TJC 2007 P1 Q4
Relative to an origin O, position vectors of A, B and C are a, b and c respectively, where a, b and c are non-parallel vectors. M is the mid-point of AC and P is on the line AB produced such that AB:BP = 2:3. The line PM meets the line BC at a point S. Show that the position vector of S is ⅛ (5b + 3c) .
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Answer:
By repeated use of Ratio theorem,
OM = ½(a + c) and OP = ½(5b - 3a)
OS = μOP + (1-μ) OM
= ½( μ(5b - 3a)+(1-μ)(a + c) )
= (½ - 2μ)a + (5μ/2)b + ((1-μ)/2) c
Also, by ratio theorem again (seeing that B, S and C are collinear),
OS = λc + (1-λ)b
Since a, b and c are non-parallel, we can compare coefficients of the vectors
½ - 2μ = 0 ====> μ = ¼
λ = ((1-μ)/2) = ¾/2 = ⅜
Hence, from OS = λc + (1-λ)b
OS = ⅜ c + (1-⅜) b
OS = ⅛ (5b + 3c) (shown)
