Examples On Basics Of Vectors Set-4

Go back to  'Vectors and 3-D Geometry'

Example – 13

In a quadrilateral PQRS, \(\overrightarrow {PQ}  = \vec a,\,\overrightarrow {QR}  = \vec b\,\,{\text{and}}\,\,\overrightarrow {SP}  = \vec a - \vec b.\) 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: 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,

\[\begin{align}& \qquad \;M \equiv \frac{{\vec a + \left( {\vec a + \vec b} \right)}}{2} = \vec a + \frac{{\vec b}}{2} \hfill \\\\&\Rightarrow  \quad X \equiv \frac{{4 \times \left( {\vec a + \frac{{\vec b}}{2}} \right) + 1 \times \left( {\vec b - \vec a} \right)}}{{4 + 1}} = \frac{{3\vec a + 3\vec b}}{5} \hfill \\\\&  \Rightarrow  \quad X \equiv \frac{3}{5}\left( {\vec a + \vec b} \right) \hfill \\ \end{align} \]


\[\overrightarrow {PX}  = \frac{3}{5}\overrightarrow {PR} \]

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:

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


(a) \(\overrightarrow {OA}  + \overrightarrow {OB}  + \overrightarrow {OC}  = \,\overrightarrow {OA}  + \left( {\overrightarrow {OB}  + \overrightarrow {OC} } \right)\)

\(\begin{align}&= \,\overrightarrow {OA}  + 2\overrightarrow {OD} \left( {Since\;D\;is\;BC's{\text{ }}mid - point} \right) \hfill \\\\& = 3\,\overrightarrow {OG} \left( {Since\;G\;lies{\text{ }}on\;AD\;and{\text{ }}divides{\text{ }}it{\text{ }}in{\text{ }}the{\text{ }}ratio{\text{ }}2{\text{ }}:{\text{ }}1} \right) \hfill \\\\&= \overrightarrow {OH} \left( {Since\;O,G\;and\;H\;are{\text{ }}collinear{\text{ }}and\;OH = {\text{ }}3OG} \right) \hfill \\ \end{align} \)

(b) \(\overrightarrow {HA}  + \overrightarrow {HB}  + \overrightarrow {HC}  = \overrightarrow {HA}  + \left( {\overrightarrow {HB}  + \overrightarrow {HC} } \right)\)

\(\begin{align}&= \overrightarrow {HA}  + 2\overrightarrow {HD}  \hfill \\\\& = 3\overrightarrow {HG} \left( {Same{\text{ }}logic{\text{ }}as{\text{ }}above} \right) \hfill \\\\&= 3 \times \frac{2}{3}\overrightarrow {HO} \left( {again,{\text{ }}same{\text{ }}as{\text{ }}above} \right) \hfill \\\\&= 2\overrightarrow {HO}  \hfill \\ 
\end{align} \)

(c) For any arbitrary point P in the plane of \(\Delta ABC,\) we have

\[\begin{align}& \overrightarrow {PA}  + \overrightarrow {PB}  + \overrightarrow {PC}  = \overrightarrow {PA}  + \left( {\overrightarrow {PB}  + \overrightarrow {PC} } \right) \hfill \\\\&\quad\qquad\qquad\qquad= \overrightarrow {PA}  + 2\overrightarrow {PD}  \hfill \\\\&\quad\qquad\qquad\qquad  = 3\overrightarrow {PG}  \hfill \\\end{align} \]

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

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