In the multivariable calculus class the teacher showed us the formula of the cross product
$ \vec{A} \times \vec{B} =\begin{vmatrix}\hat{\imath}& \hat{\jmath}& \hat{k} \\ a_1 & a_2 & a_3 \\b_1 & b_2 & b_3 \end{vmatrix}$
And formula for determinant in two dimensions which can calculate the area of parallelogram in two dimensions by
$\det(\vec{A},\vec{B}) =\begin{vmatrix}a_1 & a_2 \\b_1 & b_2 \\\end{vmatrix}$
Then teacher talked about the area of a parallelogram also being equal to the length of $\vec{A} \times \vec{B}$, that is $|\vec{A} \times \vec{B}|$, but gave no proof. I wanted to check this, so I used $a_3=0,b_3=0$ just to have the $3 \times 3$ in the form that could be compared to $\det(\vec{A},\vec{B})$ form. When I expand the calculation, I do end up with $|\hat{k}(a_1b_2 - a_2b_1)|$, and that equals to $(a_1b_2 - a_2b_1)$ The two forms are equal. Is this reasoning correct?