Generating a $n$-th dimensional vector orthogonal to $n-1$ linearly-independent vectors Let us have $n-1$ linearly independent vectors $\vec{v}_{1},\dots,\vec{v}_{n-1}\in\mathbb{R}^{n}$, define the vector $\vec{w}$ as follows:
$$\vec{w}=\begin{pmatrix}\det(\vec{v}_{1},\dots,\vec{v}_{n-1},\vec{e}_{1}) \\ \vdots \\ \det(\vec{v}_{1},\dots,\vec{v}_{n-1},\vec{e}_{n})\end{pmatrix}$$
This vector is apparantly orthogonal to all vectors $v_{i}$, $i\in\{1,\dots,n-1\}$; however I am not able to prove that this is the case and I don't understand the intuition behind it either, so I would be grateful if someone was able to prove this fact and perhaps allow me to grasp it more intuitively.
 A: Clearly this vector is orthogonal to all vectors $v_i$, because: if we suppose $\vec{v}_i=v_{i1}e_1+\dots+v_{in}e_n$, then
$$\vec{v}_i\cdot\vec{w}=(v_{i1},\dots,v_{in})\cdot\begin{pmatrix}\det(\vec{v}_{1},\dots,\vec{v}_{n-1},\vec{e}_{1}) \\ \vdots \\ \det(\vec{v}_{1},\dots,\vec{v}_{n-1},\vec{e}_{n})\end{pmatrix}=\sum_{k=1}^{n}v_{ik}det(\vec{v}_{1},\dots,\vec{v}_{n-1},\vec{e}_{k})=det(\vec{v}_{1},\dots,\vec{v}_{n-1},\vec{v}_{i})$$
so the matrix $(\vec{v}_{1},\dots,\vec{v}_{n-1},\vec{v}_{i})$ has two same colomns, so $\vec{v}_i\cdot\vec{w}=0$.
I believe that this is a method of developing an another base from one any vector. For example, one base of the vector space is $\{e_i\}$, and now we have one any vector $\vec{v}=\sum_{i=1}^nv_ie_i$ (there is at least one $i$ that $v_i\neq0$). We will make another base from this vector $\vec{v}$ and we note it $\vec{v}_1$. Your equation is the way to do this.
we suppose that $\vec{v}_2=\begin{pmatrix}\det(\vec{v}_{1},\vec{e}_{1},\dots,\vec{e}_{n-1}) \\ \det(\vec{v}_{1},\vec{e}_{2},\dots,\vec{e}_{n})\\ \vdots \\ \det(\vec{v}_{1},\vec{e}_{n},\dots,\vec{e}_{n-2})\end{pmatrix}$, we can prove that $\vec{v}_2\neq0$ and $\vec{v}_1\cdot\vec{v}_2=0$. 
Similarly, if we have already $\vec{v}_1,\vec{v}_2,\dots,\vec{v}_i$, we can make $\vec{v}_{i+1}$:
$$\vec{v}_{i+1}=\begin{pmatrix}\det(\vec{v}_{1},\dots,\vec{v}_{i},\vec{e}_{1},\dots,\vec{e}_{n-i}) \\ \det(\vec{v}_{1},\dots,\vec{v}_{i},\vec{e}_{2},\dots,\vec{e}_{n-i+1})\\ \vdots \\ \det(\vec{v}_{1},\dots,\vec{v}_{i},\vec{e}_{n},\dots,\vec{e}_{n-i-1})\end{pmatrix}$$
we can prove that for any $1\leq k\leq i$, we have $\vec{v}_{i+1}\neq0$ and $\vec{v}_k\cdot\vec{v}_{i+1}=0$.
your equation is the last step to make $\vec{v}_{n}$:
$$\vec{v}_{n}=\begin{pmatrix}\det(\vec{v}_{1},\dots,\vec{v}_{n-1},\vec{e}_{1}) \\ \vdots \\ \det(\vec{v}_{1},\dots,\vec{v}_{n-1},\vec{e}_{n})\end{pmatrix}$$
The vectors {$\vec{v}_i$} are orthogonal for each and each other. So in a space of dimension $n$, this is a new base.
A: If you write out the dot product and expand each determinant along the last column, you'll see that $$\vec{v}_i\cdot \vec{w} = \det(\vec{v}_1, \vec{v}_2,\ldots, \vec{v}_{n-1},\vec{v}_i),$$ and since the right hand side is the determinant of a matrix with two equal columns, it is zero. I'm not sure if there is geometric intuition.
A: You complicate matters needlessly by talking about a dot product. What you want to describe is a nonzero linear form on the vector space that vanishes on all vectors $\vec v_i$ (the dot product with $\vec w$ is such a linear form). Now on one hand any linear form$~\alpha$ is given by a $1\times n$ matrix $(c_1~c_2~\ldots~c_n)$ where $c_j=\alpha(e_j)$ is the value of the linear form on the standard basis vector$~e_j$ (the ususal intepretation of matrix coefficiencts); transposing that matrix gives a column vector such that $\alpha$ is dot product by it. On the other hand there is an obvious linear form that vanishes on all vectors $\vec v_i$, nonzero by the independance assumption, namely $\alpha:\vec x\mapsto\det(\vec v_1,\ldots,\vec v_{n-1},\vec x)$. Now evaluating $\alpha(e_j)$ for all $j$ and putting things together gives the statement of your question.
