5. Cross Product

b. Cross Product and Triple Product

3. Properties of Cross Products

We here list the algebraic properties of the cross product and its relation to vector addition, scalar multiplication and the dot product. The most important property is the last one, the Pythagorean Identity. All vectors are in \(R^3\).

First, there are three properties involving just the cross product.

Let \(\vec u\) and \(\vec v\) be arbitrary vectors. Then,

  1. The Cross Product is Anti-Commutative: \(\qquad \vec u\times\vec v=-\,\vec v\times\vec u\)  

This follows from the antisymmetry property of determinants. \[ \vec u\times\vec v =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} =-\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ v_1 & v_2 & v_3 \\ u_1 & u_2 & u_3 \end{vmatrix} =-\,\vec v\times\vec u \]

  1. The Cross Product of a vector with itself is zero: \(\qquad \vec v\times\vec v=\vec 0\)  

This follows from the two equal rows property of determinants. \[ \vec v\times\vec v =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ v_1 & v_2 & v_3 \\ v_1 & v_2 & v_3 \end{vmatrix} =\vec 0 \]

  1. The standard vectors \(\hat\imath\), \(\hat\jmath\) and \(\hat k\) form a Right Handed Triplet:   \(\hat\imath\times\hat\jmath=\hat k\)   \(\hat\jmath\times\hat k=\hat\imath\)   \(\hat k\times\hat\imath=\hat\jmath\)  

For the first equation, we compute: \[\begin{aligned} \hat\imath\times\hat\jmath &=\left\langle 1,0,0\right\rangle\times\left\langle0,1,0\right\rangle =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{vmatrix} \\ &=\hat\imath(0)-\hat\jmath(0)+\hat k(1) =\left\langle 0,0,1\right\rangle =\hat k \end{aligned}\] Similarly, \[ \hat\jmath\times\hat k =\left\langle 0,1,0\right\rangle\times\left\langle 0,0,1\right\rangle =\left\langle 1,0,0\right\rangle =\hat\imath \] \[ \hat k\times\hat\imath =\left\langle 0,0,1\right\rangle\times\left\langle 1,0,0\right\rangle =\left\langle 0,1,0\right\rangle =\hat\jmath \]

Second, there are two properties relating the cross product and vector addition.

Let \(\vec u\), \(\vec v\) and \(\vec w\) be arbitrary vectors. Then,

  1. The Cross Product Distributes over Vector Addition:   \(\vec u\times(\vec v+\vec w) =\vec u\times\vec v+\vec u\times\vec w\)  

This follows from the additive property of determinants. \[\begin{aligned} &\vec u\times(\vec v+\vec w) =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ u_1 & u_2 & u_3 \\ v_1+w_1 & v_2+w_2 & v_3+w_3 \end{vmatrix} \\[5pt] &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} +\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ u_1 & u_2 & u_3 \\ w_1 & w_2 & w_3 \end{vmatrix} =\vec u\times\vec v+\vec u\times\vec w \end{aligned}\]

  1. The Cross Product of \(\vec 0\) and any vector is \(\vec 0\):   \(\vec 0\times\vec v=\vec 0\)  

We expand on the second row: \[ \vec 0\times\vec v= \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 0 & 0 & 0 \\ v_1 & v_2 & v_3 \end{vmatrix} =\vec 0 \]

Third, we have two relations between the cross product and scalar multiplication.

Let \(a\) be an arbitrary scalar and \(\vec u\) and \(\vec v\) be arbitrary vectors. Then,

  1. The Cross Product Associates with Scalar Multiplication:   \((a\vec u)\times\vec v=a(\vec u\times\vec v)\)  

This follows from the multiple of a row property of determinants. \[ (a\vec u)\times\vec v= \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ au_1 & au_2 & au_3 \\ v_1 & v_2 & v_3 \end{vmatrix} =a \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} =a(\vec u\times\vec v) \]

