I wish to find the area of a parallelogram formed by two vectors in three dimensions.
Computing this is simple, taking the absolute value (modulus) of the cross product of the vectors, where the cross product is easily found with a determinant.
I am not having any difficulty with the computation. But I am having difficulty with the notation.
At the moment my calculation looks like this:
$$
\begin{align}
|\vec A\times \vec B|
&= |\langle 0,1,1 \rangle \times \langle 1,0,1\rangle|\\
&= \left|\begin{vmatrix}
\hat{\boldsymbol{\imath}}
& \hat{\boldsymbol{\jmath}}
& \hat{\boldsymbol{k}}\\
0&1&1\\1&0&1
\end{vmatrix}\right|\\
&= \left|(1-0)\hat{\boldsymbol{\imath}}
-(0-1)\hat{\boldsymbol{\jmath}}
+(0-1)\hat{\boldsymbol{k}}\right|\\
&= \left|\hat{\boldsymbol{\imath}}
+\hat{\boldsymbol{\jmath}}
-\hat{\boldsymbol{k}}\right|\\
&= |\langle 1,1,-1\rangle|\\
&= \left|\sqrt{1^2+1^2+(-1)^2}\right|\\
&= \sqrt{3}
\end{align}
$$
Mathematically, this is the result I am looking for. But notationally it looks quite horrible.
The second line is visually confusing. The absolute value bars appear exactly as the vertical lines for the determinant, and combined they look like a norm.
I could replace the second line with
$$\left|\det\begin{pmatrix}
\hat{\boldsymbol{\imath}}
& \hat{\boldsymbol{\jmath}}
& \hat{\boldsymbol{k}}\\
0&1&1\\
1&0&1
\end{pmatrix}\right|$$
or even
$$\operatorname{abs}\begin{vmatrix}
\hat{\boldsymbol{\imath}}
& \hat{\boldsymbol{\jmath}}
& \hat{\boldsymbol{k}}\\
0&1&1\\
1&0&1
\end{vmatrix}$$
But neither of these options appeal to me (I would then want to use one these notations for all other occurrences of absolute values or determinants in a large document I am composing).
Is there an alternative way I can represent the absolute value of a determinant?
Best Answer
What about $$\left|\;\begin{vmatrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\boldsymbol{k}}\\ 0&1&1\\1&0&1 \end{vmatrix}\;\right|$$ ? You can add space between your outer $|$ and the inner $|$, e.g. with
\;. You could also assign a different length, e.g. here with using\Bigg|: $$\Bigg|\begin{vmatrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\boldsymbol{k}}\\ 0&1&1\\1&0&1 \end{vmatrix}\Bigg|$$ Both combined: $$\Bigg|\;\begin{vmatrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\boldsymbol{k}}\\ 0&1&1\\1&0&1 \end{vmatrix}\;\Bigg|$$Apart from that, I usually use the $|\det A |$ variant, mainly because I almost always use $\det A$ when working with determinants.
In the spirit of Bill's comment, you can also just calculate $\det A$ to be $a$ and then immediately follow $|\det A|=|a|$.