Question

In a triangle OAB, E is the mid-point of OB and D is a point on AB such that AD: DB = 2 : 1. If OD and AE intersect at P, determine the ratio OP : PD using vector methods.

• A. $OP:PD=2:3$
• B. $OP:PD=3:2$
• C. $OP:PD=1:3$
• D. $OP:PD=3:1$

Answer 1

Answer: (B)

$OP:PD=3:2$

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 $\frac{b}{2}$
Again, since $AD:DB=2:1$, the position vector of $D$ is $\frac{1\cdot a+2b}{1+2}=\frac{a+2b}{3}$
Equation of $OD$ and $AE$ are $r=t\frac{a+2b}{3}$ ...(1)

and $r=a+s(\frac{b}{2}-a)$ or $r=(1-s)a+s\frac{b}{2}$ ...(2)
If they intersect at $p$, then we will have identical values of $r$.

Hence comparing the coefficients of $a$ and $b$, we get
$\frac{t}{3}=1-s,\frac{2t}{3}=\frac{s}{2}\therefore t=\frac{3}{5}$ or $s=\frac{4}{5}.$
Putting for t in (1) or for s in (2), we get the position vector of point of intersection $P$ as $\frac{a+2b}{5}$ ...(3)
Now let $P$ divide $OD$ in the ratio $\lambda :1.$

Hence by ratio formula the $P.V.$ of $P$ is $\frac{\frac{\lambda (a+2b)}{3}+1.0}{\lambda +1}=\frac{\lambda }{3(\lambda +1)}(a+2b)$ ....(4)
Comparing (3) and (4), we get $\frac{\lambda }{3(\lambda +1)}=\frac{1}{5}$$\Rightarrow 5\lambda =3\lambda +3\Rightarrow 2\lambda =3$$\Rightarrow \lambda =\frac{3}{2}$

$\therefore OP:PD=3:2$