========================================================= Orthogonal projection When you project "vector y" onto "vector x", "projected vector y" becomes vector which has same direction with "vector x" What will be "projected vector y's length"? "True projected vector y" is $$$u_{x} \times $$$ (length of projected vector y) So, length of projected vector y can be calculated by $$$y= y^{T}u_{x}$$$ In conclusion, projected vector y = $$$\langle y^{T},u_{x} \rangle u_{x} = (||y||\cos{\theta})u_{x}$$$ $$$||y||\cos{\theta}$$$ = (length of projected y) = (length of y)*($$$\cos{\theta}$$$) ========================================================= Example vector $$$y=[2 \;\; 4]^{T}$$$ vector $$$x=[4 \;\; 3]^{T}$$$ Calculate projection y onto x? $$$u_{x}=\frac{1}{||x||}x=\frac{1}{\sqrt{16+9}} [4\;3] = \begin{bmatrix} \frac{4}{5} \; \frac{3}{5} \end{bmatrix} $$$ $$$\begin{pmatrix} [2\;4]\begin{bmatrix} \frac{4}{5} \\ \frac{3}{5} \end{bmatrix} \end{pmatrix} \begin{bmatrix} \frac{4}{5} \\ \frac{3}{5} \end{bmatrix}$$$ $$$4*\begin{bmatrix} \frac{4}{5} \\ \frac{3}{5} \end{bmatrix}$$$ ========================================================= If "inner product" is 0, 2 vectors are "orthogonal" relationship. If "inner product" is 0 AND $$$||x||=||y||=1$$$, 2 vectors are "orthonoarmal" relationship