Scalar and Vector Triple Product

SCALAR TRIPLE PRODUCT

It is defined for three vectors $\bar{a},\bar{b},\bar{c}$in that order as the scalar $(\bar{a}\times \bar{b})\cdot \bar{c}$ which can also be written simply as $\bar{a}\times \bar{b}\cdot \bar{c}$. It denotes the volume of the parallelopiped formed by taking a, b, c as the co-terminus edges.

i.e. V = magnitude of $\bar{a}\times \bar{b}\cdot \bar{c}=|\bar{a}\times \bar{b}\cdot \bar{c}|.$

The value of scalar triple product depends on the cyclic order of the vectors and is independent of the position of the dot and cross. These may be interchange at pleasure. However and anti-cyclic permutation of the vectors changes the value of triple product in sign but not a magnitude.

Properties:

$\bullet$ If $\bar{a},b,\bar{c}$ are given as $\bar{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k},$ etc., then a \times b \cdot c = \left| {\,\,\begin{array}{*{20}{c}} {{a_1}}{{a_2}}{{a_3}} \\  {{b_1}}{{b_2}}{{b_3}} \\  {{c_1}}{{c_2}}{{c_3}} \end{array}\,\,} \right|

$\bullet$ $\bar{a}\times \bar{b}\cdot \bar{c}=\bar{a}\cdot \bar{b}\times \bar{c}$ i.e. position of dot and cross can be interchanged without altering the product. Hence it is also represented by $[\bar{a}\bar{b}\bar{c}]$

$\bullet$$[\bar{a}\bar{b}\bar{c}]=[\bar{b}\bar{c}\bar{a}]=[\bar{c}\bar{a}\bar{b}]$$\bullet$ $[\bar{a}\bar{b}\bar{c}]=-[\bar{b}\bar{a}\bar{c}]$;;;;;;;;;;$\bullet$ $[k\bar{a}\bar{b}\bar{c}]=k[\bar{a}\bar{b}\bar{c}]$$\bullet$ $[\bar{a}+\bar{b}\bar{c}\bar{d}]=[\bar{a}\bar{c}\bar{d}]+[\bar{b}\bar{c}\bar{d}]$

Take the scalar product of this equation with a. We get

0 = $l(\bar{a}\cdot \bar{b})+m(\bar{a}\cdot \bar{c})$ $\Rightarrow$ $\bar{a}\times (\bar{b}\times \bar{c})=\lambda [(\bar{a}\cdot \bar{c})\bar{b}-(\bar{a}\cdot \bar{b})\bar{c}]$

If we choose the coordinate axes in such a way that

$a=a_1\hat{i},b=b_1\hat{i}+b_2\hat{j},andc=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$, it is easy to show that $\lambda$ = 1. Hence

$\bar{a}\times (\bar{b}\times \bar{c})=(\bar{a}\cdot \bar{c})\bar{b}-(\bar{a}\cdot \bar{b})\bar{c}$

In general, $\bar{a}\times (\bar{b}\times \bar{c})\ne (\bar{a}\times \bar{b})\times \bar{c}$ ( Vector triple product is not associative ) .

$\bar{a}\times (\bar{b}\times \bar{c})=(\bar{a}\times \bar{b})\times \bar{c}$, if some or all of $\bar{a},\bar{b},\bar{c}$are zero vectors or $\bar{a}and\bar{c}$ are collinear.

Illustration 2: Let $\bar{a},\bar{b},\bar{c}$ be three mutually perpendicular vectors of the same magnitude. If the vector $\bar{x}$ satisfy the equation

$\overline{a}\times \{\bar{x}-\bar{b})\times \bar{a}\}+\bar{b}\times \{(\bar{x}-\bar{c})\times \bar{b}\}+\bar{c}\times \{(\bar{x}-\bar{a})\times \bar{c}\}=0$ then find $\bar{x}$

Solution: Here $(\bar{a}.\bar{a})(\bar{x}-\bar{b})-\{\bar{a}.(\bar{x}-\bar{b})\}.\bar{a}+(\bar{b}.\bar{b})(\bar{x}-\bar{c})$ -

{$\bar{b}(\bar{x}-\bar{c})\}\bar{b}+(\bar{c}.\bar{c})(\bar{x}-\bar{a})-\{\bar{c}.(\bar{x}-\bar{a})\}\bar{c}=0$

or $\lambda$² $(\bar{x}-\bar{b}+\bar{x}-\bar{c}+\bar{x}-\bar{a})=\{\bar{a}.(\bar{x}-\bar{b})\}\bar{a}+\{\bar{b}.(\bar{x}-\bar{c})\}\bar{b}+\{\bar{c}.(\bar{x}-\bar{a})\}\bar{c}$

where $|\bar{a}|=|\bar{b}|=|\bar{c}|=\lambda$

$\lambda$² { $3\bar{x}-(\bar{a}+\bar{b}+\bar{c})\}=(\bar{a}.\bar{x})\bar{a}+(\bar{b}.\bar{x})\bar{b}+(\bar{c}.\bar{x})\bar{c}.$

let $\bar{x}=\alpha \bar{a}+\beta \bar{b}$+ $\gamma \bar{c}$ then $\bar{a}.\bar{x}=\alpha |\bar{a}{|}^2=\alpha {\lambda }^2$,$\bar{b}.\bar{x}=\beta {\lambda }^2$and $\bar{c}.\bar{x}=\gamma {\lambda }^2$.

$\Rightarrow$$\lambda$² {3$\bar{x}-(\bar{a}+\bar{b}+\bar{c})\}={\lambda }^2\bar{x}$ $\Rightarrow$3$\bar{x}-(\bar{a}+\bar{b}+\bar{c})=\bar{x}$

Hence $\bar{x}=\frac{\bar{a}+\bar{b}+\bar{c}}{2}$ .