The position vectors of three points A,B,C relative to an origin O are a, b and ka respectively. The point P lies on AB and is such that AP=3PB. The point Q lies on BC and is such that CQ=5QB. Given that OPQ is a straight line, what is the value of k?
- Geeganage WLv 59 months ago
Given, AP:PB = 3:1 and CQ:QB = 5:1
Therefore, Vector OP = (a+3b)/4 and Vector OQ = (ka+5b)/6
Given that OPQ is a straight line and hence OQ = t.OP, where t is a scalar.
(ka+5b)/6 = t(a+3b)/4
2(ka+5b) = 3t(a+3b)
Comparing coefficient of vector b, t = 10/9
Comparing coefficient of vector a, 2k = 3t = 10/3,
k = 5/3