Orthogonal project of a vector $\vec{y}$ that sit in the same plane (aka span($u_1, u_2$)) such that $\vec{y},\vec{u_1}, \vec{u_2} \in \mathbb{R^3}$

Question

Confirmed that $u_1,u_2$ are indeed orthogonal by taking their dot product which equals zero. Correct answer = $\begin{pmatrix}-1\\ -1\\ 6\end{pmatrix}$.

Geographically, mapped out in geogebra: we see that it all sits on the same plane, therefore y clearly projects onto itself, ie. $\vec{y}=$ projection of $\vec{y}$. However...

Question 1: How can one find the orthogonal projection of y onto the span of $\vec{u_1}, \vec{u_2}$ arithmetically?

Using RREF of {$u_1, u_2$}, they are indeed a plane but I had no idea without mapping it out geometrically to it see clearly. Therefore, I've done what I was taught, to find the coefficients of y projected onto each vector $\vec{u_1}, \vec{u_2}$ then add them up to get a projection of y.

Incorrectly: $projection\:of\:\vec{y\:}=\frac{\left(\:\vec{y}\cdot u_1\right)}{u_1\cdot u_1}u_1+\frac{\left(\:\vec{y}\cdot \:u_2\right)}{u_2\cdot \:u_2}u_2$

Intuitively w/o knowing y sits on the same plane, adding them up should give a projection of y, but it isn't the right answer for some reason, I'd like to understand why.

Question 2: why doesn't the above work? And the below one does?

But this works,(trial an error on geogebra) when I minus projection of y on $\vec{u_2}$ - projection of y onto $\vec{u_1}$ = correct answer: Correct answer, don't understand why: $projection\:of\:\vec{y\:}=\frac{\left(\:\vec{y}\cdot \:\:u_2\right)}{u_2\cdot \:\:u_2}u_2-\frac{\left(\:\vec{y}\cdot u_1\right)}{u_1\cdot u_1}u_1$

Question 3: Is there a more effecient method to derive that projection of $\vec{y}$ is $\vec{y}$ itself?

Answer 1

Your intuition is correct. The formula for the projection of $\vec{y}$ onto $\text{span}(u_1,u_2)$ is $$ \pi(\vec{y}) = \frac{\vec{y}\cdot u_1}{u_1\cdot u_1} u_1 + \frac{\vec{y}\cdot u_2}{u_2\cdot u_2}u_2.$$ In this case, we see that $$ \pi(\vec{y}) = \frac{-18}{54}u_1 + \frac{24}{18}u_2 = -\frac 1 3 u_1 + \frac 4 3 u_2.$$ Substituting in $u_1$ and $u_2$ on the right hand side gives $\pi(\vec y) = \vec y$, so we see that $y \in \text{span}(u_1,u_2)$.

Another way you could find the projection of $\vec{y}$ onto $\text{span}(u_1,u_2)$ is to subtract off the normal component. The vector $\nu = \frac{u_1 \times u_2}{|u_1 \times u_2|}$ is the unit normal, and so the projection is given by $\pi(\vec y) = \vec y - (\vec y \cdot \nu) \nu$.