Vectors 1 Question 28
29. In a $\triangle O A B, E$ is the mid-point of $B O$ and $D$ is a point on $A B$ such that $A D: D B=2: 1$. If $O D$ and $A E$ intersect at $P$, determine the ratio $O P: P D$ using methods.
$(1989,4 M)$
Show Answer
Solution:
- Let $O$ be origin and $\overrightarrow{\mathbf{O A}}=\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{O B}}=\overrightarrow{\mathbf{b}}$
$$ \overrightarrow{\mathbf{O E}}=\frac{\overrightarrow{\mathbf{b}}}{2} $$
[since $E$ being mid-point of $\overrightarrow{\mathbf{O B}}$ ]
$$ \overrightarrow{\mathbf{O D}}=\frac{\overrightarrow{\mathbf{a}} \cdot 1+\overrightarrow{\mathbf{b}} \cdot 2}{1+2} $$
(since, $D$ divides $\overrightarrow{\mathbf{A B}}$ in the ratio of $2: 1$ )
$\Rightarrow$ Equation of $\overrightarrow{O \mathbf{D}}$ is $\overrightarrow{\mathbf{r}}=t\left(\frac{\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}}{3}\right)$
and equation of $\overrightarrow{\mathbf{A E}}$ is $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+s\left(\frac{\overrightarrow{\mathbf{b}}}{2}-\overrightarrow{\mathbf{a}}\right)$
If $\overrightarrow{OD}$ and $\overrightarrow{AE}$ intersect at $P$, then there must be some $\overrightarrow{\mathbf{r}}$ for which they are equal.
$$ \begin{aligned} & \Rightarrow \quad t\left(\frac{\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}}{3}\right)=\overrightarrow{\mathbf{a}}+s\left(\frac{\overrightarrow{\mathbf{b}}}{2}-\overrightarrow{\mathbf{a}}\right) \\ & \Rightarrow \quad \frac{t}{3}=1-s \text { and } \frac{2 t}{3}=\frac{s}{2} \\ & \Rightarrow \quad t=\frac{3}{5} \text { and } s=\frac{4}{5} \\ & \therefore \quad \text { Point } P \text { is } \frac{\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}}{5} . \end{aligned} $$
Since, $P$ divides $\overrightarrow{\mathbf{O D}}$ in the ratio of $\lambda: 1$.
$$ \therefore \quad \frac{\lambda\left(\frac{\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}}{3}\right)+1 \cdot 0}{\lambda+1}=\left(\frac{\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}}{5}\right) $$
From Eqs. (i) and (ii),
$$ \begin{array}{rlrl} & & \frac{\lambda}{3(\lambda+1)} & =\frac{1}{5} \\ \Rightarrow & & 5 \lambda & =3 \lambda+3 \\ \Rightarrow & \lambda & =\frac{3}{2} \\ \therefore & & \frac{O P}{P D} & =\frac{3}{2} \end{array} $$
