Finding Dual Basis 
In $\mathbb{R^{3}}$, let be $v_{1}=\begin{bmatrix} 1 \\2 \\0 \end{bmatrix}$i $v_{2}=\begin{bmatrix} 0 \\3 \\1 \end{bmatrix}$ and $v_{3}=\begin{bmatrix} 0 \\0 \\1 \end{bmatrix}$. If f is a linear functional on $\mathbb{R}^{3}$ such that $f(v_1)=-1$,$f(v_2)=2$ and $f(v_3)=4$ and if $v=\begin{bmatrix} x \\y \\z \end{bmatrix}$, find $f(v)$.

In addition to that question I want to find Dual Basis. But now, I will write my solution for $f(v)$.
Solution:
$$f\left(1 \cdot \begin{bmatrix} 1 \\0 \\0 \end{bmatrix}+2 \cdot \begin{bmatrix} 0 \\1 \\0 \end{bmatrix} + 0 \cdot \begin{bmatrix} 0 \\0 \\1 \end{bmatrix}    \right) = f(e_1)+2f(e_2)=-1$$
$$f\left(0 \cdot \begin{bmatrix} 1 \\0 \\0 \end{bmatrix}+3 \cdot \begin{bmatrix} 0 \\1 \\0 \end{bmatrix} + 1 \cdot \begin{bmatrix} 0 \\0 \\1 \end{bmatrix}    \right) = 3f(e_2)+f(e_3)=2$$
$$f\left(0 \cdot \begin{bmatrix} 1 \\0 \\0 \end{bmatrix}+0 \cdot \begin{bmatrix} 0 \\1 \\0 \end{bmatrix} + 1 \cdot \begin{bmatrix} 0 \\0 \\1 \end{bmatrix}    \right) = f(e_3)=4$$
And we get;
$$f(v)=\dfrac{1}{3}x-\dfrac{2}{3}y+4z$$
And I want to find Dual Basis, any help? Thanks

Let $f=(p,q,r)_{1\times 3}$, then $f(v_1)=-1 \implies p+2q=-1, f(v_2)=3q+r=2, f(v_3)=4 \implies r=4 \implies p=-2/3, q=1/3,r=4$. Finally
$$f(v)=(-2/3,1/3,4)\begin{pmatrix} x\\ y\\ z \end{pmatrix}=-2x/3+y/3+4z.$$
We can write $v_1=(1,2,0), v_2=(0,3,1), v_3=(0,0,3)$, then the dual basis is given by cross products:
$$u_1=v_1\times v_2=(2,-1,3), u_2=v_2\times v_3=(3,0,0), u_3=v_3 \times v_1=(-2,1,0).$$
