Equivalence of Definitions of Vector Cross Product

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Theorem

The following definitions of the concept of Vector Cross Product are equivalent:

Definition 1

The vector cross product, denoted $\mathbf a \times \mathbf b$, is defined as:

$\mathbf a \times \mathbf b = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ a_i & a_j & a_k \\ b_i & b_j & b_k \\ \end{vmatrix}$

where $\begin {vmatrix} \ldots \end {vmatrix}$ is interpreted as a determinant.

More directly:

$\mathbf a \times \mathbf b = \paren {a_j b_k - a_k b_j} \mathbf i - \paren {a_i b_k - a_k b_i} \mathbf j + \paren {a_i b_j - a_j b_i} \mathbf k$

Definition 2

The vector cross product, denoted $\mathbf a \times \mathbf b$, is defined as:

$\mathbf a \times \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \sin \theta \, \mathbf {\hat n}$

where:

$\norm {\mathbf a}$ denotes the length of $\mathbf a$
$\theta$ denotes the angle from $\mathbf a$ to $\mathbf b$, measured in the positive direction
$\hat {\mathbf n}$ is the unit vector perpendicular to both $\mathbf a$ and $\mathbf b$ in the direction according to the right-hand rule.


Proof

Let $\mathbf a \times \mathbf b$ be a vector cross product by definition $2$.

Then by definition:

$(1): \quad \mathbf a \times \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \sin \theta \, \mathbf {\hat n}$

By Angle Between Vectors in Terms of Dot Product:

$\cos \theta = \ds \frac {\mathbf a \cdot \mathbf b} {\norm {\mathbf a} \norm {\mathbf b} }$
Cross product equivalency circle.png

Note that $\theta$ is measured from $0$ to $\pi$.

If $\dfrac \pi 2 < \theta < \pi$, the dot product is negative and it exists in quadrant $2$.

The sine ratio is unaltered.

This gives:

\(\ds \sin \theta\) \(=\) \(\ds \frac {\sqrt {\paren {\norm {\mathbf a} \norm {\mathbf b} }^2 - \paren {\mathbf a \cdot \mathbf b}^2} } {\norm {\mathbf a} \norm {\mathbf b} }\)
\(\ds \leadsto \ \ \) \(\ds \mathbf a \times \mathbf b\) \(=\) \(\ds \norm {\mathbf a} \norm {\mathbf b} \ds \frac {\sqrt {\paren {\norm {\mathbf a} \norm {\mathbf b} }^2 - \paren {\mathbf a \cdot \mathbf b}^2} } {\norm {\mathbf a} \norm {\mathbf b} } \mathbf {\hat n}\) substituting for $\sin \theta$ in $(1)$
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \sqrt {\paren {\norm {\mathbf a} \norm {\mathbf b} }^2 - \paren {\mathbf a \cdot \mathbf b}^2} \mathbf {\hat n}\) cancelling


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Next we have:

\(\ds \paren {\norm {\mathbf a} \norm {\mathbf b} }^2\) \(=\) \(\ds \paren {\sqrt { {a_i}^2 + {a_j}^2 + {a_k}^2} \sqrt { {b_i}^2 + {b_j}^2 + {b_k}^2} }^2\)
\(\ds \) \(=\) \(\ds \paren { {a_i}^2 + {a_j}^2 + {a_k}^2} \paren { {b_i}^2 + {b_j}^2 + {b_k} ^2}\)
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds {a_i}^2 {b_i}^2 + {a_i}^2 {b_j}^2 + {a_i}^2 {b_k}^2 + {a_j}^2 {b_i}^2 + {a_j}^2 {b_j}^2 + {a_j}^2 {b_k}^2 + {a_k}^2 {b_i}^2 + {a_k}^2 {b_j}^2 + {a_k}^2 {b_k}^2\)


Then:

\(\ds \paren {\mathbf a \cdot \mathbf b}^2\) \(=\) \(\ds \paren {a_i b_i + a_j b_j + a_k b_k}^2\) Definition of Dot Product
\(\text {(4)}: \quad\) \(\ds \) \(=\) \(\ds {a_i}^2 {b_1}^2 + 2 a_i b_i a_j b_j + {a_j}^2 {b_j}^2 + 2 a_i b_i a_k b_k + {a_k}^2 {b_k}^2 + 2 a_j b_j a_k b_k\) multiplying out

Hence:

