# Properties Of Scalar Triple Product Of Vectors

Let us see some more significant properties of the STP:

(i) The STP of three vectors is zero if any two of them are parallel. This implies as a corollary that $$[\vec a\;\;\vec a\;\;\vec b] = 0$$(always)

(ii) For any $$\lambda \in \mathbb{R}$$ ,

$[\lambda \vec a\;\;\vec b\;\;\vec c] = \lambda [\vec a\;\;\vec b\;\;\vec c]$

(iii) $$\boxed{[(\vec a + \vec b)\,\,\,\,\vec c\;\;\vec d] = [\vec a\;\;\vec c\;\;\vec d] + [\vec b\;\;\vec c\;\;\vec d]}$$

This property is very important and is used extremely frequently. The justification is straight forward:

\begin{align}&[(\vec a + \vec b)\;\vec c\;\;\vec d] = (\vec a + \vec b) \cdot (\vec c \times \vec d) \hfill \\\\&\qquad\qquad\qquad= \vec a \cdot (\vec c \times \vec d) + \vec b \cdot (\vec c \times \vec d)\left\{ {Dot{\text{ }}product{\text{ }}is{\text{ }}distributive{\text{ }}over{\text{ }}vector{\text{ }}addition} \right\} \hfill \\\\&\qquad\qquad\qquad = [\vec a\;\;\vec c\;\;\vec d] + [\vec b\;\;\vec c\;\;\vec d] \hfill \\ \end{align}

(iv) Three vector are coplanar if and only if their STP is zero. This is because the volume of the parallelopiped formed by the three vectors becomes zero if they are coplanar.

You are urged to rigorously prove the other way implication, i.e, prove that if $$[\vec a\;\;\vec b\;\;\vec c] = 0$$ where $$\vec a\;\;\vec b\;\;\vec c$$ are non-zero non-collinear vectors, then  $$\vec a,\;\;\vec b,\;\;\vec c$$ must be coplanar.

(v) Let $$\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k,\;\;\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\;\;{\text{and}}\;\;\vec c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k$$. Then,

[\vec a\;\;\vec b\;\;\vec c] = \vec a \cdot (\vec b \times \vec c) = ({a_1}\hat i + {a_2}\hat j + {a_3}\hat k) \cdot \left| {\begin{align}&{\hat i}&{\hat j}&&{\hat k} \\& {{b_1}}&{{b_2}}&&{{b_3}} \\& {{c_1}}&{{c_2}}&&{{c_3}} \end{align}} \right|
$\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\;= {a_1}({b_2}{c_3} - {b_3}{c_2}) + {a_2}({b_3}{c_1} - {b_1}{c_3}) + {a_3}({b_1}{c_2} - {b_2}{c_1})$

\quad\;= \left| {\;\begin{align}& {{a_1}}&{{a_2}}&&{{a_3}} \\ &{{b_1}}&{{b_2}}&&{{b_3}} \\ &{{c_1}}&{{c_2}}&&{{c_3}} \end{align}\;} \right|

This relation is quite useful and is worth remembering.

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