Vectors 1 Question 29
30. 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.
(1988, 3M)
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Solution:
- $O A C B$ is a parallelogram with $O$ as origin. Let
$$ \overrightarrow{OA}=\overrightarrow{a}, \overrightarrow{OB}=\overrightarrow{b}, \overrightarrow{OC}=\overrightarrow{a}+\overrightarrow{b} $$
$$ \text { and } \quad \overrightarrow{OD}=\frac{\overrightarrow{\mathbf{a}}}{2} $$
$\overrightarrow{\mathbf{C O}}$ and $\overrightarrow{\mathbf{B D}}$ meets at $P$.
$$ \begin{array}{ll} \therefore & \overrightarrow{\mathbf{O P}}=\frac{\lambda \cdot 0+1(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})}{\lambda+1} \\ \Rightarrow & \overrightarrow{\mathbf{O P}}=\frac{\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}}{\lambda+1} \\ \text { Again, } & \overrightarrow{\mathbf{O P}}=\frac{\mu\left(\overrightarrow{\mathbf{a}} \frac{2}{2}\right)+1(\overrightarrow{\mathbf{b}})}{\mu+1} \\ \Rightarrow & \overrightarrow{\mathbf{O P}}=\frac{\mu \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}}{2(\mu+1)} \end{array} $$
[along $\overrightarrow{\mathbf{O C}}$ ]
[along $\overrightarrow{\mathbf{B D}}$ ] From Eqs. (i) and (ii),
$\frac{\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}}{\lambda+1}=\frac{\mu \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{a}}}{2(\mu+1)} \Rightarrow \frac{1}{\lambda+1}=\frac{\mu}{2(\mu+1)}$ and $\frac{1}{\lambda+1}=\frac{1}{\mu+1}$
On solving, we get $\mu=\lambda=2$
Thus, required ratio is $2: 1$.
