Chapter 11 Problem 26

Problem

Show that the cross product of two vectors ${\vec {A}}=A_{x}{\hat {i}}+A_{y}{\hat {j}}+A_{z}{\hat {k}}$, and ${\vec {B}}=B_{x}{\hat {i}}+B_{y}{\hat {j}}+B_{z}{\hat {k}}$ is

${\vec {A}}\times {\vec {B}}=\left(A_{y}B_{z}-A_{z}B_{y}\right){\hat {i}}+\left(A_{z}B_{x}-A_{x}B_{z}\right){\hat {i}}+\left(A_{x}B_{y}-A_{y}B_{x}\right){\hat {k}}$

Then show that the cross product can be written

${\vec {A}}\times {\vec {B}}=\left|{\begin{array}{ccc}{\hat {i}}&{\hat {j}}&{\hat {k}}\\A_{x}&A_{y}&A_{z}\\B_{x}&B_{y}&B_{z}\end{array}}\right|$

where we use the rules for evaluating a determinant.

Solution

(a)

${\vec {A}}\times {\vec {B}}=\left(A_{x}{\hat {i}}+A_{y}{\hat {j}}+A_{z}{\hat {k}}\right)\times \left(B_{x}{\hat {i}}+B_{y}{\hat {j}}+B_{z}{\hat {k}}\right)$ $=A_{x}B_{x}\left({\hat {i}}\times {\hat {i}}\right)+A_{x}B_{y}\left({\hat {i}}\times {\hat {j}}\right)+A_{x}B_{z}\left({\hat {i}}\times {\hat {k}}\right)$ $+A_{y}B_{x}\left({\hat {j}}\times {\hat {i}}\right)+A_{y}B_{y}\left({\hat {j}}\times {\hat {j}}\right)+A_{y}B_{z}\left({\hat {j}}\times {\hat {k}}\right)$ $+A_{z}B_{x}\left({\hat {k}}\times {\hat {i}}\right)+A_{z}B_{y}\left({\hat {k}}\times {\hat {j}}\right)+A_{z}B_{z}\left({\hat {k}}\times {\hat {k}}\right)$

${\vec {A}}\times {\vec {B}}=A_{x}B_{x}\left(0\right)+A_{x}B_{y}\left({\hat {k}}\right)+A_{x}B_{z}\left(-{\hat {j}}\right)$ $+A_{y}B_{x}\left(-{\hat {k}}\right)+A_{y}B_{y}\left(0\right)+A_{y}B_{z}\left({\hat {i}}\right)$ $+A_{z}B_{x}\left({\hat {j}}\right)+A_{z}B_{y}\left(-{\hat {i}}\right)+A_{z}B_{z}\left(0\right)$

$=\left(A_{y}B_{z}-A_{z}B_{y}\right){\hat {i}}+\left(A_{z}B_{x}-A_{x}B_{z}\right){\hat {i}}+\left(A_{x}B_{y}-A_{y}B_{x}\right){\hat {k}}$

(b)

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

${\vec {A}}\times {\vec {B}}=\left|{\begin{array}{ccc}{\hat {i}}&{\hat {j}}&{\hat {k}}\\A_{x}&A_{y}&A_{z}\\B_{x}&B_{y}&B_{z}\end{array}}\right|$ $=\left(A_{y}B_{z}-A_{z}B_{y}\right){\hat {i}}+\left(A_{z}B_{x}-A_{x}B_{z}\right){\hat {i}}+\left(A_{x}B_{y}-A_{y}B_{x}\right){\hat {k}}$