Finding the cross product of two vectors

Building the formula for the cross product of two vectors

To take the cross product of two vectors

$a\langle a_1,a_2,a_3\rangle$

$b\langle b_1,b_2,b_3\rangle$

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we’ll create a matrix in the form

$\begin{vmatrix}\bold i&\bold j&\bold k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}$

As always, we’ll use the sign matrix

$\begin{vmatrix}+&-&+\\-&+&-\\+&-&+\end{vmatrix}$

to determine the signs for our top row. We’ll expand the matrix to

$\begin{vmatrix}\bold i&\bold j&\bold k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}=\bold i\begin{vmatrix}a_2&a_3\\b_2&b_3\end{vmatrix}-\bold j\begin{vmatrix}a_1&a_3\\b_1&b_3\end{vmatrix}+\bold k\begin{vmatrix}a_1&a_2\\b_1&b_2\end{vmatrix}$

$=\bold i(a_2b_3-a_3b_2)-\bold j(a_1b_3-a_3b_1)+\bold k(a_1b_2-a_2b_1)$

and then take the coefficients on $i$, $j$ and $k$ to form the cross product vector $c\langle{c_1},c_2,c_3\rangle$, where

$c_1=a_2b_3-a_3b_2$

$c_2=a_1b_3-a_3b_1$

$c_3=a_1b_2-a_2b_1$

If you can remember the formula for

$\bold i(a_2b_3-a_3b_2)-\bold j(a_1b_3-a_3b_1)+\bold k(a_1b_2-a_2b_1)$

then you can skip the matrices and go straight to this step. If not, just use the matrix approach.

How to find the cross product of two vectors

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Using the 3x3 matrix to find the cross product of two vectors

Example

Find the cross product of the vectors.

$a\langle2,-4,1\rangle$

$b\langle-2,5,7\rangle$

For the sake of this example, we’ll assume we can’t remember the formula for

$\bold i(a_2b_3-a_3b_2)-\bold j(a_1b_3-a_3b_1)+\bold k(a_1b_2-a_2b_1)$

and use the matrix.

If you can remember the formula,

then you can skip the matrices and go straight to this step. If not, just use the matrix approach.

Plugging the values from the given vectors into our $3\times3$ matrix, we get

???\overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\bold i & \bold j & \bold k \\ 2 & -4& 1 \\ -2& 5 & 7\end{vmatrix}???

???\overrightarrow{a}\times\overrightarrow{b}=\bold i\begin{vmatrix}-4 & 1 \\ 5 & 7\end{vmatrix}-\bold j\begin{vmatrix}2 & 1 \\ -2 & 7\end{vmatrix}+\bold k\begin{vmatrix}2 & -4 \\ -2 & 5\end{vmatrix}???

???\overrightarrow{a}\times\overrightarrow{b}=\bold i\left[(-4)(7)-(1)(5)\right]-\bold j\left[(2)(7)-(1)(-2)\right]+\bold k\left[(2)(5)-(-4)(-2)\right]???

???\overrightarrow{a}\times\overrightarrow{b}=\bold i(-28-5)-\bold j(14+2)+\bold k(10-8)???

???\overrightarrow{a}\times\overrightarrow{b}=-33\bold i-16\bold j+2\bold k???

???\overrightarrow{a}\times\overrightarrow{b}=\langle-33,-16,2\rangle???

This is the cross product of the vectors ???a??? and ???b???.

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