Question

Let $O A C B$ be a parallelogram with $O$ at the origin and $O C$ a diagonal. Let $D$ be the mid-point of $O A$. Using vector methods prove that $B D$ and $C O$ intersect in the same ratio. Determine this ratio.

Answer 1

$O A C B$ is a parallelogram with $O$ as origin. Let
$\overrightarrow{ OA }=\overrightarrow{ a }, \overrightarrow{ OB }=\overrightarrow{ b }, \overrightarrow{ OC }=\overrightarrow{ a }+\overrightarrow{ b }$

and
$\overrightarrow{ OD }=\frac{\overrightarrow{ a }}{2}$
$\overrightarrow{ C O }$ and $\overrightarrow{ B D }$ meets at $P$
$\therefore \overrightarrow{ O P }=\frac{\lambda \cdot 0+1(\overrightarrow{ a }+\overrightarrow{ b })}{\lambda+1} $ [along $\overrightarrow{ O C } ]$
$\Rightarrow \overrightarrow{ O P }=\frac{\overrightarrow{ a }+\overrightarrow{ b }}{\lambda+1} .....$(i)
Again, $ \overrightarrow{ O P }=\frac{\mu\left(\frac{\overrightarrow{ a }}{2}\right)+1(\overrightarrow{ b })}{\mu+1} $ [along $ \overrightarrow{BD} $ ] ......(ii)
From Eqs. (i) and (ii),
$\frac{\overrightarrow{ a }+\overrightarrow{ b }}{\lambda+1}=\frac{\mu \overrightarrow{ a }+2 \overrightarrow{ a }}{2(\mu+1)} \Rightarrow \frac{1}{\lambda+1}=\frac{\mu}{2(\mu+1)}$
and $ \frac{1}{\lambda+1}=\frac{1}{\mu+1}$
On solving, we get $\mu=\lambda=2$
Thus, required ratio is $2: 1$.