# Thread: Orthogonal projection onto span of vectors using weighted inner product

1. ## Orthogonal projection onto span of vectors using weighted inner product

Find the orthogonal projection of
$\displaystyle v=\begin{pmatrix}1 \\ 2 \\ -1 \\ 2 \end{pmatrix}$
onto the span of $\displaystyle \begin{pmatrix}1 \\ -1 \\ 2 \\ 5 \end{pmatrix}$ and $\displaystyle \begin{pmatrix}2 \\ 1 \\ 0 \\ -1 \end {pmatrix}$
using the weighted inner product $\displaystyle <v,w>=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$

2. Originally Posted by JCS007
Find the orthogonal projection of
$\displaystyle v=\begin{pmatrix}1 \\ 2 \\ -1 \\ 2 \end{pmatrix}$
onto the span of $\displaystyle \begin{pmatrix}1 \\ -1 \\ 2 \\ 5 \end{pmatrix}$ and $\displaystyle \begin{pmatrix}2 \\ 1 \\ 0 \\ -1 \end {pmatrix}$
using the weighted inner product $\displaystyle <v,w>=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$
Let:

$\displaystyle b_1=\begin{pmatrix}1 \\ -1 \\ 2 \\ 5 \end{pmatrix}$

and:

$\displaystyle b_2=\begin{pmatrix}2 \\ 1 \\ 0 \\ -1 \end {pmatrix}$

and:

$\displaystyle u_1=\frac{b_1}{||b_1||}$

$\displaystyle u_2=\frac{b_2-\langle b_2,u_1 \rangle u_1}{||b_2-\langle b_2,u_1 \rangle u_1||}$

Then the orthogonal projection is:

$\displaystyle p=\langle v,u_1\rangle u_1 + \langle v,u_2 \rangle u_2$

Where $\displaystyle ||x||=\langle x,x \rangle^{\frac{1}{2}}$

RonL