Property of a basis Let $V:=\Bbb R^3$ and $\{e_1,e_2,e_3\}$ be its standard basis.
Let $v_1=e_1+3e_2+4e_3$ and $v_2=7e_2+2e_3$
Without using the concept of dimension, show that $(v_1,v_2)$ is not a basis of $V$. Find a vector which is not in $\text{span}(v_1,v_2)$.
 A: One way to do it is to directly use definition of span: $v\in\operatorname{span}\{v_1,v_2\}$ iff $v = \alpha v_1+\beta v_2$, for some scalars $\alpha,$ $\beta$. That means that $\operatorname{span}\{v_1,v_2\} = \{(\alpha,3\alpha+7\beta,4\alpha+2\beta) \mid \alpha,\beta\in\mathbb R\}.$ You can use this parametrization to find vectors not in span. Notice that giving first two coordinates of vector in the span completely determines $\alpha$ and $\beta$, which in turn completely determines the third coordinate. So, if $(1,10,?)$ is in the span, $\alpha = \beta = 1$, so the third coordinate is $6$. That means that $(1,10,z)$ is not in the span for any $z\neq 6$.
Another way to do it is to notice that $(x,y,z)$ is in the span if and only if $$\det
\begin{pmatrix}
x & y & z \\
1 & 3 & 4 \\
0 & 7 & 2 
\end{pmatrix} = 0.$$
This is because if that determinant weren't $0$, the matrix would be invertible, which is equivalent to being of full rank, meaning that rows are linearly independent. Conversely, by determinant being $0$ we have that the matrix is not of full rank and consequently that $(x,y,z)$ is in the span.
If you calculate the determinant, you get that $(x,y,z)$ is in the span if and only if $$24x+2y-7z = 0.$$
All you have to do now is find a triplet that is not a solution.
A: You can calculate $v_1\times v_2$.
