Question

Let ABC be an acute scalene triangle, O and H be its circumcentre and orthocentre respectively. Further, let N be the midpoint of OH. Find the value of the vector sum $\overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}$?
(a) $\overrightarrow{0}$ (zero vector)
(b) $\overrightarrow{HO}$,
(c) $\dfrac{1}{2}\overrightarrow{HO}$,
(d) $\dfrac{1}{2}\overrightarrow{OH}$.

Answer 1

Hint: We start solving the problem by assigning a variable vector for the origin vector. We then use the formula of the vector of line segments in the given sum. We then find the point N in terms of O and H using the midpoint formula. We recall the definition of centroid and use the fact that the centroid divides circumcentre and orthocentre in a ratio of $2:1$ to find the centroid in terms of O and H. We substitute all the obtained relations in the vector sum and make necessary calculations to get the required result.

Complete step-by-step solution
According to the problem, we have an acute scalene triangle ABC and O, H be its circumcentre and orthocentre. We need to find the value of the $\overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}$ if N is the midpoint of OH.
Let us draw the given information to get a better view.

Let us assume $\overrightarrow{E}$ be the origin vector. We know that the vector equation of the segment $\overrightarrow{AB}$ is defined as $\overrightarrow{EB}-\overrightarrow{EA}$.
Let us consider the vector sum $\overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}$.
So, we have $\overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}=\left( \overrightarrow{EA}-\overrightarrow{EN} \right)+\left( \overrightarrow{EB}-\overrightarrow{EN} \right)+\left( \overrightarrow{EC}-\overrightarrow{EN} \right)$.
$\overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}=\overrightarrow{EA}+\overrightarrow{EB}+\overrightarrow{EC}-3\overrightarrow{EN}$ ---(1).
According to the problem we have given that N is the midpoint of OH i.e., the vector $\overrightarrow{EN}$ is the mid-point of the vectors $\overrightarrow{EO}$ and $\overrightarrow{EH}$.
We know that the midpoint of the two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given as $\dfrac{1}{2}\times \left( \overrightarrow{a}+\overrightarrow{b} \right)$. We use this to find the vector $\overrightarrow{EN}$.
So, we have $\overrightarrow{EN}=\dfrac{1}{2}\times \left( \overrightarrow{EO}+\overrightarrow{EH} \right)$. Let us use this in equation (1).
$\Rightarrow \overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}=\overrightarrow{EA}+\overrightarrow{EB}+\overrightarrow{EC}-\left( 3\times \dfrac{1}{2}\times \left( \overrightarrow{EO}+\overrightarrow{EH} \right) \right)$.
$\Rightarrow \overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}=\overrightarrow{EA}+\overrightarrow{EB}+\overrightarrow{EC}-\dfrac{3}{2}\times \left( \overrightarrow{EO}+\overrightarrow{EH} \right)$.
$\Rightarrow \overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}=\overrightarrow{EA}+\overrightarrow{EB}+\overrightarrow{EC}-\dfrac{3}{2}\overrightarrow{EO}-\dfrac{3}{2}\overrightarrow{EH}$ ---(2).
We know that the centroid of the triangle ABC is defined as $\overrightarrow{EG}=\dfrac{\overrightarrow{EA}+\overrightarrow{EB}+\overrightarrow{EC}}{3}$.
$\Rightarrow \overrightarrow{EA}+\overrightarrow{EB}+\overrightarrow{EC}=3\overrightarrow{EG}$. We substitute this in equation (2).
$\Rightarrow \overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}=3\overrightarrow{EG}-\dfrac{3}{2}\overrightarrow{EO}-\dfrac{3}{2}\overrightarrow{EH}$ ---(3).
We know that the centroid divides the orthocentre and circumcentre in the ratio $1:2$.
From section formula, we know that if the point divides the points A and B in the ratio $m:n$ is \[\dfrac{mB+nA}{m+n}\].
So, we get $\overrightarrow{EG}=\dfrac{2\overrightarrow{EH}+\overrightarrow{EO}}{2+1}$.
$\Rightarrow \overrightarrow{EG}=\dfrac{2\overrightarrow{EH}+\overrightarrow{EO}}{3}$.
$\Rightarrow 3\overrightarrow{EG}=2\overrightarrow{EH}+\overrightarrow{EO}$. We substitute this in equation (3).
$\Rightarrow \overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}=2\overrightarrow{EH}+\overrightarrow{EO}-\dfrac{3}{2}\overrightarrow{EO}-\dfrac{3}{2}\overrightarrow{EH}$.
$\Rightarrow \overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}=\dfrac{1}{2}\overrightarrow{EH}-\dfrac{1}{2}\overrightarrow{EO}$.
$\Rightarrow \overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}=\dfrac{1}{2}\left( \overrightarrow{EH}-\overrightarrow{EO} \right)$.
$\Rightarrow \overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}=\dfrac{1}{2}\overrightarrow{OH}$.
So, we have found the value of the vector sum $\overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}$ as $\dfrac{1}{2}\overrightarrow{OH}$.
∴ The correct option for the given problem is (c).

Note: We can also assume orthocentre as the origin in order to reduce the calculation time. We should not write the answer as $\dfrac{1}{2}\overrightarrow{HO}$ as we need to take care of the direction of the vector while calculating the sum of the vectors. We should keep in mind that the vector sum is different from the scalar sum and should not be confused between both. We can also find the magnitude of the vector $\dfrac{1}{2}\overrightarrow{OH}$ if the vector points of the vertices are given.