ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 13- Worked Out Examples

Friday, 8 August 2014

CHAPTER 13- Worked Out Examples –3

Example: 10

Show by vector methods that the angular bisectors of a triangle are concurrent and find the position of the point of concurrency in terms of the position vectors of the vertices.

Solution: 10

Let the vertices of the triangle by $A\left( {\vec a} \right),\,\,B\,(\vec b)$ and $C(\vec c)$ . We use the geometrical fact that an angle bisector divides the opposite side in the ratio of the sides containing the angle.

We thus have,

$\dfrac{{BD}}{{DC}} = \dfrac{c}{b}$

Thus, $D$ is given by (the internal division formula):

$D \equiv \dfrac{{z\vec c + y\vec b}}{{z + y}}$

In $\Delta ABD$ , since $BI$ is the angle bisector, we have

$\dfrac{{DI}}{{IA}} = \dfrac{{BD}}{{BA}} = \dfrac{{\dfrac{z}{{z + y}}x}}{z} = \dfrac{x}{{z + y}}$

Thus, we now have the position vectors of $A$ and $D$ we know what ratio $I$ divides $AD$ in. $I$ can now be easily determined using the internal division formula:

	$I \equiv \dfrac{{x\vec a + \left( {x + y} \right)\left( {\dfrac{{z\vec c + y\vec b}}{{z + y}}} \right)}}{{x + z + y}}$
	$= \dfrac{{x\vec a + y\vec b + z\vec c}}{{x + y + z}}$	$\ldots(1)$

The symmetrical nature of this expression proves that the bisector of $C$ will also pass through $I$ . The angle bisectors will therefore be concurrent at $I$ , called the incentre. The position vector of the incentre is given by $(1).$

Example: 11

In a parallelogram $ABCD$ , let $M$ be the mid-point of $AB$ . $AC$ and $MD$ meet in $E$ . Prove that both $AC$ and $MD$ are trisected at $E$ .

Solution: 11

There’s no loss of generality in assuming $A$ to be the origin $(\vec 0)$ . Let $B$ and $C$ have the position vectors $\vec b$ and $\vec c$ . The position vector of $D$ is given by $\overrightarrow {AD}$ where

	$\overrightarrow {AD} = \overrightarrow {AC} + \overrightarrow {CD}$
	$= \vec c - \overrightarrow {AB}$
	$= \overrightarrow c - \overrightarrow b$

$D$ is therefore the point $\overrightarrow c - \overrightarrow b$ .

Since $M$ is the mid-point of $AB$ , $M$ ‘s position vector is $\dfrac{{\vec b}}{2}$ . Let us first find the position vector of a point $E$ ‘ which lies on $AC$ and trisects it. We’ll then show that same point lies on $DM$ and trisects on $DM$ and trisects it too, proving the stated assertion.

$E' \equiv \dfrac{{1 \times \vec c + 2 \times \vec 0}}{3} = \dfrac{{\vec c}}{3}$