Question

Let p and q be the position vectors of P and Q respectively then respect O and $\left| p \right| = p\& \left| q \right| = q$. The points R and S divide PQ internally and externally in the ratio 2:3 respectively. If OR and OS are perpendicular then
A.$9{p^2} = 4{q^2}$
B.$4{p^2} = 9{q^2}$
C.$9p = 4q$
D.$4p = 9q$

Answer 1

Hint : Here we are given two position vectors such that they are divided internally and externally both. We will simply use the section formula for internal and external division here along with the property that when two vectors are perpendicular their dot product is zero. This will help to find the correct relation between p and q.

Complete step-by-step answer :
Given that, p and q be the position vectors of P and Q respectively.
Now point R divides PQ internally in the ratio 2:3
So we will use formulas for internal division.
$OR = \dfrac{{3p + 2q}}{{3 + 2}}$
Now point S divides PQ externally in the same ratio.
So we will use formulas for external division.
$OS = \dfrac{{3p - 2q}}{{3 - 2}}$
Now given that OR and OS are perpendicular
So we can write that,
$OR.OS = 0$
Putting the values we get,
$\left( {\dfrac{{3p + 2q}}{{3 + 2}}} \right)\left( {\dfrac{{3p - 2q}}{{3 - 2}}} \right) = 0$
$\left( {\dfrac{{3p + 2q}}{5}} \right)\left( {\dfrac{{3p - 2q}}{1}} \right) = 0$
Transposing the denominators we get,
$\left( {3p + 2q} \right)\left( {3p - 2q} \right) = 0$
This on simplification gives,
${\left( {3p} \right)^2} - {\left( {2q} \right)^2} = 0$
Taking the squares,
$9{p^2} - 4{q^2} = 0$
Transposing a term we get,
$9{p^2} = 4{q^2}$
Thus this is the relation.
So option A is the correct answer.
So, the correct answer is “Option A”.

Note : Here note that generally we deal with the section formula used for segments with two endpoints fixed and with their coordinates written where we use formula as,
If the ratio of division is m:n then formula given below gives the coordinates of the point that divides the segment,
Case 1: for internal division, $\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}$
Case 2: for external division, $\dfrac{{m{x_2} - n{x_1}}}{{m - n}},\dfrac{{m{y_2} - n{y_1}}}{{m - n}}$
But in the case above we used the vectors instead of the endpoints of segments. Is the only difference.
Note that, when we find the relation at the end we can further simplify that with taking root on both sides; but on that it is not available in options.