Loomis and Sternberg - Orthogonality and Scalar Product

Solution 1:

First it wants to convince by applying the Pythagorean theorem twice that for any vector $x=\overrightarrow{OX}=(x_1,x_2,x_3)$ its length squared is $$\|x\|^2\ =\ \|\overrightarrow{OX}\|^2\ =\ {x_1}^2+{x_2}^2+{x_3}^2\ =\ (x,x)$$ Then, using this for the vector $\overrightarrow{XY}=(y-x)$, we obtain $$\|\overrightarrow{XY}\|^2\ =\ \|y-x\|^2\ =\ (y-x,\,y-x)$$ We can spell it out by the coordinate definition of the inner product (nothing else is used): $$(y-x,\,y-x)\ =\ \sum_{i=1}^3(y-x)_i\cdot (y-x)_i\ =\ \sum_{i=1}^3(y_i-x_i)^2\,.$$

If we use distributivity of the inner product (the property that makes it a 'product') and symmetry, we get $$(y-x,\,y-x)\ =\ (y,y)-(x,y)\,-(y,x)+(x,x)\ =\ (x,x)+(y,y)\ -2(x,y)$$ which is equal to $\|OX\|^2+\|OY\|^2$, i.e. to $(x,x)+(y,y)$, if and only if $(x,y)=0$.