Examples On Vector Triple Product Of Vectors Set-1

Go back to  'Vectors and 3-D Geometry'

Example – 33

For three arbitrary vectors \(\vec a,\vec b,\vec c,\), prove that the vectors

\[\begin{align}&{{\vec r}_1} = \vec a \times (\vec b \times \vec c) \hfill \\\\&{{\vec r}_2} = \vec b \times (\vec c \times \vec a) \hfill \\\\&{{\vec r}_3} = \vec c \times (\vec a \times \vec b) \hfill \\& \end{align} \]

are coplanar.

Solution: Using the expansion rule for the vector triple product obtained in the proceeding discussion, we have

\[\begin{align}&{{\vec r}_1} = (\vec a \cdot \vec c)\vec b - (\vec a \cdot \vec b)\vec c \hfill \\\\& {{\vec r}_2} = (\vec b \cdot \vec a)\vec c - (\vec b \cdot \vec c)\vec a \hfill \\\\&{{\vec r}_3} = (\vec c \cdot \vec b)\vec a - (\vec c \cdot \vec a)\vec b \hfill \\ &\end{align} \]

This gives

\[{\vec r_1} + {\vec r_2} + {\vec r_3} = \vec 0\]

which implies that  \({\vec r_1},\;\;{\vec r_2},\;\;{\vec r_3}\) must be the sides of a triangle, and hence must be coplanar.

Example – 34

For three arbitrary vectors \(\vec a,\;\;\vec b,\;\;\vec c,\) prove that

\[[\vec a \times \vec b \quad \vec b \times \vec c  \quad\vec c \times \vec a] = {[\vec a\;\;\vec b\;\;\vec c]^2}\]

Solution: The left hand side, upon expansion gives,

\[(\vec a \times \vec b) \cdot \left\{ {(\vec b \times \vec c) \times (\vec c \times \vec a)} \right\}\]


\[\begin{align}& = (\vec a \times \vec b) \cdot \left\{ {[\vec a\;\;\vec b\;\;\vec c]\;\vec c} \right\} \hfill \\\\& = \left\{ {(\vec a \times \vec b) \cdot \vec c} \right\}\;[\vec a\;\;\vec b\;\;\vec c] \hfill \\\\&= {[\vec a\;\;\vec b\;\;\vec c]^2} \hfill \\ \end{align} \]

This also proves as a consequence that \((\vec a \times \vec b),\;\;(\vec b \times \vec c)\;\;{\text{and}}\;\;(\vec c \times \vec a)\) are coplanar iff  \(\vec a,\;\;\vec b\;\;{\text{and}}\;\;\vec c\) are coplanar, a fact that can intuitively be expected.

Example – 35

Show that  \((\vec a \times \vec b) \times \vec c = \vec a \times (\vec b \times \vec c)\) if and only if \(\vec a\;\;{\text{and}}\;\;\vec c\) are collinear.

Solution: Let us first assume that

\[\begin{align}&\quad\quad (\vec a \times \vec b) \times \vec c = \vec a \times (\vec b \times \vec c) \hfill \\\\& \Rightarrow  \quad  - \;\vec c \times (\vec a \times \vec b) = \vec a \times (\vec b \times \vec c) \hfill \\\\&\Rightarrow  \quad( - \vec c \cdot \vec b)\vec a - ( - \vec c \cdot \vec a)\vec b = (\vec a - \vec c)\vec b - (\vec a \cdot \vec b)\vec c \hfill \\\\&\Rightarrow  \quad (\vec b \cdot \vec c)\vec a = (\vec a \cdot \vec b)\vec c \hfill \\\\&\Rightarrow  \quad \vec a = \left( {\frac{{\vec a \cdot \vec b}}{{\vec b \cdot \vec c}}} \right)\vec c \hfill \\\\&\qquad\quad= \lambda \vec c\,where\,\lambda  \in \mathbb{R} \hfill \\ 
\end{align} \]

Since \(\vec a\)  is a scalar multiple of \(\vec c\), this proves that  \(\vec a\;\;{\text{and}}\;\;\vec c\) are collinear.

To prove the reverse implication is left to the reader as an exercise.

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