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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 xyz 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 xyand z in (1)  proves the stated assertion
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