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Question
In Δ OAB, E is the midpoint of OB and D is the point on AB such that AD : DB = 2 : 1. If OD and AE intersect at P, then determine the ratio OP : PD using vector methods.
Solution
Let A, B, D, E, P have position vectors `bar"a", bar"b", bar"d", bar"e", bar"p"` respectively
w.r.t. O.
∵ AD : DB = 2 : 1
∴ D divides AB internally in the ratio 2 : 1.
Using section formula for internal division, we get
`bar"d" = (2bar"b" + bar"a")/(2 + 1)`
∴ `3bar"d" = 2bar"b" + bar"a"` ...(1)
Since E is the midpoint of OB, `bar"e" = bar"OE" = 1/2 bar"OB" = bar"b"/2`
∴ `bar"b" = 2bar"e"` ....(2)
∴ from (1),
`3bar"d" = 2(2bar"e") + bar"a"` ...[By(2)]
`= 4bar"e" + bar"a"`
∴ `(3bar"d" + 2.bar0)/(3 + 2) = (4bar"e" + bar"a")/(4 + 1)`
LHS is the position vector of the point which divides OD internally in the ratio 3 : 2.
RHS is the position vector of the point which divides AE internally in the ratio 4 : 1.
But OD and AE intersect at P
∴ P divides OD internally in the ratio 3 : 2.
Hence, OP : PD = 3 : 2.
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