Consider the following vectors in $\mathbb{R}^3$
$$ u_1=\left(\begin{array}{l} {1 / 9} \\ {4 / 9} \\ {8 / 9} \end{array}\right), \quad u_2=\left(\begin{array}{c} {8 / 9} \\ {-4 / 9} \\ {1 / 9} \end{array}\right), \quad u_3=\left(\begin{array}{c} {4 / 9} \\ {7 / 9} \\ {-4 / 9} \end{array}\right) $$
(a) Prove that the family $u=(u_1,u_2,u_3)$ is orthonormal with respect to standard inner product on $\mathbb R^3$.
My answer
For all the three vectors I got $<-,-> = 1$. The norm is $1$.
I have also proved that they are orthogonal.
(b) Given an arbitary vector $x$ $$ x=\left(\begin{array}{l} {x_{1}} \\ {x_{2}} \\ {x_{3}} \end{array}\right) \in \mathbb{R}^{3}, $$ determine $c_1,c_2,c_3 \in \mathbb R^3$ such that $$ x = u_1c_1+u_2c_2+u_3c_3 $$
My answer
I actually do not have one. In the solution the first one $c_1$ is given as $c_1 = <u_1,x>$, and so on. I do not understand the thinking behind this.
Since $u_1$, $u_2$ and $u_3$ are an orthonormal set, we can write any vector $x \in \mathbb{R}^3$ as linear comibation of those 3 as you have shown:
$$x=c_1 u_1 + c_2 u_2 + c_3 u_3$$
We can take the Inner Product of the right and left hand sides with respect to $u_1$, which should be equal:
$$\left(x,u_1\right)= \left(c_1 u_1,u_1\right) + \left(c_2 u_2,u_1\right) + \left(c_3 u_3,u_1\right)$$ Apply Linearity Property of inner product: $$\left(x,u_1\right)= c_1 \left( u_1,u_1\right) + c_2 \left( u_2,u_1\right) + c_3 \left( u_3,u_1\right)$$
Now recall that $\left( u_1,u_1\right)=1$, $\left( u_1,u_2\right)=0$ and $\left( u_1,u_3\right)=0$ because of the orthonormality of $u_1$, $u_2$ and $u_3$, so:
$$\left(x,u_1\right)= c_1 $$
Repeat the same process with $u_2$ and $u_3$ to find $c_2$ and $c_3$.