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I have to calculate the orthogonal projection of the vector $$ v = \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix} $$ on the $\text{span}(u_1,u_2)$ where $$ u_1 = \begin{pmatrix} 1/2 \\ 0 \\ 1/2 \end{pmatrix}, u_2 = \begin{pmatrix} 3/4 \\ 1/4 \\ -1/4 \end{pmatrix} $$ My professor has provided the answers and says that the orthogonal projection can be calculated as $$ p = \langle v,u_1 \rangle u_1 + \langle v,u_2 \rangle u_2 $$ however, normally, I would use the formula $$ p = \frac{\langle v,u_1 \rangle u_1}{||u_1||^2} + \frac{\langle v,u_2 \rangle u_2}{||u_2||^2} $$ so I don't understand the reason behind the provided answers. Do you mind explaining why? I do not know if it makes any difference but $u_1,u_2$ is an orthonormal basis which I have created by using the Gram-Schmidt Process for the basis $V = (v_1,v_2,v_3)$ where $v_1 = (1,0,1)^T, v_2 = (5,1,1)^T, v_3 = (4,-1,0)^T$

Mathias
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    The first formula is valid if the vector onto you are projecting is a unit vector, is indeed a special case of the second formula you wrote. – Fabrizio Jun 05 '20 at 10:29
  • But $u_1,u_2$ are not units vectors? So is the professor's answer wrong? I do not get the same result if I use the last formula as $||u_1||^2 \neq 1$ and $||u_2||^2 \neq 1$ – Mathias Jun 05 '20 at 10:32
  • Just use the second formula and you get the right result. – Fabrizio Jun 05 '20 at 17:45
  • So my professors answers using the first formula is wrong? – Mathias Jun 05 '20 at 17:46
  • Yes, i think... if the resut is different your prof is wrong :) – Fabrizio Jun 05 '20 at 17:51

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