Friday 8 August 2014

CHAPTER 21 -Properties of Cross Product

We now note some important properties of the cross product:
(i) If \vec a and \vec b are parallel, their cross product is zero, i.e.
\vec a \times \vec b = \vec 0
since \sin {\rm{\theta }} = 0. Conversely, if \vec a \times \vec b = \vec 0, then \vec a and \vec b must be parallel.
(ii) The cross product is not commutative. In fact,
\vec b \times \vec a =  - \vec a \times \vec b
This is because the direction of \vec a \times \vec b was defined so that \vec a\vec b and \vec a \times \vec b form a right handed system
(iii) The cross product is distributive over vector addition:
\vec a \times (\vec b + \vec c) = \vec a \times \vec b + \vec a \times \vec c
and (\vec a + \vec b) \times \vec c = \vec a \times \vec c + \vec b \times \vec c
(iv) \hat i \times \hat i = \hat j \times \hat j = \hat k \times \hat k = \vec 0
\hat i \times \hat j = \hat k,\;\;\hat j \times \hat k = \hat i,\;\;\hat k \times \hat i = \hat j
These relations can be remembered as
Going in the reverse direction, we have
\hat j \times \hat i =  - \hat k,\;\;\hat i \times \hat k =  - \hat j,\;\;\hat k \times \hat j =  - \hat i
Thus, for two vectors \vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k and \vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k we have
\vec a \times \vec b = ({a_1}\hat i + {a_2}\hat j + {a_3}\hat k) \times ({b_1}\hat i + {b_2}\hat j + {b_3}\hat k)
 = {a_1}\;{b_2}\hat k - {a_1}{b_3}\hat j - {a_2}{b_1}\hat k + {a_2}{b_3}\hat i + {a_3}{b_1}\hat j - {a_3}{b_2}\hat i
 = \hat i({a_2}{b_3} - {a_3}{b_2}) + \hat j({a_3}{b_1} - {a_1}{b_3}) + \hat k({a_1}{b_2} - {a_2}{b_1})
This can be written concisely in determinant notation as
\vec a \times \vec b = \left| {\begin{array}{*{20}{c}}  {\hat i}\,\,\,\,{\hat j}\,\,\,\,{\hat k}\\  {{a_1}}\,\,\,\,{{a_2}}\,\,\,\,{{a_3}}\\  {{b_1}}\,\,\,\,{{b_2}}\,\,\,\,{{b_3}}  \end{array}} \right|
(v) The unit vector(s) \hat r normal to the plane of \vec a and \vec b can be written as
\hat r =  \pm \dfrac{{\vec a \times \vec b}}{{\left| {\vec a \times \vec b} \right|}}

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