# 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  $$\vec a$$ .

## TRY YOURSELF - V

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.

Vectors
grade 11 | Questions Set 1
Vectors