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In Triangle OAB, E is the mid point of AB and F is a point on OA such that OF = 2FA. If C is the point of intersection of OE and BF, then find the ratios OC : CE and BC: CF. |
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Answer» Answer: With O as origin let a and b be the position vectors of A and B respectively. Then the position VECTOR of E, the mid-point of OB, is 2 b
Again, since AD:DB=2:1, the position vector of D is 1+2 1⋅a+2b
= 3 a+2b
EQUATION of OD and AE are r=t 3 a+2b
...(1) and r=a+s( 2 b
−a) or r=(1−s)a+s 2 b
...(2) If they intersect at p, then we will have identical VALUES of r. Hence comparing the coefficients of a and b, we get 3 t
=1−s, 3 2t
= 2 s
∴t= 5 3
or s= 5 4
. PUTTING for t in (1) or for s in (2), we get the position vector of point of intersection P as 5 a+2b
...(3) Now let P divide OD in the ratio λ:1. Hence by ratio formula the P.V. of P is λ+1 3 λ(a+2b)
+1.0
= 3(λ+1) λ
(a+2b) ....(4) Comparing (3) and (4), we get 3(λ+1) λ
= 5 1
⇒5λ=3λ+3⇒2λ=3⇒λ= 2 3
∴OP:PD=3:2 |
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