its meaning.

Question

Prove $|x + y|^2 + |x - y|^2 = 2|x|^2 + 2|y|^2$ if $x, y \in \mathbb{R}^k$.

Interpret this geometrically, as a statement about parallelograms.

I've shown that the expression given equates to $x \cdot x + 2x \cdot y + y \cdot y + x \cdot x - 2x \cdot y + y \cdot y$. But what does this have to do with parallelograms?

Answer 1

Remember that $\mathbf{x}$ and $\mathbf{y}$ are vectors. Then if you form a parallelogram with sides $\mathbf{x}$ and $\mathbf{y}$, then $\mathbf{x-y}$ is the short diagonal and $\mathbf{x+y}$ is the long diagonal (assuming you labelled the sides sensibly)

Answer 2

Hint: The points, $0$, $x$, $y$, $x+y$ form a parallelogram.

Answer 3

Hint: Given a parallelogram where one vertex is at the origin and the adjacent sides are given by the vectors $x$ and $y$, $|x+y|$ and $|x-y|$ will be the lengths of the two diagonals (why?).

Answer 4

It is easy to prove this if you have a good insight about vectors assume X and Y to be two vectors.

Then $ | X | $ and $ | Y | $ denote their magnitudes $X+Y $ and $X-Y$ denote the two diagonals of the parallelogram formed(as in parallelogram law of vector addition).

U simply need to prove that 2 times sum of squares of two adjacent sides of a parallelogram = sum of squares of two diagonals.

Now let ABCD be a $parallelogram$ with E as point of intersection of diagonals.

USING THE PROPERTY THAT

$1)$ diagonals bisect each other in a parallelogram .

$2)$ Opposite sides are equal.

$3)$ Using $Appolonius$ $ Theorem$

$$AB^2+BC^2=2AE^2+2BE^2$$

AND IN THE OTHER TRIANGLE

$$CD^2+BC^2=2AE^2+2BE^2$$

Add both the equations u get

$$2AB^2+2BC^2=AC^2+BD^2$$

HENCE PROVED