Examples On Basics Of Vectors Set-2

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Example – 9   SECTION FORMULA

Let \(A\left( {\vec a} \right)\,\,{\text{and}}\,B\left( {\vec b} \right)\)  be two fixed points. Find the position vectors of the points lying on the (extended) line AB which divide the segment internally and externally in the ratio m : n.

Solution: We consider internal division; the external division case follows analogously.

Let \(C\left( {\vec c} \right)\) be the point which divides AB internally in the ratio m : n.

We have,

\[\begin{align}&\quad\qquad\overrightarrow {AC}  = \frac{m}{{m + n}}\,\,\,\overrightarrow {AB}  \hfill \\\\&\Rightarrow  \quad \left( {\overrightarrow {OC}  - \overrightarrow {OA} } \right) = \frac{m}{{m + n}}\,\,\left( {\,\overrightarrow {OB}  - \overrightarrow {OA} } \right) \hfill \\\\&  \Rightarrow \quad  \vec c - \vec a = \frac{m}{{m + n}}\,\,\left( {\vec b - \vec a} \right) \hfill \\\\& \Rightarrow  \quad \boxed{\vec c = \frac{{m\vec b + n\vec a}}{{m + n}}\,} \hfill \\ \end{align} \]

Similarly, the point \(D(\vec d)\) which divides AB externally in the ratio m : n is given by

\[\vec d = \frac{{m\vec b - n\vec a}}{{m - n}}\]

A particular case of internal division is the mid-point of \(A(\vec a)\,and\,B(\vec b)\): the mid-point is \(\begin{align}\frac{{\vec a + \vec b}}{2},\end{align}\)

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: Let the vertices of the triangle by \(A\left( {\vec a} \right),\,\,B\,(\vec b)\,{\text{and}}\,C(\vec c)\). 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,

\[\frac{{BD}}{{DC}} = \frac{c}{b}\]

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

\[D \equiv \frac{{z\vec c + y\vec b}}{{z + y}}\]

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

\[\frac{{DI}}{{IA}} = \frac{{BD}}{{BA}} = \frac{{\frac{z}{{z + y}}x}}{z} = \frac{x}{{z + y}}\]

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:

\[\begin{align}&\;\;I \equiv \frac{{x\vec a + \left( {x + y} \right)\left( {\frac{{z\vec c + y\vec b}}{{z + y}}} \right)}}{{x + z + y}} \hfill \\\\&\quad= \frac{{x\vec a + y\vec b + z\vec c}}{{x + y + z}}...{\text{ }}\left( 1 \right) \hfill \\ \end{align} \]

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).

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