\(\ds \paren {\norm {\mathbf a} \norm {\mathbf b} }^2 - \paren {\mathbf a \cdot \mathbf b}^2\) \(=\) \(\ds \paren { {a_i}^2 {b_i}^2 + {a_i}^2 {b_j}^2 + {a_i}^2 {b_k}^2 + {a_j}^2 {b_i}^2 + {a_j}^2 {b_j}^2 + {a_j}^2 {b_k}^2 + {a_k}^2 {b_i}^2 + {a_k}^2 {b_j}^2 } - \paren { {a_i}^2 {b_1}^2 + 2 a_i b_i a_j b_j + {a_j}^2 {b_j}^2 + 2 a_i b_i a_k b_k + {a_k}^2 {b_k}^2 + 2 a_j b_j a_k b_k}\) subtracting $(4)$ from $(3)$
\(\ds \) \(=\) \(\ds \paren { {a_i}^2 {b_j}^2 - 2 a_i b_i a_j b_j + {a_j}^2 {b_i}^2} + \paren { {a_k}^2 {b_i}^2 - 2 a_j b_j a_k b_k + {a_i}^2 {b_k}^2} + \paren { {a_j}^2 {b_k}^2 - 2 a_j b_j a_k b_k + {a_k}^2 {b_j}^2}\) rearranging
\(\ds \) \(=\) \(\ds \paren {a_i b_j - a_j b_i}^2 + \paren {a_k b_i - a_i b_k}^2 + \paren {a_j b_k - a_k b_j}^2\) Square of Difference

To simplify subsequent algebra, let us define $u$:

\(\ds u\) \(:=\) \(\ds \paren {\norm {\mathbf a} \norm {\mathbf b} }^2 - \paren {\mathbf a \cdot \mathbf b}^2\)
\(\text {(5)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \mathbf a \times \mathbf b\) \(=\) \(\ds \sqrt u \, \mathbf {\hat n}\) substituting back into $(2)$


Let $\mathbf {\hat n} = \begin {bmatrix} n_i \\ n_j \\ n_k \end {bmatrix}$

Now, by definition $2$ of the cross product:

\(\ds \norm {\mathbf {\hat n} }\) \(=\) \(\ds 1\) by hypothesis
\(\ds \mathbf {\hat n} \cdot \mathbf a\) \(=\) \(\ds 0\) Dot Product of Perpendicular Vectors
\(\ds \mathbf {\hat n} \cdot \mathbf b\) \(=\) \(\ds 0\) Dot Product of Perpendicular Vectors


Hence by definition of dot product:

\(\text {(6)}: \quad\) \(\ds \sqrt { {n_i}^2 + {n_j}^2 + {n_k}^2}\) \(=\) \(\ds 1\)
\(\text {(7)}: \quad\) \(\ds n_i a_i + n_j a_j + n_k a_k\) \(=\) \(\ds 0\)
\(\text {(8)}: \quad\) \(\ds n_i b_i + n_j b_j + n_k b_k\) \(=\) \(\ds 0\)
\(\text {(9)}: \quad\) \(\ds \leadsto \ \ \) \(\ds n_i\) \(=\) \(\ds \frac {-n_j a_j - n_k a_k} {a_i}\) from $(7)$
\(\ds \leadsto \ \ \) \(\ds \frac {-n_j a_j b_i} {a_i} - \frac {n_k a_k b_i} {a_i} + n_j b_j + n_k b_k\) \(=\) \(\ds 0\) substituting into $(8)$
\(\ds \leadsto \ \ \) \(\ds n_j \paren {\frac {a_i b_j - a_j b_i} {a_i} }\) \(=\) \(\ds n_k \paren {\frac {a_k b_i - a_i b_k} {a_i} }\) splitting fractions and distributing $b_i$
\(\text {(10)}: \quad\) \(\ds \leadsto \ \ \) \(\ds n_j\) \(=\) \(\ds n_k \paren {\frac {a_k b_i - a_i b_k} {a_i b_j - a_j b_i} }\) multiplying both sides by $\dfrac {a_i} {a_i b_j - a_j b_i}$
\(\ds \leadsto \ \ \) \(\ds n_i\) \(=\) \(\ds \frac {-n_k \paren {\dfrac {a_k b_i - a_i b_k} {a_i b_j - a_j b_i} } a_j - n_k a_k} {a_i}\) substituting into $(9)$
\(\ds \) \(=\) \(\ds n_k \paren {\dfrac {a_j a_i b_k - a_j b_i a_k - a_k a_i b_j + a_k b_i a_j} {a_i b_j - a_j b_i} } \paren {\frac 1 {a_i} }\)
\(\ds \) \(=\) \(\ds n_k \paren {\dfrac {a_i \paren {a_j b_k - a_k b_j} } {a_i b_j - a_j b_i} } \paren {\frac 1 {a_i} }\)
\(\text {(11)}: \quad\) \(\ds \leadsto \ \ \) \(\ds n_i\) \(=\) \(\ds n_k \paren {\frac {a_j b_k - a_k b_j} {a_i b_j - a_j b_i} }\)


