Consider the following vectors in R3 Consider the following vectors in $\Bbb{R}^3$:
${\bf v}_1= \frac{1}{\sqrt{3}}(1,1,-1)$, ${\bf v}_2 = \frac{1}{\sqrt{2}}(1,-1,0),$ and ${\bf v}_3 = \frac{1}{\sqrt{6}}(1,1,2)$
a) Show that $\{{\bf v}_1, {\bf v}_2, {\bf v}_3\}$ form a basis of $\Bbb{R}^3$. (hint: compute their inner products)
b) Work out the coordinates of a vector ${\bf x} = (x_1,x_2,x_3)$ in the basis $\{{\bf v}_1, {\bf v}_2, {\bf v}_3\}$; that is, find the numbers $c_1, c_2, c_3 \in \Bbb{R}$ such that ${\bf x} = c_1{\bf v}_1 + c_2{\bf v}_2 +c_3{\bf v}_3$.
This is one example in  my finals review for linear algebra. I do not know how to tackle this problem especially for part b.

A set of any 3 linearly independent vectors in $R^3$ is a basis for $R^3.$ If $x_1,x_2,x_3$ are  3 non-zero pairwise-orthogonal vectors then they are linearly independent. Because if $0=a_1x_+a_2x_2+a_3x_3 $, then for $j\in \{1,2,3\}$ we have $0=0\cdot x_j=(a_1x_1+a_2x_2+a_3x_3)\cdot x_j=a_j(x_j\cdot x_j)\implies a_j=0.$ And for every vector $v$ we have $v=\sum_{j=1}^{j=3} x_j(x_j\cdot v)/(x_j\cdot x_j).$
