ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 15-Worked Out Examples 5

Friday, 8 August 2014

CHAPTER 15-Worked Out Examples 5

Example: 13

In a quadrilateral $PQRS$ , $\overrightarrow {PQ} = \vec a,\,\overrightarrow {QR} = \vec b$ and $\overrightarrow {SP} = \vec a - \vec b$ . If $M$ is the mid-point of $QR$ and $X$ is a point on $SM$ such that $SX : SM = 4 : 5$ , prove that $P$ , $X$ and $R$ collinear.

Solution: 13

Since no position vectors have been specified in the question (only the sides have been specified), there is no loss of generality in assuming that $P$ is the origin $\vec 0$ .

We have,

	$M \equiv \dfrac{{\vec a + \left( {\vec a + \vec b} \right)}}{2} = \vec a + \dfrac{{\vec b}}{2}$
	$\Rightarrow \,\,\,\, X \equiv \dfrac{{4 \times \left( {\vec a + \dfrac{{\vec b}}{2}} \right) + 1 \times \left( {\vec b - \vec a} \right)}}{{4 + 1}} = \dfrac{{3\vec a + 3\vec b}}{5}$
	$\Rightarrow \,\,\,\, X \equiv \dfrac{3}{5}\left( {\vec a + \vec b} \right)$

Thus,

$\overrightarrow {PX} = \dfrac{3}{5}\overrightarrow {PR}$

implying that $P$ , $X$ and $R$ are collinear.

Example: 14

It is known that in a $\Delta ABC$ with centroid $G$ , circumcentre $O$ and orthocentre $H$ ,

$OG:GH = 1:2$

Let $P$ be any point in the plane of $\Delta ABC$ . Prove the following assertions:

	(a) $\overrightarrow {OG} + \overrightarrow {OB} + \overrightarrow {OC} = \overrightarrow {OH}$
	(b) $\overrightarrow {HA} + \overrightarrow {HB} + \overrightarrow {HC} = 2\,\overrightarrow {HO}$
	(c) $\overrightarrow {PA} + \overrightarrow {PB} + \overrightarrow {PC} = 3\,\overrightarrow {PG}$

Solution: 14-(a)

	$\overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} = \,\overrightarrow {OA} + \left( {\overrightarrow {OB} + \overrightarrow {OC} } \right)$
	$= \,\overrightarrow {OA} + 2\overrightarrow {OD}$ (since $D$ is $BC$ ‘s mid-point)
	$= 3\,\overrightarrow {OG}$ (Since $G$ lies on $AD$ and divides it in the ratio $2:1$
	$= \overrightarrow {OH}$ (Since $O$ , $G$ and $H$ are collinear and $OH=3OG$

Solution: 14-(b)

	$\overrightarrow {HA} + \overrightarrow {HB} + \overrightarrow {HC} = \overrightarrow {HA} + \left( {\overrightarrow {HB} + \overrightarrow {HC} } \right)$
	$= \overrightarrow {HA} + 2\overrightarrow {HD}$
	$= 3\overrightarrow {HG}$ (Same logic as above)
	$= 3 \times \dfrac{2}{3}\overrightarrow {HO}$ (again, same as above)
	$= 2\overrightarrow {HO}$

Solution: 14-(c)

For any arbitrary point $P$ in the plane of $\Delta ABC,$ we have

	$\overrightarrow {PA} + \overrightarrow {PB} + \overrightarrow {PC} = \overrightarrow {PA} + \left( {\overrightarrow {PB} + \overrightarrow {PC} } \right)$
	$= \overrightarrow {PA} + 2\overrightarrow {PD}$
	$= 3\overrightarrow {PG}$

Go over the solution again if you find any part of it confusing.

Friday, 8 August 2014

CHAPTER 15-Worked Out Examples 5

No comments:

https://www.youtube.com/TarunGehlot