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???
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.
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???.