Question

Find the position vectors of a point R which divides the line joining two points P and Q whose position vectors are $2\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-3\overrightarrow{b}$ respectively, externally in the ratio 1:2. Also, show that P is the midpoint of the line segment RQ.

Answer 1

Hint:First of all, draw a line segment PQ in which R is dividing externally in the ratio 1:2. Now use the formula, $\dfrac{{{m}_{1}}\left( {{P}_{2}} \right)-{{m}_{2}}\left( {{P}_{1}} \right)}{{{m}_{1}}-{{m}_{2}}}$ to find the position vector of R where ${m}_1$ and ${{m}_2}$ are ratio which R divides externally. Now, find the midpoint of RQ by using the formula, $\dfrac{{{P}_{1}}+{{P}_{2}}}{2}$ and verify the given result.

Complete step-by-step answer:
Here, we are given that the position vectors of a point R which divides the line joining two points P and Q whose position vectors are $2\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-3\overrightarrow{b}$ respectively, externally in the ratio 1:2. We have to find the position vector of point R. Also, we have to show that P is the midpoint of the line segment RQ.
Let us first draw the line segment PQ such that R divides it externally in the ratio 1:2.

Here, we are given that position vectors of P and Q are $2\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-3\overrightarrow{b}$ respectively. We know that the position vector of any point A is given by $\overrightarrow{OA}$. So, we get,
$\overrightarrow{OP}=2\overrightarrow{a}+\overrightarrow{b}$
$\overrightarrow{OQ}=\overrightarrow{a}-3\overrightarrow{b}$
We know that when any point (say A) divides a line segment externally. So, its position vector is given by the sectional formula as,
$\overrightarrow{OA}=\dfrac{{{m}_{1}}\left( {{P}_{2}} \right)-{{m}_{2}}\left( {{P}_{1}} \right)}{{{m}_{1}}-{{m}_{2}}}$
From the diagram, we can see that ${{m}_{1}}=1,{{m}_{2}}=2,{{P}_{1}}=\overrightarrow{OP}=2\overrightarrow{a}+\overrightarrow{b}$ and ${{P}_{2}}=\overrightarrow{OQ}=\overrightarrow{a}-3\overrightarrow{b}$
So, we get,
$\overrightarrow{OR}=\dfrac{{{m}_{1}}\left( \overrightarrow{OQ} \right)-{{m}_{2}}\left( \overrightarrow{OP} \right)}{{{m}_{1}}-{{m}_{2}}}$
$=\dfrac{1\left( \overrightarrow{a}-3\overrightarrow{b} \right)-2\left( 2\overrightarrow{a}+\overrightarrow{b} \right)}{1-2}$
$=\dfrac{\overrightarrow{a}-3\overrightarrow{b}-4\overrightarrow{a}-2\overrightarrow{b}}{-1}$
$=\dfrac{-3\overrightarrow{a}-5\overrightarrow{b}}{-1}$
$=3\overrightarrow{a}+5\overrightarrow{b}$
So, we get the position vector of R as $3\overrightarrow{a}+5\overrightarrow{b}$.
Now, let us find the midpoint of RQ.

We know that the position vector of the midpoint (say A) of any line segment is given by:
$\overrightarrow{OA}=\dfrac{\overrightarrow{{{P}_{1}}}+\overrightarrow{{{P}_{2}}}}{2}$
Here, from the above diagram, we can see that,
$\overrightarrow{{{P}_{1}}}=\overrightarrow{OR}=3\overrightarrow{a}+5\overrightarrow{b}$
$\overrightarrow{{{P}_{2}}}=\overrightarrow{OQ}=\overrightarrow{a}-3\overrightarrow{b}$
So, we get,
$\overrightarrow{OM}=\dfrac{\overrightarrow{OR}+\overrightarrow{OQ}}{2}$
$\overrightarrow{OM}=\dfrac{3\overrightarrow{a}+5\overrightarrow{b}+\overrightarrow{a}-3\overrightarrow{b}}{2}$
$\overrightarrow{OM}=\dfrac{4\overrightarrow{a}+2\overrightarrow{b}}{2}$
$\overrightarrow{OM}=2\overrightarrow{a}+\overrightarrow{b}$
This is equal to the position vector of P that is $\overrightarrow{OP}=2\overrightarrow{a}+\overrightarrow{b}$.
So, we have proved that the midpoint of the line segment RQ is point P.

Note: In these types of questions, students often make mistakes while applying sectional formula by taking the wrong values of ${{P}_{1}}$ and ${{P}_{2}}$ or reversing their value. So, this must be taken care of. Also, take special notice, whether that point is dividing the line externally or internally. For external division use, $\dfrac{{{m}_{1}}\left( {{P}_{2}} \right)-{{m}_{2}}\left( {{P}_{1}} \right)}{{{m}_{1}}-{{m}_{2}}}$ while for internal division, use $\dfrac{{{m}_{1}}\left( {{P}_{2}} \right)+{{m}_{2}}\left( {{P}_{1}} \right)}{{{m}_{1}}+{{m}_{2}}}$.