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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