Original Definition
Given two vectors u \mathbf{u} u v \mathbf{v} v
u = u 1 i + u 2 j + u 3 k \mathbf{u} =u_1\mathbf{i}+u_2\mathbf{j}+u_3\mathbf{k} u = u 1 i + u 2 j + u 3 k v = v 1 i + v 2 j + v 3 k \mathbf{v} =v_1\mathbf{i}+v_2\mathbf{j}+v_3\mathbf{k} v = v 1 i + v 2 j + v 3 k The cross product u × v \mathbf{u}\times \mathbf{v} u × v vector that is orthogonal to the plane created by u \mathbf{u} u v \mathbf{v} v
∥ u × v ∥ = ∥ u ∥ ∥ v ∥ sin θ \|\mathbf{u}\times \mathbf{v}\|=\|\mathbf{u}\|\|\mathbf{v}\|\sin{\theta} ∥ u × v ∥ = ∥ u ∥∥ v ∥ sin θ The direction (right hand rule):
Alternative Definition
Sometimes, you might encounter a definition of the cross product like this:
u × v = ( u 2 v 3 − u 3 v 2 ) i − ( u 1 v 3 − u 3 v 1 ) j + ( u 1 v 2 − u 2 v 1 ) k \mathbf{u}\times \mathbf{v}=(u_2v_3-u_3v_2)\mathbf{i}-(u_1v_3-u_3v_1)\mathbf{j}+(u_1v_2-u_2v_1)\mathbf{k} u × v = ( u 2 v 3 − u 3 v 2 ) i − ( u 1 v 3 − u 3 v 1 ) j + ( u 1 v 2 − u 2 v 1 ) k We will try to Derive this. Here are some useful properties:
u × v = − v × u u × ( v + w ) = ( u × v ) + ( u × w ) c ( u × v ) = ( c u ) × v = u × ( c v ) u × 0 = 0 × u = 0 u × u = 0 u ⋅ ( v × w ) = ( u × v ) ⋅ w i × j = k j × k = i k × i = j \begin{align}
\mathbf{u} \times\mathbf{v}&=-\mathbf{v} \times\mathbf{u}\\
\mathbf{u}\times (\mathbf{v}+\mathbf{w})&=(\mathbf{u}\times\mathbf{v})+(\mathbf{u}\times\mathbf{w})\\
c(\mathbf{u}\times \mathbf{v})&=(c\mathbf{u})\times \mathbf{v}=\mathbf{u}\times (c\mathbf{v})\\
\mathbf{u}\times 0&= 0\times \mathbf{u}=0\\
\mathbf{u}\times\mathbf{u}&=0\\
\mathbf{u}\cdot(\mathbf{v}\times \mathbf{w})&=(\mathbf{u}\times \mathbf{v})\cdot \mathbf{w}\\
\mathbf{i}\times \mathbf{j}=\mathbf{k} \qquad
\mathbf{j}\times &\mathbf{k}=\mathbf{i} \qquad
\mathbf{k}\times \mathbf{i}=\mathbf{j}
\end{align} u × v u × ( v + w ) c ( u × v ) u × 0 u × u u ⋅ ( v × w ) i × j = k j × = − v × u = ( u × v ) + ( u × w ) = ( c u ) × v = u × ( c v ) = 0 × u = 0 = 0 = ( u × v ) ⋅ w k = i k × i = j Derivation u × v = ( u 1 i + u 2 j + u 3 k ) × ( v 1 i + v 2 j + v 3 k ) = ( u 1 i × v 1 i ) + ( u 1 i × v 2 j ) + ( u 1 i × v 3 k ) + ( u 2 j × v 1 i ) + ( u 2 j × v 2 j ) + ( u 2 j × v 3 k ) + (Property 2) ( u 3 k × v 1 i ) + ( u 3 k × v 2 j ) + ( u 3 k × v 3 k ) = u 1 v 1 ( i × i ) + u 1 v 2 ( i × j ) + u 1 v 3 ( i × k ) + u 2 v 1 ( j × i ) + u 2 v 2 ( j × j ) + u 2 v 3 ( j × k ) + (Property 3) u 3 v 1 ( k × i ) + u 3 v 2 ( k × j ) + u 3 v 3 ( k × k ) = u 1 v 2 ( i × j ) + u 1 v 3 ( i × k ) + u 2 v 1 ( j × i ) + u 2 v 3 ( j × k ) + u 3 v 1 ( k × i ) + u 3 v 2 ( k × j ) (Property 5) = u 1 v 2 ( i × j ) − u 2 v 1 ( i × j ) + u 2 v 3 ( j × k ) − u 3 v 2 ( j × k ) + (Property 1) u 3 v 1 ( k × i ) − u 1 v 3 ( k × i ) = ( u 1 v 2 − u 2 v 1 ) k + ( u 2 v 3 − u 3 v 2 ) i + ( u 3 v 1 − u 1 v 3 ) j (Property 7) = ( u 2 v 3 − u 3 v 2 ) i − ( u 1 v 3 − u 3 v 1 ) j + ( u 1 v 2 − u 2 v 1 ) k (Rearrange) \begin{align*}
\mathbf{u}\times\mathbf{v} =& (u_1\mathbf{i}+u_2\mathbf{j}+u_3\mathbf{k})\times (v_1\mathbf{i}+v_2\mathbf{j}+v_3\mathbf{k}) \\
=&(u_1\mathbf{i}\times v_1\mathbf{i})+(u_1\mathbf{i}\times v_2\mathbf{j})+(u_1\mathbf{i}\times v_3\mathbf{k})+\\
&(u_2\mathbf{j}\times v_1\mathbf{i})+(u_2\mathbf{j}\times v_2\mathbf{j})+(u_2\mathbf{j}\times v_3\mathbf{k})+
&&\text{(Property 2)}\\
&(u_3\mathbf{k}\times v_1\mathbf{i})+(u_3\mathbf{k}\times v_2\mathbf{j})+(u_3\mathbf{k}\times v_3\mathbf{k})\\
=&u_1v_1(\mathbf{i}\times \mathbf{i})+
u_1v_2(\mathbf{i}\times \mathbf{j})+
u_1v_3(\mathbf{i}\times \mathbf{k})+\\
&u_2v_1(\mathbf{j}\times \mathbf{i})+
u_2v_2(\mathbf{j}\times \mathbf{j})+
u_2v_3(\mathbf{j}\times \mathbf{k})+
&&\text{(Property 3)}\\
&u_3v_1(\mathbf{k}\times \mathbf{i})+
u_3v_2(\mathbf{k}\times \mathbf{j})+
u_3v_3(\mathbf{k}\times \mathbf{k})\\
=&u_1v_2(\mathbf{i}\times \mathbf{j})+
u_1v_3(\mathbf{i}\times \mathbf{k})+
u_2v_1(\mathbf{j}\times \mathbf{i})+\\
&u_2v_3(\mathbf{j}\times \mathbf{k})+
u_3v_1(\mathbf{k}\times \mathbf{i})+
u_3v_2(\mathbf{k}\times \mathbf{j})
&&\text{(Property 5)}\\
=&u_1v_2(\mathbf{i}\times \mathbf{j})-u_2v_1(\mathbf{i}\times \mathbf{j})+\\
&u_2v_3(\mathbf{j}\times \mathbf{k})-u_3v_2(\mathbf{j}\times \mathbf{k})+&&\text{(Property 1)}\\
&u_3v_1(\mathbf{k}\times \mathbf{i})-u_1v_3(\mathbf{k}\times \mathbf{i})\\
=&(u_1v_2-u_2v_1)\mathbf{k}+(u_2v_3-u_3v_2)\mathbf{i}+(u_3v_1-u_1v_3)\mathbf{j}
&&\text{(Property 7)}\\
=&(u_2v_3-u_3v_2)\mathbf{i}-(u_1v_3-u_3v_1)\mathbf{j}+(u_1v_2-u_2v_1)\mathbf{k}
