# Equivalence of Definitions of Vector Cross Product

## 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:

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

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

 $\ds \norm {\mathbf {\hat n} }$ $=$ $\ds 1$ 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.

$\blacksquare$