Question

Show that the sum and difference of two perpendicular vectors of equal lengths are also perpendicular and of some lengths ?

Answer 1

Question

Let takes that we have two vectors: $x = (x_{1},y_{1})$ and $y = (x_{2},y_{2})$ . These vectors are perpendicular, so we have: $\cos \alpha = \frac{x_1x_2 + y_1y_2}{\sqrt{x_1^2 + y_1^2}\cdot\sqrt{x_2^2 + y_2^2}} = 0\Rightarrow x_1x_2 + y_1y_2 = 0$ and as they have equal length, then $\sqrt{x_1^2 + y_1^2} = \sqrt{x_2^2 + y_2^2}\Rightarrow x_1^2 +y_1^2 = x_2^2 +y_2^2$ .

Then their sum and difference will be equal: $x + y = \left(x_{1} + x_{2},y_{1} + y_{2}\right)$ and $x - y = \left(x_{1} - x_{2},y_{1} - y_{2}\right)$ . And in this case we have:

$$\begin{array}{l} \cos \beta = \frac {\left(x _ {1} + x _ {2}\right) \cdot \left(x _ {1} - x _ {2}\right) + \left(y _ {1} + y _ {2}\right) \cdot \left(y _ {1} - y _ {2}\right)}{\sqrt {x _ {1} ^ {2} + y _ {1} ^ {2}} \cdot \sqrt {x _ {2} ^ {2} + y _ {2} ^ {2}}} = \frac {x _ {1} ^ {2} - x _ {2} ^ {2} + y _ {1} ^ {2} - y _ {2} ^ {2}}{\sqrt {x _ {1} ^ {2} + y _ {1} ^ {2}} \cdot \sqrt {x _ {2} ^ {2} + y _ {2} ^ {2}}} = \\ = \frac {\left(x _ {1} ^ {2} + y _ {1} ^ {2}\right) - \left(x _ {2} ^ {2} + y _ {2} ^ {2}\right)}{\sqrt {x _ {1} ^ {2} + y _ {1} ^ {2}} \cdot \sqrt {x _ {2} ^ {2} + y _ {2} ^ {2}}} = \left| x _ {1} ^ {2} + y _ {1} ^ {2} = x _ {2} ^ {2} + y _ {2} ^ {2} \right| = 0 \Rightarrow \beta = 9 0 {}^ {\circ} \Rightarrow \\ \Rightarrow (x + y) \perp (x - y). \\ \end{array}$$

Proved.

Answer: Proved.