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 - Mathematics and Statistics

Sum

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"   ......∴ 15bar"d" - 9bar"b" = 6bar"c"

From (2), 5bar"e" - 2bar"a" = 3bar"c"   ......∴ 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.

Concept: Three Dimensional (3-D) Coordinate System
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