# Examples On Scalar Triple Product Of Vectors Set-1

Go back to  'Vectors and 3-D Geometry'

Example – 27

For four points with position vectors  $$\vec a,\;\;\vec b,\;\;\vec c,\;\;\vec d$$ prove that if  $$\left[ {\vec d\,\vec b\,\vec c} \right] + \left[ {\vec d\,\vec c\,\vec a} \right] + \left[ {\vec d\,\vec a\,\vec b} \right] = \left[ {\vec a\,\vec b\,\vec c} \right]$$ then the four points must be coplanar.

Solution: Upon expansion, this relation gives

\begin{align}&\quad\quad\quad\vec d \cdot \left( {\vec b \times \vec c} \right) + \vec d \cdot \left( {\vec c \times \vec a} \right) + \vec d \cdot \left( {\vec a \times \vec b} \right) = \vec a \cdot \left( {\vec b \times \vec c} \right) \hfill \\\\&\Rightarrow \quad \vec d \cdot \left\{ {\left( {\vec b \times \vec c} \right) + \left( {\vec c \times \vec a} \right) + \left( {\vec a \times \vec b} \right)} \right\} = \vec a \cdot \left( {\vec b \times \vec c} \right) \hfill \\\\& \Rightarrow \quad \vec d \cdot \left\{ {\left( {\vec b - \vec a} \right) \times \left( {\vec c - \vec a} \right)} \right\} = \vec a \cdot \left( {\vec b \times \vec c} \right)\quad\qquad\qquad...\left( 1 \right) \hfill \\ \end{align}

Note that the RHS of (1) can also be written as

$\vec a \cdot \left( {\vec b \times \vec c} \right) = \vec a \cdot \left\{ {\left( {\vec b - \vec a} \right) \times \left( {\vec c - \vec a} \right)} \right\}$

You must verify why this can be done. Using this in (1), we have

$\left( {\vec d - \vec a} \right) \cdot \left\{ {\left( {\vec b - \vec a} \right) \times \left( {\vec c - \vec a} \right)} \right\} = 0$

This implies that the vector $$\left( {\vec d - \vec a} \right)$$ is perpendicular to the cross product of the vectors $$\left( {\vec b - \vec a} \right)$$ and $$\left( {\vec c - \vec a} \right)$$ which in turn means that $$\left( {\vec d - \vec a} \right)$$ must lie in the plane of $$\left( {\vec b - \vec a} \right)$$ and $$\left( {\vec c - \vec a} \right)$$

\begin{align}& \Rightarrow \quad \left( {\vec b - \vec a} \right),\left( {\vec c - \vec a} \right)\,\,and\left( {\vec d - \vec a} \right)\,are{\text{ }}coplanar{\text{ }}vectors \hfill \\\\& \Rightarrow \quad \vec a,\,\vec b,\,\vec c,\,\vec d\,\,are{\text{ }}coplanar{\text{ }}points. \hfill \\ \end{align}

Example – 28

Prove that, for any three vectors $$\vec a,\,\vec b,\,\vec c,$$

(a) $$\left[ {\left( {\vec a + \vec b} \right)\,\,\left( {\vec b + \vec c} \right)\,\,\,\left( {\vec c + \vec a} \right)} \right] = 2\left[ {\vec a\,\vec b\,\vec c} \right]$$

(b) $$\left[ {\left( {\vec a - \vec b} \right)\,\,\,\left( {\vec b - \vec c} \right)\,\,\,\left( {\vec c - \vec a} \right)} \right] = 0$$

Solution:(a) The left hand side is

\begin{align}& \left( {\vec a + \vec b} \right) \cdot \left\{ {\left( {\vec b + \vec c} \right) \times \left( {\vec c + \vec a} \right)} \right\} \hfill \\\\&= \left( {\vec a + \vec b} \right) \cdot \left\{ {\vec b \times \vec c + \vec b \times \vec a + \vec c \times \vec a} \right\} \hfill \\ \end{align}

$$= 2\left[ {\vec a\,\vec b\,c} \right]$$

This relation incidentally proves that $$\vec a + \vec b,\,\vec b + \vec c$$and $$\vec c + \vec a$$ are coplanar if  only if $$\vec a,\,\vec b$$ and $$\vec c$$are coplanar.

(b) Since $$\left( {\vec a - \vec b} \right) + \left( {\vec b - \vec c} \right) + \left( {\vec c - \vec a} \right) = \vec 0,$$ these three vectors are the sides of a triangle, implying that they are coplanar vectors. Thus, their STP must be zero.

Example – 29

For any three non-coplanar vectors $$\vec a,\,\vec b,\,\vec c$$ and any other vector $$\vec r$$ prove that the relation

$\vec r = \frac{{\left[ {\vec r\,\vec b\,\vec c} \right]}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}}\vec a + \frac{{\left[ {\vec r\,\vec c\,\vec a} \right]}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}}\vec b + \frac{{\left[ {\vec r\,\vec a\,\vec b} \right]}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}}\vec c$

is satisfied.

Solution: Assume $$\vec r = x\vec a + y\vec b + z\vec c$$ where  $$x,\,y,\,z \in \mathbb{R}\quad\qquad\qquad...(1)$$

This can be done since $$\vec a,\,\vec b,\,\vec c$$ are non-coplanar vectors and hence any vector in space is expressible as their linear combination. To find x, y, z we do the following:

Take the dot product on both sides of (1) with $$\left( {\vec b \times \vec c} \right):$$

\begin{align}&\Rightarrow \quad [\vec r\;\vec b\;\vec c] = x[\vec a\;\vec b\;\vec c]\\&\Rightarrow \quad x = \frac{{[\vec r\;\;\vec b\;\,\vec c]}}{{[\vec a\;\;\vec b\;\,\vec c]}}\end{align}

Similarly, y and z can be determined. Now substituting the values of x, y and z in (1) proves the stated assertion.