Question

Find the position vector of the midpoint of the vector joining the points $P(2,2,4)$ and $Q(4,1, - 2)$

Answer 1

Hint:Find the position vectors \[\overrightarrow {OP} \]and\[\overrightarrow {OQ} \]using the fact:
The position vector of $A({x_1},{y_1},{z_1})$ is given by $ \overrightarrow {OA} = {x_1}\widehat i + {y_1}\widehat j + {z_1}\widehat k $
Compute the position vector of the midpoint of P and Q using the formula:
If$ C(x,y,z) $ is the midpoint of segment $ \overrightarrow {AB} $, then we have the position vector of C
$ = \overrightarrow {OC} = \dfrac{{\overrightarrow {OA} + \overrightarrow {OB} }}{2} = (\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2}) $

Complete step by step solution:
We are given two points $ P(2,2,4)$and$Q(4,1, - 2) $.
We are required to find the position vector of the midpoint of the vector joining these two points $ P(2,2,4)$ and $Q(4,1, - 2) $
For a point $ A(x,y,z) $, its position vector from the origin $ O(0,0,0) $is denoted by $\overrightarrow {OA} $ and is given by $ \overrightarrow {OA} = x\widehat i + y\widehat j + z\widehat k $
Where
$ \widehat i = $unit vector (vector with magnitude 1) along the X-axis
$ \widehat j = $unit vector along the Y-axis
$ \widehat k = $unit vector along the Z-axis

Consider the origin $ O(0,0,0) $. Suppose we are given two points $A({x_1},{y_1},{z_1})$and$B({x_2},{y_2},{z_2})$.
Then the vector joining the points A and B is denoted by $ \overrightarrow {AB} $

Therefore, the vector joining the given points $ P(2,2,4 )$and$ Q(4,1, - 2) $will be denoted by $ \overrightarrow {PQ} $ and the position vectors of P and Q are\[\overrightarrow {OP} = 2\widehat i + 2\widehat j + 4\widehat k\] and $\overrightarrow {OQ} = 4\widehat i + \widehat j - 2\widehat k $respectively.
Let the midpoint of this vector $ \overrightarrow {PQ} $ be denoted by R.
The question says that we need to find the position vector of this point R.
Consider the formula for the position vector of the midpoint of the segment formed by joining the points $ A({x_1},{y_1},{z_1}) $ and $ B({x_2},{y_2},{z_2}) $.
Let $ O(0,0,0) $ be the origin.
Then the position vectors of A and B are $ \overrightarrow {OA} = {x_1}\widehat i + {y_1}\widehat j + {z_1}\widehat k$and$\overrightarrow {OB} = {x_2}\widehat i + {y_2}\widehat j + {z_2}\widehat k $
If$C(x,y,z)$is the midpoint of segment$\overrightarrow {AB} $, then we have the position vector of C
$ = \overrightarrow {OC} = \dfrac{{\overrightarrow {OA} + \overrightarrow {OB} }}{2} = (\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2})$
Using this formula, we get the position vector of the midpoint R of the segment formed by joining the points $ P(2,2,4) $and $ Q(4,1, - 2) $as
$
   = \overrightarrow {OR} \\
   = \dfrac{{\overrightarrow {OP} + \overrightarrow {OQ} }}{2} \\
   = (\dfrac{{2 + 4}}{2},\dfrac{{2 + 1}}{2},\dfrac{{4 + ( - 2)}}{2}) \\
   = (\dfrac{6}{2},\dfrac{3}{2},\dfrac{{ - 2}}{2}) \\
   = (3,\dfrac{3}{2}, - 1) \\
 $
We can write this position vector of R as $\overrightarrow {OR} = 3\widehat i + \dfrac{3}{2}\widehat j - \widehat k $
Hence $ \overrightarrow {OR} = 3\widehat i + \dfrac{3}{2}\widehat j - \widehat k $ is the required answer.

Note: Some students tend to find the equation of the segment PQ for such questions. However, it is not needed here.