Then:

\(\ds \sqrt {\paren {n_k \paren {\frac {a_j b_k - a_k b_j} {a_i b_j - a_j b_i} } }^2 + \paren {n_k \paren {\frac {a_k b_i - a_i b_k} {a_i b_j - a_j b_i} } }^2 + \paren {n_k \paren {\frac {a_i b_j - a_j b_i} {a_i b_j - a_j b_i} } }^2}\) \(=\) \(\ds 1\) substituting from $(10)$ and $(11)$ into $(6)$
\(\ds \leadsto \ \ \) \(\ds \sqrt { {n_k}^2 \paren {\frac {\paren {a_j b_k - a_k b_j}^2 + \paren {a_k b_i - a_i b_k}^2 + \paren {a_i b_j - a_j b_i}^2} {\paren {a_i b_j - a_j b_i}^2} } }\) \(=\) \(\ds 1\) rearranging
\(\ds \leadsto \ \ \) \(\ds n_k \paren {\frac {\sqrt u} {a_i b_j - a_j b_i} }\) \(=\) \(\ds 1\) as the numerator is $u$
\(\text {(12)}: \quad\) \(\ds \leadsto \ \ \) \(\ds n_k\) \(=\) \(\ds \frac {a_i b_j - a_j b_i} {\sqrt u}\)


Substituting from $(12)$ into $(11)$:

\(\ds n_i\) \(=\) \(\ds \frac {a_i b_j - a_j b_i} {\sqrt u} \paren {\frac {a_j b_k - a_k b_j} {a_i b_j - a_j b_i} }\)
\(\text {(13)}: \quad\) \(\ds \) \(=\) \(\ds \frac {a_j b_k - a_k b_j} {\sqrt u}\)

and from $(12)$ into $(10)$:

\(\ds n_j\) \(=\) \(\ds \frac {a_i b_j - a_j b_i} {\sqrt u} \paren {\frac {a_k b_i - a_i b_k} {a_i b_j - a_j b_i} }\)
\(\text {(14)}: \quad\) \(\ds \) \(=\) \(\ds \frac {a_k b_i - a_i b_k} {\sqrt u}\)


Hence from $(12)$, $(13)$ and $(14)$ we have the components of $\mathbf {\hat n}$:

\(\ds \mathbf {\hat n}\) \(=\) \(\ds \begin {bmatrix} \dfrac {a_j b_k - a_k b_j} {\sqrt u} \\ \dfrac {a_k b_i - a_i b_k} {\sqrt u} \\ \dfrac {a_i b_j - a_j b_i} {\sqrt u} \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt u} \begin {bmatrix} a_j b_k - a_k b_j \\ a_k b_i - a_i b_k \\ a_i b_j - a_j b_i \end {bmatrix}\)


Substituting back into $(5)$:

\(\ds \mathbf a \times \mathbf b\) \(=\) \(\ds \sqrt u \frac 1 {\sqrt u} \begin {bmatrix} a_j b_k - a_k b_j \\ a_k b_i - a_i b_k \\ a_i b_j - a_j b_i \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} a_j b_k - a_k b_j \\ a_k b_i - a_i b_k \\ a_i b_j - a_j b_i \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \paren {a_j b_k - a_k b_j} \mathbf i + \paren {a_k b_i - a_i b_k} \mathbf j + \paren {a_i b_j - a_j b_i} \mathbf k\) in component form
\(\ds \) \(=\) \(\ds \paren {a_j b_k - a_k b_j} \mathbf i - \paren {a_i b_k - a_k b_i} \mathbf j + \paren {a_i b_j - a_j b_i}\mathbf k\)

Hence the result.

$\blacksquare$


Sources

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