4.4 Expression of a Vector as the Linear Combination of a Few Vectors

4.4 Expression of a Vector as the Linear Combination of a Few Vectors
1. Polygon Law for Vectors
PQ = PU + UT + TS + SR + RQ

2. To prove that two vectors are parallel, we must express one of the vectors as a scalar multiple of the other vector.
For example, AB =k CD  or  CD =h AB .  

3. To prove that points P, Q and R are collinear, prove one of the following.
   PQ =k QR  or  QR =h PQ    PR =k PQ  or  PQ =h PR    PR =k QR  or  QR =h PR

Example:
Diagram below shows a parallelogram ABCD. Point Q lies on the straight line AB and point S lies on the straight line DC. The straight line AS is extended to the point T such that AS = 2ST.
It is given that AQ : QB = 3 : 1, DS : SC = 3 : 1, AQ =6 a ˜  and  AD = b ˜   
(a) Express, in terms of a ˜  and  b ˜ :   
     (i)  AS                (ii)  QC
(b) Show that the points Q, C and T are collinear.

Solution:
(a)(i)  AS = AD + DS              = AD + AQ AQ:QB= 3:1 and  DS:SC= 3:1 AQ = DS              = b ˜ +6 a ˜              =6 a ˜ + b ˜

(a)(ii)  QC = QB + BC           = 1 3 AQ + AD AQ:QB= 3:1 AQ QB = 3 1 QB= 1 3 AQ and for parallelogram,  BC//AD, BC=AD            = 1 3 ( 6 a ˜ )+ b ˜           =2 a ˜ + b ˜

(b)  QT = QA + AT           = QA + 3 2 AS AS=2ST AT=3ST= 3 2 AS           =6 a ˜ + 3 2 ( 6 a ˜ + b ˜ )           =3 a ˜ + 3 2 b ˜           = 3 2 ( 2 a ˜ + b ˜ )           = 3 2 QC Points Q, C and T are collinear.