In the book "Linear Algebra as an Introduction to Abstract Mathematics", the latter part of Chapter 9 (Page 107) introduces the idea of the Orthogonal Projection $P_U$ of $V$ onto $U$, where $V$ is an inner product space and $U$ is a subspace of $V$. For clarification, $P_U$ is defined as follows:
$$ P_U: V \to V$$ $$v \mapsto u$$
After proving the proposition that $\lVert v - P_Uv \rVert \leq \lVert v -u \rVert$, the chapter concludes with an example problem that reads as follows:
Consider the plane $U \subset \mathbb R^3$ through $0$ and perpendicular to the vector $u = (1,1,1)$. Using the standard norm on $\mathbb R^3$, we can calculate the distance of the point $v = (1,2,3)$ to $U$ using [aforementioned proposition]. In particular, the distance $d$ between $v$ and $U$ is given by $d = \lVert v - P_u v \rVert$. Let $(\frac{1}{\sqrt3}u,u_1,u_2)$ be a basis for $\mathbb R^3$ such that $(u_1,u_2)$ is an orthonormal basis of $U$. Then we have:
$$v - P_Uv = \left(\color{red}{\frac{1}{3}}\langle v,u \rangle u + \langle v, u_1 \rangle u_1 + \langle v, u_2 \rangle u_2\right) - \left(\langle v, u_1\rangle u_1 + \langle v, u_2\rangle u_2\right)$$
The example continues simplifying until an answer of $v - P_Uv = (2,2,2)$ is provided. I have highlighted in $\color{red}{\text{red}}$ the source of my confusion. How exactly did the author arrive at this coefficient? I am still pretty green when it comes to this material, so I am unsure if this is a typo or if I am simply deriving these coefficients incorrectly.
Here is what I did...arriving at a different coefficient:
Given the basis $\left(\frac{1}{\sqrt3}u,u_1,u_2\right)$, we know that $v$ can be expressed as $v= \eta \frac{1}{\sqrt3}u + \alpha_1u_1+\alpha_2u_2$.
From orthogonal decomposition, and the knowledge that $u_1$ and $u_2$ are orthonormal, $v$ can be expressed as $v = \langle v,u_1 \rangle u_1 + \langle v, u_2 \rangle u_2 + \big(v-(\langle v, u_1 \rangle u_1 + \langle v,u_2 \rangle u_2) \big)$.
Noting that $u$ is orthogonal to the plane $U$ spanned by $(u_1, u_2)$, it is clear then that $\alpha_1 = \langle v, u_1 \rangle$ and $\alpha_2 = \langle v, u_2 \rangle$. Thus, we now have:
$$v = \eta \frac{1}{\sqrt3} u + \langle v, u_1 \rangle u_1+\langle v, u_2 \rangle u_2$$
So we only need to solve for $\eta$.
To do this, consider $\langle v, \frac{1}{\sqrt 3}u\rangle$
$$ \left\langle v, \frac{1}{\sqrt 3}u\right\rangle = \left\langle \eta \frac{1}{\sqrt3} u + \langle v, u_1 \rangle u_1+\langle v, u_2 \rangle u_2, \frac{1}{\sqrt 3}u\right\rangle $$
Because of linearity of the inner product and $u \bot u_1,u_2$, this simplifies to:
$$\left\langle v, \frac{1}{\sqrt 3}u\right\rangle=\left\langle \eta \frac{1}{\sqrt3} u,\frac{1}{\sqrt 3}u\right\rangle = \frac{1}{3}\eta \langle u, u \rangle$$
Recalling that $u=(1,1,1)$, we have $\langle u,u \rangle = 1^2+1^2+1^2=3$
So we arrive at:
$$ \left\langle v, \frac{1}{\sqrt 3}u\right\rangle = \frac{1}{3}\eta (3) = \eta$$
Taking the $\frac{1}{\sqrt 3}$ out of the inner product: $\eta = \frac{1}{\sqrt 3} \langle v,u \rangle$
If anyone could offer what I did wrong (i.e. arriving at $\eta = \frac{1}{\sqrt 3} \langle v,u \rangle$ instead of $\eta = \frac{1}{3} \langle v,u \rangle$ ), I would greatly appreciate it.
Cheers~
Edit: Silly mistake on my end. I stopped prematurely. If you continue on, noting that $\eta = \frac{1}{\sqrt 3} \langle v, u \rangle$ simply multiply it by the vector $\frac{1}{\sqrt 3} u$ and you get the correct coefficient of $\frac{1}{3} \langle v, u \rangle$
It should be $1/3$, because we get $\langle v,1/\sqrt3u\rangle1/\sqrt3u=1/3\langle v,u\rangle u$, as the first term in the sum.