Question

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.​

Answer 1

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

Answer 2

Given:

In Triangle OAB, E is the mid point of AB and F is a point on OA such that OF = 2FA

$\text { Let } \overrightarrow{O A}=a, \quad \overrightarrow{O B}=b \text {. }$

$\text { then } E \text { mid point of } A B$

$\overrightarrow{O E}=\frac{a+b}{2}$

$O F=2 F A$

$OA = OF + FA$

$OA = 2FA + FA$

$OA = 3FA$

$a = 3FA$

$FA =a/3$

$Lets\ C \ divides\ E \ in \ the \ ratio \ \lambda: 1$

$\frac{\lambda\left\frac{(a+b)}{2}+1(0)\right.}{\lambda+1}$

$=\frac{\lambda}{2(\lambda+1)}(a+b)$     -(1)

$\text { Let }{ }^{\prime} \mathrm{C}^{\prime} \text { divides } B F\text { in the ration } \mu: 1$

$\frac{2 \mu a / 3+b}{\mu+1}$

$\frac{2 \mu a+b}{3(\mu+1)}$       -(2)

$\text { from } a \ and \ b \quad 2 \mu=3 \Rightarrow \mu=3 / 2$

$\overrightarrow{OC}=\frac{2}{5}(a+b)=\frac{x}{2(\lambda+1)}$

$\begin{array}{l}\frac{\lambda}{2(\lambda+1)}=\frac{2}{5} \\\lambda=4\end{array}$

$\text { OC: } C E=\lambda:1=4: 1$

$B C: C F=\mu: 1 \Rightarrow \frac{3}{2} : 1 = 3:2$

Hence the ratios OC : CE is 4:1 and BC: CF is 3:2​