## 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.