Examples On Vector Triple Product Of Vectors Set-2

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Example – 36

For arbitrary vector  \(\vec a,\;\;\vec b,\;\;\vec c,\;\;\vec d,\) prove the following relations.

(i) \((\vec a \times \vec b) \times (\vec c \times \vec d) = [\vec a\;\;\vec b\;\;\vec d]\,\vec c - [\vec a\;\;\vec b\;\;\vec c]\,\vec d\)

(ii) \(\vec a \times (\vec b \times (\vec c \times \vec d)) = (\vec b \cdot \vec d)(\vec a \times \vec c) - (\vec b \cdot \vec c)(\vec a \times \vec d)\)

(iii) \((\vec a \times \vec b) \cdot (\vec c \times \vec d) = \left| {\begin{array}{*{20}{c}}  {\vec a \cdot \vec c}&{\vec a \cdot \vec d} \\   {\vec b \cdot \vec c}&{\vec b \cdot \vec d} \end{array}} \right|\)

Solution:(i) This relation is simply obtained by expanding the left hand side:

\((\mathop {\vec a \times \vec b}\limits_{\text{I}} ) \times (\mathop {\mathop {\vec c}\limits_{{\text{II}}}  \times \mathop {\vec d}\limits_{{\text{III}}} }\limits_{} ) = (\vec a \times \vec b) \cdot \vec d)\,\vec c - ((\vec a \times \vec b) \cdot \vec c)\vec d\)

  \(\quad\;\; = [\vec a\;\;\vec b\;\;\vec d]\,\vec c - [\vec a\;\;\vec b\;\;\vec c]\,\vec d\,\)

(ii)\(\vec a \times (\mathop {\vec b}\limits_{\text{I}}  \times (\mathop {\vec c}\limits_{{\text{II}}}  \times \mathop {\vec d}\limits_{{\text{III}}} )) = \vec a \times ((\vec b \cdot \vec d)\vec c - (\vec b \cdot \vec c)\vec d)\)

 \( = (\vec b \cdot \vec d)(\vec a \times \vec c) - (\vec b \cdot \vec c)(\vec a \times \vec d)\)

(iii)  \((\vec a \times \vec b) \cdot (\vec c \times \vec d) = \vec a \cdot (\vec b \times (\vec c \times \vec d))\) (why ?)

 \(= \vec a \cdot ((\vec b \cdot \vec d)\vec c - (\vec b \cdot \vec c)\vec d)\)

 \(= (\vec a \cdot \vec c)(\vec b \cdot \vec d) - (\vec a \cdot \vec d)(\vec b \cdot \vec c)\)

 \(= \left| {\begin{align}&{\vec a \cdot \vec c}&&{\vec a \cdot \vec d} \\& {\vec b \cdot \vec c}&&{\vec b \cdot \vec d} \end{align}}   \right|\)

Example – 37

If  \(\vec a,\;\;\vec b,\;\;\vec c\) are vectors such that  \(\left| {\vec b} \right| = \left| {\vec c} \right|\), prove that

\[\left\{ {\left\{ {(\vec a + \vec b) \times (\vec a + \vec c)} \right\} \times (\vec b \times \vec c)} \right\} \cdot (\vec b + \vec c) = 0\]

Solution: The left hand side equals

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

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

The last step uses the result of part (i) in the previous example. The expression now reduces to

\[\begin{align}&( - [\vec a\;\;\vec c\;\;\vec b]\,\vec c + [\vec b\;\;\vec a\;\;\vec c]\,\vec b) \cdot (\vec b + \vec c) \hfill \\\\& = [\vec a\;\;\vec b\;\;\vec c](\vec c - \vec b) \cdot (\vec c + \vec b) \hfill \\\\&  = [\vec a\;\;\vec b\;\;\vec c]({\left| {\vec c} \right|^2} - {\left| {\vec b} \right|^2}) \hfill \\\\&= 0\qquad\qquad\qquad{\text{ }}\left( {since\,\,\left| {\vec c} \right| = \left| {\vec b} \right|{\text{ }}} \right) \hfill \\\end{align} \]

Example – 38

If \(\vec a,\;\;\vec b,\;\;\vec c\) are non-coplanar unit vectors such that

\[\vec a \times (\vec b \times \vec c) = \frac{{\vec b + \vec c}}{{\sqrt 2 }}\]

and \(\vec b\) and  \(\vec c\) are non-collinear, find the angle \({\theta _1}\) and  \({\theta _2}\) which \(\vec a\)  makes with \(\vec b\) and \(\vec c\)respectively.

