ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 12- Worked Out Examples

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,

	$\overrightarrow {AC} = \dfrac{m}{{m + n}}\,\,\,\overrightarrow {AB}$
	$\Rightarrow \,\,\,\, \left( {\overrightarrow {OC} - \overrightarrow {OA} } \right) = \dfrac{m}{{m + n}}\,\,\left( {\,\overrightarrow {OB} - \overrightarrow {OA} } \right)$
	$\Rightarrow \,\,\,\, \vec c - \vec a = \dfrac{m}{{m + n}}\,\,\left( {\vec b - \vec a} \right)$
	$\Rightarrow\,\,\,\,{\vec c = \dfrac{{m\vec b + n\vec a}}{{m + n}}}$