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## 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