3.11 *Vector Triple Product*

(a×b) ×c is perpendicular to both c and a×b, so lies in the plane of a and b. Basic result obtained easily, is,

NB. The order and the brackets must not be changed , if we do this will alter the result. If c is normal to the plane of a and b then (a×b) ×c = 0 (Why?)

Example 3.13: 

Find (a×b) ×c and a× (b×c) given a = i− 2j −k, b = 2i−j −k, c = i + 3j + 2k.

Solution: 

\begin{eqnarray*} (a×b) ×c& =& (a⋅c)b− (b⋅c)a%& \\ & =& 5b + 3a %& \\ & =& 13i + j − 8k, %& \\ \end{eqnarray*}

\begin{eqnarray*} a× (b×c)& =& −(b⋅c)a %& \\ & =& −[(b⋅a)c− (c⋅a)b]%& \\ & =& (a⋅c)b− (a⋅b)c %& \\ & =& 5b−c %& \\ & =& 9i− 8j − 7k %& \\ \end{eqnarray*}