4
$\begingroup$

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?

  • 0
    [Related](htt$p$://math.stackexchange.com/questions/61526)...2011-09-07

2 Answers 2

1

Anton and Rorres in their Elem Linear Algebra with Applications 9th Ed seem to use something similar to what I described in my question. Their wording is "the key to the proof is to use [cross prod theorem], however that theorem applies to vectors in 3-space whereas our vectors are in 2-space. To circumvent this "dimension problem" we shall view u and v as vectors in xy-plane of an xyz-coordinate system (figure below) in which case the vectors are described as $u = (u_1,u_2,0)$ and $v = (v_1,v_2,0)$".

enter image description here

1

This post here gives an excellent explanation regarding the similarities between cross products and determinants