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Friday, 8 August 2014

CHAPTER 7 - Worked Out Examples

    Example: 4      

Suppose that for three non-zero vectors \vec a,\vec b,\vec c, any two of them are non-collinear. If the vectors \left( {\vec a + 2\vec b} \right) and \vec c are collinear and the vectors \left( {\vec b + 3\vec c} \right) and \vec a are collinear, prove that
\vec a + 2\vec b + 6\vec c = \vec 0
Solution: 4

We must have some \lambda ,\mu  \in\mathbb{R}  such that
\vec a + 2\vec b = \lambda \vec c\ldots(1)
\vec b + 3\vec c = \mu \vec a\ldots(2)
From (1), we have
\vec c = \dfrac{1}{\lambda }\left( {\vec a + 2\vec b} \right)\ldots(3)
We use this in (2) :
\vec b + \dfrac{3}{\lambda }\left( {\vec a + 2\vec b} \right) = \mu \vec a
 \Rightarrow  \,\,\,\,\, \left( {\dfrac{3}{\lambda } - \mu } \right)\vec a + \left( {1 + \dfrac{6}{\lambda }} \right)\vec b = \vec 0
Since \vec a and \vec b are non-collinear, their linear combination can be zero if and only if the two scalars are zero.
This gives
\dfrac{3}{\lambda } - \mu  = 0
1 + \dfrac{6}{\lambda } = 0
 \Rightarrow  \,\,\,\,\, \lambda  =  - 6,\,\,\mu  =  - \dfrac{1}{2}
Using the value of \lambda  in (3), we have
\vec a + 2\vec b + 6\vec c = 0
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