Linear Algebra: V Cross Product and Triple Scalar Product

Cross Product and Triple Scalar Product


Consider two $2$ dimensional vectors:


$$ \vec{u}= \begin{bmatrix} a \\ c \end{bmatrix}, \quad \vec{v}= \begin{bmatrix} b \\ d \end{bmatrix} $$

The area of the parallelogram formed by these vectors is:


$$ \text{Area} = \|\vec{u}\| \, \|\vec{v}\| \sin \theta $$

where $\|\vec{u}\| = \sqrt{a^2 + c^2}$, $\|\vec{v}\| = \sqrt{b^2 + d^2}$, and $\theta$ is the angle between these vectors.


Recall:


$$ \vec{u}^\top \vec{v} = \|\vec{u}\| \, \|\vec{v}\| \cos \theta $$

The dot product is:


$$ \vec{u}^\top\vec{v} = ab + cd $$

Therefore:


$$ \cos \theta = \frac{ab + cd}{\sqrt{a^2 + c^2}\sqrt{b^2 + d^2}} $$

Since $\sin^2 \theta = 1 - \cos^2 \theta$, then:


$$ \sin^2 \theta = 1 - \frac{(ab + cd)^2}{(a^2 + c^2)(b^2 + d^2)} = \frac{(a^2 + c^2)(b^2 + d^2)-(ab + cd)^2}{(a^2 + c^2)(b^2 + d^2)} $$

Thus:


$$ \text{Area}^2 = (a^2 + c^2)(b^2 + d^2)\sin^2 \theta = (a^2 + c^2)(b^2 + d^2) - (ab + cd)^2 $$
$$ = a^2b^2 + a^2d^2 + c^2b^2 + c^2d^2 - (a^2b^2 + 2abcd + c^2d^2) $$
$$ = a^2d^2 + c^2b^2 - 2abcd = (ad - bc)^2 $$

Since the area scaled by the column vectors of $ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $ is $|ad-bc|$, then:


$$ |\det(A)| = |ad-bc| $$

Consider two $3$ dimensional vectors:


$$ \vec{a} = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} $$

The cross product is defined as:


$$ \vec{a} \times \vec{b} = \begin{bmatrix} \det\begin{bmatrix} a_2 & a_3 \\ b_2 & b_3 \end{bmatrix} \\[1em] -\det \begin{bmatrix} a_1 & a_3 \\ b_1 & b_3 \end{bmatrix} \\[1em] \det \begin{bmatrix} a_1 & a_2 \\ b_1 & b_2 \end{bmatrix} \end{bmatrix} = \begin{bmatrix} a_2 b_3 - a_3 b_2 \\[1em] -(a_1 b_3 - a_3 b_1) \\[1em] a_1 b_2 - a_2 b_1 \end{bmatrix} $$

Its magnitude is the area of the parallelogram formed by $\vec{a}$ and $\vec{b}$:


$$ \|\vec{a} \times \vec{b}\|^2 = (a_2b_3 - a_3b_2)^2 +(a_1b_3-a_3b_1)^2 + (a_1b_2 - a_2b_1)^2 $$
$$ = (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2) - (a_1b_1+a_2b_2+a_3b_3)^2 $$
$$ = \|\vec{a}\|^2 \|\vec{b}\|^2 - (\vec{a} \cdot \vec{b})^2 $$
$$ = \|\vec{a}\|^2 \|\vec{b}\|^2 - \left(\|\vec{a}\| \|\vec{b}\| \cos\theta\right)^2 $$
$$ = \|\vec{a}\|^2 \|\vec{b}\|^2 (1 - \cos^2\theta) = \|\vec{a}\|^2 \|\vec{b}\|^2 \sin^2\theta = \text{Area}^2 $$

The cross product is orthogonal to both vectors:


$$ \vec{a} \perp \vec{a} \times \vec{b} \quad \text{and} \quad \vec{b} \perp \vec{a} \times \vec{b} $$

Since


$$ \vec{a}^\top (\vec{a} \times \vec{b}) = a_1(a_2 b_3 - a_3 b_2) + a_2(a_3 b_1 - a_1 b_3) + a_3(a_1 b_2 - a_2 b_1) $$
$$ = a_1 a_2 b_3 - a_1 a_3 b_2 + a_2 a_3 b_1 - a_2 a_1 b_3 + a_3 a_1 b_2 - a_3 a_2 b_1 = 0 $$

And


$$ \vec{b}^\top (\vec{a} \times \vec{b}) = b_1(a_2 b_3 - a_3 b_2) + b_2(a_3 b_1 - a_1 b_3) + b_3(a_1 b_2 - a_2 b_1) $$
$$ = b_1 a_2 b_3 - b_1 a_3 b_2 + b_2 a_3 b_1 - b_2 a_1 b_3 + b_3 a_1 b_2 - b_3 a_2 b_1 = 0 $$

Consider three $3$ dimensional vectors:


$$ \vec{a} = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}, \quad \vec{c} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} $$

Since $\vec{b} \times \vec{c}$ is orthogonal to both $\vec{b}$ and $\vec{c}$, the parallelepiped with base formed by $\vec{b}$ and $\vec{c}$ has height equal to the scalar projection of $\vec{a}$ onto $\vec{b} \times \vec{c}$:


$$ \text{Comp}_{\vec{b} \times\vec{c}}(\vec{a}) = \frac{\vec{a}^\top (\vec{b} \times \vec{c})}{\|\vec{b} \times\vec{c}\|} $$

Since the magnitude of $\vec{b} \times \vec{c}$ is the area of the base, the volume of the parallelepiped formed by $\vec{a}$, $\vec{b}$, and $\vec{c}$ is:


$$ \frac{\vec{a}^\top (\vec{b} \times \vec{c})}{\|\vec{b} \times\vec{c}\|}\|\vec{b} \times\vec{c}\| = \vec{a}^\top (\vec{b} \times \vec{c}) $$

This is called the scalar triple product. Since volume is the same for three different bases:


$$ \vec{a}^\top (\vec{b} \times \vec{c}) = \vec{b}^\top (\vec{a} \times \vec{c}) = \vec{c}^\top (\vec{a} \times \vec{b}) $$

Consider a $3$ by $3$ matrix with these column vectors:


$$ A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} $$

Since the volume scaled by the column vectors is $\vec{a}^\top (\vec{b} \times \vec{c})$, then:


$$ \det(A) = \vec{a}^\top (\vec{b} \times \vec{c}) = a_1(b_2 c_3 - b_3 c_2) - a_2(b_1 c_3 - b_3 c_1) + a_3 (b_1 c_2 - b_2 c_1) $$

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