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Friday, 8 August 2014

CHAPTER 12- Worked Out Examples – 2

   Example: 8     

Let be three non-coplanar vectors. Prove that the points A\left( {2\vec a + 3\vec b - \vec c} \right),\,B\left( {\vec a - 2\vec b - 3\vec c} \right),\, C\left( {3\vec a + 4\vec b - 2\vec c} \right) and D\left( {\vec a - 6\vec b + 6\vec c} \right) are coplanar.
Solution: 8

As in the previous example, we first draw a visual picture to determine when four points can be coplanar.
Thus, as explained in the figure, we must have some scalars \lambda ,\mu  \in \mathbb{R} for which
\overrightarrow {AC}  = \lambda \overrightarrow {AB}  + \mu \overrightarrow {AD}
 \Rightarrow  \,\,\,\, \left( {\overrightarrow {OC}  - \overrightarrow {OA} } \right) = \lambda \left( {\overrightarrow {OB}  - \overrightarrow {OA} } \right) + \mu \left( {\overrightarrow {OD}  - \overrightarrow {OA} } \right){O is the origin}
 \Rightarrow  \,\,\,\, \vec a + \vec b - \vec c = \lambda \left( { - \vec a - 5\vec b + 4\vec c} \right) + \mu \left( { - \vec a - 9\vec b + 7\vec c} \right)
 \Rightarrow  \,\,\,\, \left( {1 + \lambda  + \mu } \right)\vec a + \left( {1 + 5\lambda  + 9\mu } \right)\vec b + \left( { - 1 - 4\lambda  - 7\mu } \right)\vec c = \vec 0
Since \vec a,\vec b and \,\vec c are non-coplanar, we must have
1 + \lambda  + \mu  = 0
1 + 5\lambda  + 9\mu  = 0
1 + 4\lambda  + 7\mu  = 0
As can be easily verified, this system has the solution \lambda  =  - 2,\mu  = 1,implying \overrightarrow {AB} ,\,\,\overrightarrow {AC}  and \overrightarrow {AD}  are indeed coplanar.
Thus, the points ABC and D are coplanar.

SECTION FORMULA

     Example: 9      

Let A\left( {\vec a} \right) 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: 9

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,
\overrightarrow {AC}  = \dfrac{m}{{m + n}}\,\,\,\overrightarrow {AB}
 \Rightarrow  \,\,\,\, \left( {\overrightarrow {OC}  - \overrightarrow {OA} } \right) = \dfrac{m}{{m + n}}\,\,\left( {\,\overrightarrow {OB}  - \overrightarrow {OA} } \right)
 \Rightarrow  \,\,\,\, \vec c - \vec a = \dfrac{m}{{m + n}}\,\,\left( {\vec b - \vec a} \right)
\Rightarrow\,\,\,\,{\vec c = \dfrac{{m\vec b + n\vec a}}{{m + n}}}
Similarly, the point D(\vec d) which divides AB externally in the ratio m : n  is given by
\vec d = \dfrac{{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 \dfrac{{\vec a + \vec b}}{2},
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