  1. The Cross Product Commutes with Scalar Multiplication:   \((a\vec u)\times\vec v=\vec u\times(a\vec v)\)  

This follows from the multiple of a row property of determinants used twice. \[ (a\vec u)\times\vec v= \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ au_1 & au_2 & au_3 \\ v_1 & v_2 & v_3 \end{vmatrix} =a \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ u_1 & u_2 & u_3 \\ av_1 & av_2 & av_3 \end{vmatrix} =\vec u\times(a\vec v) \]

It is also useful to note that the cross product itself is not associative. An easy counter-example can be found using the above formulas: \[ (\hat\imath\times\hat\imath)\times\hat k =\vec 0\times\hat k =\vec 0 \quad \text{but} \quad \hat\imath\times(\hat\imath\times\hat k) =\hat\imath\times(-\hat\jmath) =-\hat k. \]

Finally we look at the properties relating the dot product and cross product. Since the triple product is constructed from these, all properties of the triple product are in this catagory. However, there is one more property that is the most important property on this page.

Let \(\vec u\) and \(\vec v\) be arbitrary vectors. Then,

  1. Pythagorean Identity for Dot and Cross Products:   \((\vec u\cdot\vec v)^2+|\vec u\times\vec v|^2 =|\vec u|^2|\vec v|^2\)

The reason for the name Pythagorean Identity will become clear on the page on the geometric interpretation of the cross product.

It is essential that you fill in the details of this proof by hand on paper. Everyone needs to see the miraculous cancellation for themself.

Write out each of the following quantities in terms of the components of \(\vec u=\left\langle u_1,u_2,u_3\right\rangle\) and \(\vec v=\left\langle v_1,v_2,v_3\right\rangle\), no numbers, just symbols.

  1. With \(\vec u\cdot\vec v=u_1v_1+u_2v_2+u_3v_3\), multiply out \((\vec u\cdot\vec v)^2\), producing \(6\) terms.
  2. With \(\vec u\times\vec v =\left\langle u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1\right\rangle\), write out \(|\vec u\times\vec v|^2\), producing \(9\) terms.
  3. Write out \((\vec u\cdot\vec v)^2 +|\vec u\times\vec v|^2\), producing \(9\) terms.   (\(3\) terms from \((\vec u\cdot\vec v)^2\) cancel \(3\) terms from \(|\vec u\times\vec v|^2\).)
  4. With \(|\vec u|^2=(u_1)^2+(u_2)^2+(u_3)^2\) and \(|\vec v|^2=(v_1)^2+(v_2)^2+(v_3)^2\), multiply out \(|\vec u|^2|\vec v|^2\), producing \(9\) terms.
  5. Check that the result of step 3, \((\vec u\cdot\vec v)^2+|\vec u\times\vec v|^2\), is equal to the result of step 4, \(|\vec u|^2|\vec v|^2\). This is the Pythagorean Identity.

the Pythagorean Identity for Dot and Cross Products for the vectors \(\vec a=\left\langle 2,5,1\right\rangle\) and \(\vec b=\left\langle -3,2,4\right\rangle\). In other words, compute \((\vec a\cdot\vec b)^2\), \(|\vec a\times\vec b|^2\), and \(|\vec a|^2|\vec b|^2\) to see the first two add up to the third.

Verify means to check that a property works for specific vectors. It is not a proof!

\((\vec a\cdot\vec b)^2=64\),   \(|\vec a\times\vec b|^2=806\),   \(|\vec a|^2|\vec b|^2=870\),   \(64+806=870\)

\[ (\vec a\cdot\vec b)^2 =(-6+10+4)^2=64 \] \[ \vec a\times\vec b =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 2 & 5 & 1 \\ -3 & 2 & 4 \end{vmatrix} =\left\langle 18,-11,19\right\rangle \] \[ |\vec a\times\vec b|^2 =|\left\langle 18,-11,19\right\rangle|^2 =18^2+11^2+19^2=806 \] \[ (\vec a\cdot\vec b)^2+|\vec a\times\vec b|^2 =64+806=870 \] \[ |\vec a|^2 =4+25+1=30 \] \[ |\vec b|^2 =9+4+16=29 \] \[ |\vec a|^2|\vec b|^2 =30\cdot29=870 \]

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