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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 PX 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 PX 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 OG 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.
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