Let A = (a1, a2, ..., an) and B = (b1, b2, ..., bn) be two points in n-dimensional Euclidean space, and let P be the orthogonal projection of B onto the line passing through A and B. Then, the sum of squares of the projection in coordinate axes is square length of AB.
To prove this, we can start by defining the vector AB = B - A. The length of AB is given by:
|AB|^2 = (b1-a1)^2 + (b2-a2)^2 + ... + (bn-an)^2
Next, we can define the vector AP = P - A. Since P is the projection of B onto the line passing through A and B, AP is orthogonal to AB. Thus, the dot product of AP and AB is zero:
AP · AB = 0
Expanding this dot product, we get:
(a1-p1)(b1-a1) + (a2-p2)(b2-a2) + ... + (an-pn)(bn-an) = 0
where p1, p2, ..., pn are the coordinates of P. Rearranging this equation, we get:
p1(b1-a1) + p2(b2-a2) + ... + pn(bn-an) = a1b1 + a2b2 + ... + anbn
Now, the projection of B onto the x-axis is given by (b1, 0, ..., 0), and the projection of B onto the y-axis is given by (0, b2, ..., 0), and so on. Therefore, the sum of squares of the projection in coordinate axes is given by:
(b1-a1)^2 + (b2-a2)^2 + ... + (bn-an)^2
which is exactly the same as the square length of AB. This can be seen by expanding the expression for |AB|^2 and collecting like terms:
|AB|^2 = (b1-a1)^2 + (b2-a2)^2 + ... + (bn-an)^2
Thus, we have proved that the sum of squares of the projection in coordinate axes is square length of AB.