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Question
D and E divides sides BC and CA of a triangle ABC in the ratio 2 : 3 respectively. Find the position vector of the point of intersection of AD and BE and the ratio in which this point divides AD and BE.
Solution
Let AD and BE intersect at P.
Let A, B, C, D, E, P have position vectos `bar"a", bar"b", bar"c", bar"d", bar"e", bar"p"` respectively.
D and E divide segments BC and CA internally in the ratio 2 : 3.
By the section formula for internal division,
`bar"d" = (2bar"c" + 3bar"b")/(2 + 3)`
∴ `5bar"d" = 2bar"c" + 3bar"b"` ....(1)
and `bar"e" = (2bar"a" + 3bar"c")/(2 + 3)`
∴ `5bar"e" = 2bar"a" + 3bar"c"` ....(2)
From (1), `5bar"d" - 3bar"b" = 2bar"c"` ...[both side multiply by 3]
∴ `15bar"d" - 9bar"b" = 6bar"c"`
From (2), `5bar"e" - 2bar"a" = 3bar"c"` ...[both side multiply by 2]
∴ `10bar"e" - 4bar"a" = 6bar"c"`
Equating both values of `6bar"c"`, we get
`15bar"d" - 9bar"b" = 10bar"e" - 4bar"a"`
∴ `15bar"d" + 4bar"a" = 10bar"e" + 9bar"b"`
∴ `(15bar"d" + 4bar"a")/(15 + 4) = (10bar"e" + 9bar"b")/(10 + 9)`
L.H.S is the position vector of the point which divides segment AD internally in the ratio 15 : 4
R.H.S is the position vector of the point which divides segment BE internally in the ratio 10 : 9
But P is the point of intersection of AD and BE
∴ P divides AD internally in the ratio 15 : 4 and P divides BE internally in the ratio 10 : 9
Hence, the position vector of the point of interaction of AD and BE is
`bar"p" = (15bar"d" + 4bar"a")/19 = (10bar"e" + 9bar"b")/19`
and it divides AD internally in the ratio 15 : 4 and BE internally in the ratio 10 : 9.
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