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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?

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    $\begingroup$ This is not an answer. Here's a wikipedia link for the formula: en.wikipedia.org/wiki/Determinant#2-by-2_matrices $\endgroup$ – Srivatsan Sep 7 '11 at 10:52
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    $\begingroup$ I can't resist pointing to Dick Palais's question on MO (even if it is only tangentially related). $\endgroup$ – t.b. Sep 7 '11 at 10:56
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    $\begingroup$ The most common way to find the cross product is to expand a determinant by minors where the determinant has rows of the unit vectors in the x, y, z directions. And i am guessing you know about determinants here. just expland both sides even in case of 3x3 determinants to see the proof yourself $\endgroup$ – Bhargav Sep 7 '11 at 10:57
  • $\begingroup$ Are you assuming that the area of the parallelogram is equal to the length of the cross-product? (If you point to the formula $|\mathbf a \times \mathbf b| = |\mathbf a| \cdot |\mathbf b| \cdot \sin \theta$ to prove this, then you still need to justify that this formula follows from the coordinate-based definition of cross-product you are working with.) $\endgroup$ – Srivatsan Sep 7 '11 at 11:02
  • $\begingroup$ Also, note that the $\det(A,B)$ gives the signed area of the parallelogram. For the "real" area, you need to put an absolute value. (I see that you aren't that consistent with your use of absolute values.) $\endgroup$ – Srivatsan Sep 7 '11 at 11:11
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This post here gives an excellent explanation regarding the similarities between cross products and determinants

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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

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