View source for Chapter 11 Problem 26

__NOTOC__
==Problem==
Show that the cross product of two vectors <math>\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}</math>, and <math>\vec{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}</math> is 

<math>\vec{A}\times\vec{B}=\left(A_yB_z-A_zB_y\right)\hat{i}+\left(A_zB_x-A_xB_z\right)\hat{i}+\left(A_xB_y-A_yB_x\right)\hat{k}</math>

Then show that the cross product can be written 

<math>\vec{A}\times\vec{B}=\left| \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
A_x & A_y & A_z \\
B_x & B_y & B_z \end{array} \right|</math>

where we use the rules for evaluating a determinant.

==Solution==
===(a)===

<math>\vec{A}\times\vec{B}=\left(A_x\hat{i}+A_y\hat{j}+A_z\hat{k}\right)\times\left(B_x\hat{i}+B_y\hat{j}+B_z\hat{k}\right)</math>
<math>=A_xB_x\left(\hat{i}\times\hat{i}\right)+A_xB_y\left(\hat{i}\times\hat{j}\right)+A_xB_z\left(\hat{i}\times\hat{k}\right)</math>
<math>+A_yB_x\left(\hat{j}\times\hat{i}\right)+A_yB_y\left(\hat{j}\times\hat{j}\right)+A_yB_z\left(\hat{j}\times\hat{k}\right)</math>
<math>+A_zB_x\left(\hat{k}\times\hat{i}\right)+A_zB_y\left(\hat{k}\times\hat{j}\right)+A_zB_z\left(\hat{k}\times\hat{k}\right)</math>

<math>\vec{A}\times\vec{B}=A_xB_x\left(0\right)+A_xB_y\left(\hat{k}\right)+A_xB_z\left(-\hat{j}\right)</math>
<math>+A_yB_x\left(-\hat{k}\right)+A_yB_y\left(0\right)+A_yB_z\left(\hat{i}\right)</math>
<math>+A_zB_x\left(\hat{j}\right)+A_zB_y\left(-\hat{i}\right)+A_zB_z\left(0\right)</math>

<math>=\left(A_yB_z-A_zB_y\right)\hat{i}+\left(A_zB_x-A_xB_z\right)\hat{i}+\left(A_xB_y-A_yB_x\right)\hat{k}</math>

===(b)===
<math>\vec{A}\times\vec{B}=\left| \begin{array}{ccc}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \end{array} \right|</math>
<math>=a_{11}\left(a_{22}a_{33}-a_{23}a_{32}\right)+a_{12}\left(a_{23}a_{31}-a_{21}a_{33}\right)+a_{13}\left(a_{21}a_{32}-a_{22}a_{31}\right)</math>

<math>\vec{A}\times\vec{B}=\left| \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
A_x & A_y & A_z \\
B_x & B_y & B_z \end{array} \right|</math>
<math>=\left(A_yB_z-A_zB_y\right)\hat{i}+\left(A_zB_x-A_xB_z\right)\hat{i}+\left(A_xB_y-A_yB_x\right)\hat{k}</math>
<math></math>