Solution: \(\begin{align}\qquad \qquad \vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c)\,\vec b - (\vec a \cdot \vec b)\,\vec c = \frac{{\vec b + \vec c}}{{\sqrt 2 }}(given)\end{align}\)

\(\qquad\qquad \qquad \Rightarrow \quad \left( {\vec a \cdot \vec c - \frac{1}{{\sqrt 2 }}} \right)\vec b - \left( {\vec a \cdot \vec b + \frac{1}{{\sqrt 2 }}} \right)\vec c = 0\)

Since \(\vec b,\;\;\vec c\) are non-collinear vectors, we must have

\(\begin{align}  \qquad \qquad \quad &\vec a \cdot \vec c = \frac{1}{{\sqrt 2 }}\;\;{\text{and}}\;\;\vec a \cdot \vec b = - \frac{1}{{\sqrt 2 }} \hfill \\& \Rightarrow \quad \cos {\theta _1} = \frac{1}{{\sqrt 2 }}\;\;{\text{and}}\;\;\cos {\theta _2} = - \frac{1}{{\sqrt 2 }} \hfill \\&\Rightarrow \quad  {\theta _1} = \frac{\pi }{4}\;\;and\;\;{\theta _2} = \frac{{3\pi }}{4} \hfill \\ \end{align} \)

Example – 39

If \(\vec b,\;\;\vec c,\;\;\vec d\) are non-coplanar vectors, then prove that the vector

\[\vec r = (\vec a \times \vec b) \times (\vec c \times \vec d) + (\vec a \times \vec c) \times (\vec d \times \vec b) + (\vec a \times \vec d) \times (\vec b \times \vec c)must{\text{ }}be{\text{ }}parallel{\text{ }}to\,\,\vec a{\text{ }}.\]

Solution: We first write the expression in a slightly modified way (the reason for this will become clear subsequently):

\[\vec r =  - (\vec c \times \vec d) \times (\vec a \times \vec b) + (\vec a \times \vec c) \times (\vec d \times \vec b) - (\vec b \times \vec c) \times (\vec a \times \vec d)\]

Thus, we have written the first and third vector products in the original expression for \(\vec r\) in reverse.

Now, expanding each vector product using the using the result of Example - 36 Part - (i), we have




\[\begin{align}&  =  - ([\vec c\;\;\vec d\;\;\vec b] + [\vec b\;\;\vec c\;\;\vec d])\,\vec a \hfill \\\\&  =  - 2[\vec b\;\;\vec c\;\;\vec d]\;\vec a \hfill \\\\&= \lambda \,\vec a\;\,where\;\,\lambda  \in \mathbb{R}\;\,and\,\;\lambda  \ne 0\,\;since\,\,\vec b,\;\;\vec c,\;\;\vec d\,\,are{\text{ }}non - coplanar \hfill \\ \end{align} \]

The reason for the manipulation done earlier should be apparent in step (1): the cancellation of the terms indicated.

Since \(\vec r\) can be written as a scalar multiple of  \(\vec a,\vec r\) must be parallel to  \(\vec a\) .


Q. 1 Let   \(\vec p,\;\;\vec q,\;\;\vec r\) be the vectors \(\vec b \times \vec c,\;\;\vec c \times \vec a\) and  \(\vec a \times \vec b\) respectively. Show that \(\vec a,\;\;\vec b\) and  \(\vec c\) are parallel to  \(\vec q \times \vec r,\;\;\;\vec r \times \vec p\) and \(\vec p \times \vec q\) respectively.

Q. 2 Show that

\[[\vec a \times \vec p  \quad \vec b \times \vec q  \quad \vec c \times \vec r] + [\vec a \times \vec q \quad  \vec b \times \vec r \quad \vec c \times \vec p] + [\vec a \times \vec r  \quad \vec b \times \vec p\quad  \vec c \times \vec q] = 0\]

Q. 3 For the vectors \(\vec a,\;\;\vec b,\;\;\vec c,\;\;\vec d\) , prove that \((\vec b \times \vec c) \cdot (\vec a \times \vec d) + (\vec c \times \vec a){\kern 1pt}  \cdot (\vec b \times \vec d) + (\vec a \times \vec b) \cdot (\vec c \times \vec d) = 0\)

Q. 4 If \(\vec a,\;\;\vec b\,\;and\;\,\vec c\) are three non-parallel non-coplanar unit vectors such that \(\vec a \times (\begin{align}\vec b \times \vec c) = \frac{1}{2}\vec b\end{align}\), prove that the angles that \(\vec a\) makes with \(\vec b\) and \(\vec c\)are \(\begin{align}\frac{\pi }{2}\,\,and\,\,\frac{\pi }{3}\end{align}\) respectively.

Q. 5 If \(\vec a,\;\;\vec b,\;\;\vec c,\;\;\vec d\) are non-zero vectors such that \((\vec a \times \vec b) \times (\vec c \times \vec d) - (\vec b \times \vec c) \times (\vec d \times \vec a) = [\vec a\;\;\vec c\;\;\vec d]\,\vec b\), show that \(\vec a,\;\;\vec b\,\,and\,\,\vec c\)  must be coplanar.

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