8.4 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.
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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=6a˜ 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˜+6a˜ =6a˜+b˜
(a)(ii) −−→QC=−−→QB+−−→BC =13−−→AQ+−−→AD←AQ:QB= 3:1AQQB=31⇒QB=13AQand for parallelogram, BC//AD, BC=AD =13(6a˜)+b˜ =2a˜+b˜
(b) −−→QT=−−→QA+−−→AT =−−→QA+32−→AS←AS=2STAT=3ST=32AS =−6a˜+32(6a˜+b˜) =3a˜+32b˜ =32(2a˜+b˜) =32−−→QC∴Points Q, C and T are collinear.