&&\text{(Rearrange)}
\end{align*} u × v = = = = = = = ( u 1 i + u 2 j + u 3 k ) × ( v 1 i + v 2 j + v 3 k ) ( u 1 i × v 1 i ) + ( u 1 i × v 2 j ) + ( u 1 i × v 3 k ) + ( u 2 j × v 1 i ) + ( u 2 j × v 2 j ) + ( u 2 j × v 3 k ) + ( u 3 k × v 1 i ) + ( u 3 k × v 2 j ) + ( u 3 k × v 3 k ) u 1 v 1 ( i × i ) + u 1 v 2 ( i × j ) + u 1 v 3 ( i × k ) + u 2 v 1 ( j × i ) + u 2 v 2 ( j × j ) + u 2 v 3 ( j × k ) + u 3 v 1 ( k × i ) + u 3 v 2 ( k × j ) + u 3 v 3 ( k × k ) u 1 v 2 ( i × j ) + u 1 v 3 ( i × k ) + u 2 v 1 ( j × i ) + u 2 v 3 ( j × k ) + u 3 v 1 ( k × i ) + u 3 v 2 ( k × j ) u 1 v 2 ( i × j ) − u 2 v 1 ( i × j ) + u 2 v 3 ( j × k ) − u 3 v 2 ( j × k ) + u 3 v 1 ( k × i ) − u 1 v 3 ( k × i ) ( u 1 v 2 − u 2 v 1 ) k + ( u 2 v 3 − u 3 v 2 ) i + ( u 3 v 1 − u 1 v 3 ) j ( u 2 v 3 − u 3 v 2 ) i − ( u 1 v 3 − u 3 v 1 ) j + ( u 1 v 2 − u 2 v 1 ) k (Property 2) (Property 3) (Property 5) (Property 1) (Property 7) (Rearrange) Hence,
u × v = ( u 2 v 3 − u 3 v 2 ) i − ( u 1 v 3 − u 3 v 1 ) j + ( u 1 v 2 − u 2 v 1 ) k ■ \mathbf{u}\times \mathbf{v}=(u_2v_3-u_3v_2)\mathbf{i}-(u_1v_3-u_3v_1)\mathbf{j}+(u_1v_2-u_2v_1)\mathbf{k} \tag*{$\blacksquare$} u × v = ( u 2 v 3 − u 3 v 2 ) i − ( u 1 v 3 − u 3 v 1 ) j + ( u 1 v 2 − u 2 v 1 ) k ■ It is obvious that according to our original definition of cross product, the magnitude of the cross product is also equal to the area of parallelogram created by u \mathbf{u} u v \mathbf{v} v
Related to parallelogram:
∥ u × v ∥ = ∥ u ∥ ∥ v ∥ sin θ = S parallelogram \|\mathbf{u}\times \mathbf{v}\|=\|\mathbf{u}\|\|\mathbf{v}\|\sin{\theta}=S_{\text{parallelogram}} ∥ u × v ∥ = ∥ u ∥∥ v ∥ sin θ = S parallelogram Interestingly, we could Derive this from our alternative definition . Recall some properties:
∥ u ∥ = u 1 2 + u 2 2 + u 3 2 u ⋅ v = u 1 v 1 + u 2 v 2 + u 3 v 3 u ⋅ v = ∣ u ∥ ∥ v ∥ cos θ u × v = ( u 2 v 3 − u 3 v 2 ) i − ( u 1 v 3 − u 3 v 1 ) j + ( u 1 v 2 − u 2 v 1 ) k cos 2 θ + sin 2 θ = 1 ( a − b ) 2 = a 2 − 2 a b + b 2 ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( a b + a c + b c ) \begin{align}
\|\mathbf{u}\|=&\sqrt{u_1^2+u_2^2+u_3^2}\\
\mathbf{u}\cdot \mathbf{v}=u_1v_1&+u_2v_2+u_3v_3\\
