Question

In a $\Delta ABO,E$ is the midpoint of $\overline{OB}$ and $D$ is a point on $AB$ such that $AD:DB=2:1$ if $\overline{OD}$ and $\overline{AE}$ intersect at $P$ then $\overline{OP}:\overline{PD}=$

Answer 1

With $O$ as origin let $a$ and $b$ the position vectors of $A$ and $B$ respectively.

Then the positoion vector of $E$, the mid-point of $OB$, is $\frac{b}{2}$

Again, since $AD:DB=2:1$, the position vector of $D$ is

$\frac{1.a+2b}{1+2}=\frac{a+2b}{3}$

Equation of $OD$ and $AE$ are $r=t\frac{a+2b}{3}...\left(1\right)$ and $r=a+s(\frac{b}{2}-a)$ or $r=(1-s)a+s\frac{b}{2}...\left(2\right)$

If the intersect at $P$, then we will have identical values of $r$.

Hence companing the coefficient 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 $\left(1\right)$ or for $S$ in $\left(2\right)$, we get the position vector of position of intersection $p$ as $\frac{a+2b}{5}..\left(3\right)$

now let $p$ divide $OD$ in the ratio $\lambda :1$.

Hence by ratio formula the $P.V$ of $P$ is

$\frac{\frac{\lambda (a+2b)}{3}+1.0}{\lambda +1}=\frac{\lambda }{3(\lambda +1)}(a+2b)...\left(4\right)$

Comparing $\left(3\right)$ and $\left(4\right)$, we get $\frac{\lambda }{3(\lambda +1)}=\frac{1}{5}\Rightarrow 5\lambda =3\lambda +3\Rightarrow 2\lambda =3\Rightarrow \lambda =\frac{3}{2}$

$\therefore OP:PD=3:2$