Done (i), got idea how to do (ii), by can't really get the way round it, please help...... :P
As shown in the diagram the points A and B have position vectors a and b with respect to the origin O.
(i) Make a sketch of the diagram, and mark the points C, D and E such that OC=2a, OD=2a + b and OE = 1/3(OD).
(ii) By expressing suitable vectors in terms of a and b, prove that E lies on the line joining A and B.
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Vectors - C4 watch
- Thread Starter
- 10-05-2005 20:59
- 10-05-2005 21:09
OE = (1/3) OD = (2a+b)/3
AE = AO + OE = -OA + OE = -a + (2a+b)/3 = (b-a)/3
BE = BO + OE = -BO + OE = -b + (2a+b)/3 = 2(b-a)/3
Since AB = (b-a), then E divides the line AB into the ratio 1:2.
- 10-05-2005 21:21
let r be the line joioning A and B
r = OA + λAB
r = a + λ(b - a)
r = a(1 - λ) + λb
OE = (1/3)OD
OE = (1/3) (2a + b)
OE = (2/3)a + (1/3)b
OE = a(1 - 1/3) + (1/3)b = a(1 - λ) + λb for λ = 1/3
i.e E lies on the line r.
A is the point on r at which λ = 0
B is the point on r at which λ = 1
Since λ = 1/3 is between λ = 0 and 1, then E lies between A and B