<s>[INST] Showing a vector belongs in the span Without using dimension, I have to show
From: question
$\overrightarrow{x} \in \mathbb{R}^3$ is also in span $\{v_1, v_2, n\}$ where $n = v_1 \times v_2$. Where $v_1, v_ 2\in \mathbb{R}^3$ as well.
We can let $x = \langle x_1, x_2, x_3 \rangle$, but that isn't very helpful.
I know the set is linearly independent, could that be used? Should I write their sum in the form of components?
 [/INST] I'll use the notation from the previous answer, since I don't like subscripts. We are given two vectors $\mathbf{u}$ and $\mathbf{v}$, and we want to show that any given vector $\mathbf{x}$ is in the span of $\{\mathbf{u}, \mathbf{v}, \mathbf{n}\}$, where $\mathbf{n} = \mathbf{u} \times \mathbf{v}$.
First let's focus on the point $\mathbf{w} = p \mathbf{u} + q \mathbf{v}$ that is the foot of the perpendicular from $\mathbf{x}$ to the plane spanned by $\mathbf{u}$ and $\mathbf{v}$. The scalars $p$ and $q$ are unknown, as yet -- we have to find them. To get perpendicularity, we need $p$ and $q$ to satisfy the equations
$$(\mathbf{x} - \mathbf{w}) \cdot \mathbf{u} = (\mathbf{x} - p \mathbf{u} - q \mathbf{v}) \cdot \mathbf{u} = 0
$$
$$
(\mathbf{x} - \mathbf{w}) \cdot \mathbf{v} = (\mathbf{x} - p \mathbf{u} - q \mathbf{v}) \cdot \mathbf{v} = 0
$$
In other words, $p$ and $q$ that are solutions of the linear system
$$
p (\mathbf{u} \cdot \mathbf{u}) + q (\mathbf{u} \cdot \mathbf{v}) = \mathbf{x} \cdot \mathbf{u}
$$
$$
p (\mathbf{u} \cdot \mathbf{v}) + q (\mathbf{v} \cdot \mathbf{v}) = \mathbf{x} \cdot \mathbf{v}
$$
A solution exists because the determinant of this system is 
$(\mathbf{u} \cdot \mathbf{v})(\mathbf{v} \cdot \mathbf{v}) - (\mathbf{u} \cdot \mathbf{v})^2$, which is positive unless $\mathbf{u}$ and $\mathbf{v}$ are linearly independent. Or, you can just see that the system has a solution by calculating it via Cramer's rule.
Since $\mathbf{x} - \mathbf{w}$ is orthogonal to both $\mathbf{u}$ and $\mathbf{v}$, it must be parallel to $\mathbf{n} = \mathbf{u} \times \mathbf{v}$. So, there is a scalar $k$ such that 
$\mathbf{x} - \mathbf{w} = k \mathbf{n}$. But then 
$$
\mathbf{x} = \mathbf{w} + k \mathbf{n} = p \mathbf{u} + q \mathbf{v} + k \mathbf{n}
$$
which means that $\mathbf{x}$ lies in the span of $\{\mathbf{u}, \mathbf{v}, \mathbf{n}\}$.
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