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## Friday, 8 August 2014

### CHAPTER 12- Worked Out Examples – 2

 Example: 8
 Let be three non-coplanar vectors. Prove that the points $A\left( {2\vec a + 3\vec b - \vec c} \right),\,$$B\left( {\vec a - 2\vec b - 3\vec c} \right),\,$ $C\left( {3\vec a + 4\vec b - 2\vec c} \right)$ and $D\left( {\vec a - 6\vec b + 6\vec c} \right)$ are coplanar.
 Solution: 8
As in the previous example, we first draw a visual picture to determine when four points can be coplanar.
Thus, as explained in the figure, we must have some scalars $\lambda ,\mu \in \mathbb{R}$ for which
 $\overrightarrow {AC} = \lambda \overrightarrow {AB} + \mu \overrightarrow {AD}$ $\Rightarrow \,\,\,\, \left( {\overrightarrow {OC} - \overrightarrow {OA} } \right) = \lambda \left( {\overrightarrow {OB} - \overrightarrow {OA} } \right) + \mu \left( {\overrightarrow {OD} - \overrightarrow {OA} } \right)$ {$O$ is the origin} $\Rightarrow \,\,\,\, \vec a + \vec b - \vec c = \lambda \left( { - \vec a - 5\vec b + 4\vec c} \right) + \mu \left( { - \vec a - 9\vec b + 7\vec c} \right)$ $\Rightarrow \,\,\,\, \left( {1 + \lambda + \mu } \right)\vec a + \left( {1 + 5\lambda + 9\mu } \right)\vec b + \left( { - 1 - 4\lambda - 7\mu } \right)\vec c = \vec 0$
Since $\vec a,\vec b$ and $\,\vec c$ are non-coplanar, we must have
 $1 + \lambda + \mu = 0$ $1 + 5\lambda + 9\mu = 0$ $1 + 4\lambda + 7\mu = 0$
As can be easily verified, this system has the solution $\lambda = - 2,\mu = 1,$implying $\overrightarrow {AB} ,\,\,\overrightarrow {AC}$ and $\overrightarrow {AD}$ are indeed coplanar.
Thus, the points $A$$B$$C$ and $D$ are coplanar.

#### SECTION FORMULA

 Example: 9
 Let $A\left( {\vec a} \right)$ and $B\left( {\vec b} \right)$ be two fixed points. Find the position vectors of the points lying on the (extended) line $AB$ which divide the segment internally and externally in the ratio $m : n$.
 Solution: 9
We consider internal division; the external division case follows analogously.
Let $C\left( {\vec c} \right)$ be the point which divides $AB$ internally in the ratio $m : n$.
We have,
 $\overrightarrow {AC} = \dfrac{m}{{m + n}}\,\,\,\overrightarrow {AB}$ $\Rightarrow \,\,\,\, \left( {\overrightarrow {OC} - \overrightarrow {OA} } \right) = \dfrac{m}{{m + n}}\,\,\left( {\,\overrightarrow {OB} - \overrightarrow {OA} } \right)$ $\Rightarrow \,\,\,\, \vec c - \vec a = \dfrac{m}{{m + n}}\,\,\left( {\vec b - \vec a} \right)$ $\Rightarrow\,\,\,\,{\vec c = \dfrac{{m\vec b + n\vec a}}{{m + n}}}$
Similarly, the point $D(\vec d)$ which divides $AB$ externally in the ratio $m : n$ is given by
 $\vec d = \dfrac{{m\vec b - n\vec a}}{{m - n}}$
A particular case of internal division is the mid-point of $A(\vec a)$ and $B(\vec b)$: the mid-point is $\dfrac{{\vec a + \vec b}}{2},$