\mathbf{u}\cdot \mathbf{v}=|\mathbf{u}\|&\|\mathbf{v}\|\cos{\theta}\\
\mathbf{u}\times \mathbf{v}=(u_2v_3-u_3v_2)\mathbf{i}-(u_1v_3&-u_3v_1)\mathbf{j}+(u_1v_2-u_2v_1)\mathbf{k}\\
\cos^2{\theta}+\sin^2&{\theta}=1\\
(a-b)^2=a^2-&2ab+b^2\\
(a+b+c)^2=a^2+b^2+c^2&+2(ab+ac+bc)
\end{align} ∥ u ∥ = u ⋅ v = u 1 v 1 u ⋅ v = ∣ u ∥ u × v = ( u 2 v 3 − u 3 v 2 ) i − ( u 1 v 3 cos 2 θ + sin 2 ( a − b ) 2 = a 2 − ( a + b + c ) 2 = a 2 + b 2 + c 2 u 1 2 + u 2 2 + u 3 2 + u 2 v 2 + u 3 v 3 ∥ v ∥ cos θ − u 3 v 1 ) j + ( u 1 v 2 − u 2 v 1 ) k θ = 1 2 ab + b 2 + 2 ( ab + a c + b c ) Since we want ∥ u ∥ ∥ v ∥ sin θ = ∥ u × v ∥ \|\mathbf{u}\|\|\mathbf{v}\|\sin{\theta} = \|\mathbf{u}\times \mathbf{v}\| ∥ u ∥∥ v ∥ sin θ = ∥ u × v ∥ Derive the squared version:
∥ u ∥ 2 ∥ v ∥ 2 sin 2 θ = ∥ u × v ∥ 2 \|\mathbf{u}\|^2\|\mathbf{v}\|^2\sin^2{\theta} = \|\mathbf{u}\times \mathbf{v}\|^2 ∥ u ∥ 2 ∥ v ∥ 2 sin 2 θ = ∥ u × v ∥ 2 Derivation ∥ u ∥ 2 ∥ v ∥ 2 sin 2 θ = ∥ u ∥ 2 ∥ v ∥ 2 ( 1 − cos 2 θ ) (Property 12) = ∥ u ∥ 2 ∥ v ∥ 2 − ( ∥ u ∥ 2 ∥ v ∥ 2 cos 2 θ ) = ∥ u ∥ 2 ∥ v ∥ 2 − ( u ⋅ v ) 2 (Property 10) = ∥ u ∥ 2 ∥ v ∥ 2 − ( u 1 v 1 + u 2 v 2 + u 3 v 3 ) 2 (Property 9) = ( u 1 2 + u 2 2 + u 3 2 ) ( v 1 2 + v 2 2 + v 3 2 ) − (Property 8) [ u 1 2 v 1 2 + u 2 2 v 2 2 + u 3 2 v 3 2 + 2 ( u 1 v 1 u 2 v 2 + u 1 v 1 u 3 v 3 + u 2 v 2 u 3 v 3 ) ] (Property 14) = u 1 2 v 1 2 + u 1 2 v 2 2 + u 1 2 v 3 2 + u 2 2 v 1 2 + u 2 2 v 2 2 + u 2 2 v 3 2 + u 3 2 v 1 2 + u 3 2 v 2 2 + u 3 2 v 3 2 − (Cancel terms) u 1 2 v 1 2 − u 2 2 v 2 2 − u 3 2 v 3 2 − 2 u 1 v 1 u 2 v 2 − 2 u 1 v 1 u 3 v 3 − 2 u 2 v 2 u 3 v 3 = ( u 2 2 v 3 2 − 2 u 2 v 2 u 3 v 3 + u 3 2 v 2 2 ) + ( u 1 2 v 3 2 − 2 u 1 v 1 u 3 v 3 + u 3 2 v 1 2 ) + (Rearrange) ( u 1 2 v 2 2 − 2 u 1 v 1 u 2 v 2 + u 2 2 v 1 2 ) = ( u 2 v 3 − u 3 v 2 ) 2 + ( u 1 v 3 − u 3 v 1 ) 2 + ( u 1 v 2 − u 2 v 1 ) 2 (Property 10) = ∥ u × v ∥ 2 \begin{align*}
\|\mathbf{u}\|^2\|\mathbf{v}\|^2\sin^2{\theta} =&\|\mathbf{u}\|^2\|\mathbf{v}\|^2(1-\cos^2{\theta})
&&\text{(Property 12)}\\
=&\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\|\mathbf{u}\|^2\|\mathbf{v}\|^2\cos^2{\theta})\\
=&\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot \mathbf{v})^2
&&\text{(Property 10)}\\
