Question

In ΔABC, D and E are points on BC and AC, respectively, such that BD = 2DC and AE = 3EC. Let P be the point of intersection of AD and BE. Find the ratio `(BP)/(PE)` using the vector method.

Answer 1

Let `bara,barb,barc` be the position vectors of A, B and C respectively with respect to some origin.

D divides BC in the ratio 2 : 1 and E divides AC in the ratio 3: 1. 

∴ `bard=(barb+2barc)/3` and `bare=(bara+3barc)/4`

Let point of intersection P of AD and BE divides BE in the ratio k : 1 and AD in the ratio m : 1, then position vectors of P in these two cases are

`(barb+k((bara  +  3barc)/4))/(k+1)` and `(bara+m((barb  +  2c)/3))/(m+1)` respectively.

Equating the position vectors of P, we get,

`k/(4(k+1))bara+1/(k+1)barb+(3k)/(4(k+1))barc=1/(m+1)bara+m/(3(m+1))barb+(2m)/(3(m+1))barc` 

∴ `k/(4(k+1))=1/(m+1)`   ...(1)

∴ `1/(k+1)=m/(3(m+1))`  ...(2)

∴ `(3k)/(4(k+1))=(2m)/(3(m+1))`   ...(3)

Dividing equation (3) by equation (2),

`((3k)/(4(k  +  1)))/(1/(k  +  1))=((2m)/(3(m  +  1)))/(m/(3(m  +  1))`

`(3k)/4 = 2`

k = `8/3`

∴ `(BP)/(PE)=k/1`

∴ `(BP)/(PE)=8/3`