ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 26 Worked Out Examples

Friday, 8 August 2014

CHAPTER 26 Worked Out Examples

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: 27

Upon expansion, this relation gives

	$\vec d \cdot \left( {\vec b \times \vec c} \right) + \vec d \cdot \left( {\vec c \times \vec a} \right) + \vec d \cdot \left( {\vec a \times \vec b} \right) = \vec a \cdot \left( {\vec b \times \vec c} \right)$
	$\Rightarrow \,\,\,\, \vec d \cdot \left\{ {\left( {\vec b \times \vec c} \right) + \left( {\vec c \times \vec a} \right) + \left( {\vec a \times \vec b} \right)} \right\} = \vec a \cdot \left( {\vec b \times \vec c} \right)$
	$\Rightarrow \,\,\,\, \vec d \cdot \left\{ {\left( {\vec b - \vec a} \right) \times \left( {\vec c - \vec a} \right)} \right\} = \vec a \cdot \left( {\vec b \times \vec c} \right)$	$\ldots(1)$

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 $\left( {\vec c - \vec a} \right).$

	$\Rightarrow \,\,\,\, \left( {\vec b - \vec a} \right),\left( {\vec c - \vec a} \right)$ and $\left( {\vec d - \vec a} \right)$ are coplanar vectors
	$\Rightarrow \,\,\,\, \vec a,\,\vec b,\,\vec c,\,\vec d$ are coplanar vectors

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: 28-(a)

The left hand side is

	$\left( {\vec a + \vec b} \right) \cdot \left\{ {\left( {\vec b + \vec c} \right) \times \left( {\vec c + \vec a} \right)} \right\}$
	$= \left( {\vec a + \vec b} \right) \cdot \left\{ {\vec b \times \vec c + \vec b \times \vec a + \vec c \times \vec a} \right\}$
	$\left[ {\mathop a\limits^ \to \mathop b\limits^ \to \mathop c\limits^ \to } \right] + \underbrace{\left[ {\mathop a\limits^ \to \mathop b\limits^ \to \mathop a\limits^ \to } \right] + \left[ {\mathop a\limits^ \to \mathop c\limits^ \to \mathop a\limits^ \to } \right] + \left[ {\mathop b\limits^ \to \mathop b\limits^ \to \mathop c\limits^ \to } \right] + \left[ {\mathop b\limits^ \to \mathop b\limits^ \to \mathop a\limits^ \to } \right]}_{\rm{Comment}} + \left[ {\mathop b\limits^ \to \mathop c\limits^ \to \mathop a\limits^ \to } \right]$
	Comment: Equal to $0$
	$= 2\left[ {\vec a\,\vec b\,c} \right]$

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

Solution: 28-(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

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

is satisfied.

Solution: 29

Assume $\vec r = x\vec a + y\vec b + z\vec c$ where $x,\,y,\,z \in\mathbb{R}$

$\ldots(1)$

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 $\left( {\vec b \times \vec c} \right):$

	$\vec r \cdot (\vec b \times \vec c) = x\vec a \cdot (\vec b \times \vec c) + y\vec b \cdot \underbrace{\left( {\mathop b\limits^ \to \times \mathop c\limits^ \to } \right)}_{\rm{Comment:1}} + z\mathop c\limits^ \to .\underbrace{\left( {\mathop b\limits^ \to \times \mathop c\limits^ \to } \right)}_{\rm{Comment:2}}$
	Comment:1: Equal to $0$
	Comment:2: Equal to $0$
	$\Rightarrow \,\,\,\, [\vec r\;\vec b\;\vec c] = x[\vec a\;\vec b\;\vec c]$
	$\Rightarrow \,\,\,\, x = \dfrac{{[\vec r\;\;\vec b\;\,\vec c]}}{{[\vec a\;\;\vec b\;\,\vec c]}}$

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

Friday, 8 August 2014

CHAPTER 26 Worked Out Examples

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