=&\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(u_1v_1+u_2v_2+u_3v_3)^2
&&\text{(Property 9)}\\
=&(u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2)-
&&\text{(Property 8)}\\
&[u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2+2(u_1v_1u_2v_2+u_1v_1u_3v_3+u_2v_2u_3v_3)]
&&\text{(Property 14)}\\
=&\cancel{u_1^2v_1^2}+u_1^2v_2^2+u_1^2v_3^2+\\
&u_2^2v_1^2+\cancel{u_2^2v_2^2}+u_2^2v_3^2+\\
&u_3^2v_1^2+u_3^2v_2^2+\cancel{u_3^2v_3^2}-&&\text{(Cancel terms)}\\
&\cancel{u_1^2v_1^2}-\cancel{u_2^2v_2^2}-\cancel{u_3^2v_3^2}-\\
&2u_1v_1u_2v_2-2u_1v_1u_3v_3-2u_2v_2u_3v_3\\
=&(u_2^2v_3^2-2u_2v_2u_3v_3+u_3^2v_2^2)+\\
&(u_1^2v_3^2-2u_1v_1u_3v_3+u_3^2v_1^2)+
&&\text{(Rearrange)}\\
&(u_1^2v_2^2-2u_1v_1u_2v_2+u_2^2v_1^2)\\
=&(u_2v_3-u_3v_2)^2+(u_1v_3-u_3v_1)^2+(u_1v_2-u_2v_1)^2
&&\text{(Property 10)}\\
=&\|\mathbf{u}\times \mathbf{v}\|^2
\tag*{\text{(Property 11 and 8)}}\\
\end{align*} ∥ u ∥ 2 ∥ v ∥ 2 sin 2 θ = = = = = = = = = ∥ u ∥ 2 ∥ v ∥ 2 ( 1 − cos 2 θ ) ∥ u ∥ 2 ∥ v ∥ 2 − ( ∥ u ∥ 2 ∥ v ∥ 2 cos 2 θ ) ∥ u ∥ 2 ∥ v ∥ 2 − ( u ⋅ v ) 2 ∥ u ∥ 2 ∥ v ∥ 2 − ( u 1 v 1 + u 2 v 2 + u 3 v 3 ) 2 ( u 1 2 + u 2 2 + u 3 2 ) ( v 1 2 + v 2 2 + v 3 2 ) − [ u 1 2 v 1 2 + u 2 2 v 2 2 + u 3 2 v 3 2 + 2 ( u 1 v 1 u 2 v 2 + u 1 v 1 u 3 v 3 + u 2 v 2 u 3 v 3 )] u 1 2 v 1 2 + u 1 2 v 2 2 + u 1 2 v 3 2 + u 2 2 v 1 2 + u 2 2 v 2 2 + u 2 2 v 3 2 + u 3 2 v 1 2 + u 3 2 v 2 2 + u 3 2 v 3 2 − u 1 2 v 1 2 − u 2 2 v 2 2 − u 3 2 v 3 2 − 2 u 1 v 1 u 2 v 2 − 2 u 1 v 1 u 3 v 3 − 2 u 2 v 2 u 3 v 3 ( u 2 2 v 3 2 − 2 u 2 v 2 u 3 v 3 + u 3 2 v 2 2 ) + ( u 1 2 v 3 2 − 2 u 1 v 1 u 3 v 3 + u 3 2 v 1 2 ) + ( u 1 2 v 2 2 − 2 u 1 v 1 u 2 v 2 + u 2 2 v 1 2 ) ( u 2 v 3 − u 3 v 2 ) 2 + ( u 1 v 3 − u 3 v 1 ) 2 + ( u 1 v 2 − u 2 v 1 ) 2 ∥ u × v ∥ 2 (Property 12) (Property 10) (Property 9) (Property 8) (Property 14) (Cancel terms) (Rearrange) (Property 10) (Property 11 and 8) Hence,
∥ u ∥ 2 ∥ v ∥ 2 sin 2 θ = ∥ u × v ∥ 2 \|\mathbf{u}\|^2\|\mathbf{v}\|^2\sin^2{\theta} = \|\mathbf{u}\times \mathbf{v}\|^2 ∥ u ∥ 2 ∥ v ∥ 2 sin 2 θ = ∥ u × v ∥ 2 ∥ u × v ∥ = ∥ u ∥ ∥ v ∥ sin θ = S parallelogram ■ \|\mathbf{u}\times \mathbf{v}\|=\|\mathbf{u}\|\|\mathbf{v}\|\sin{\theta}=S_{\text{parallelogram}} \tag*{$\blacksquare$} ∥ u × v ∥ = ∥ u ∥∥ v ∥ sin θ = S parallelogram ■ 
