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Q16:
Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2).

Solution :

Given Variables & Initial Setup

Let the points be $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in a three-dimensional Cartesian coordinate system. The given coordinates are:

  • $P(2, 3, 4) \implies \vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}$
  • $Q(4, 1, -2) \implies \vec{q} = 4\hat{i} + 1\hat{j} - 2\hat{k}$

We are required to find the position vector $\vec{r}$ of the midpoint $R$ of the line segment $PQ$. [By the Midpoint Formula in Vector Algebra, the position vector of the midpoint of a line segment connecting two points with position vectors $\vec{a}$ and $\vec{b}$ is given by $\frac{\vec{a} + \vec{b}}{2}$].

Step 1: Applying the Midpoint Formula

The position vector $\vec{r}$ of the midpoint $R$ is calculated as:

$\vec{r} = \frac{\vec{p} + \vec{q}}{2}$

Step 2: Vector Addition

Substitute the given vectors into the equation:

$\vec{r} = \frac{(2\hat{i} + 3\hat{j} + 4\hat{k}) + (4\hat{i} + 1\hat{j} - 2\hat{k})}{2}$

Group the components corresponding to the unit vectors $\hat{i}, \hat{j},$ and $\hat{k}$:

$\vec{r} = \frac{(2+4)\hat{i} + (3+1)\hat{j} + (4-2)\hat{k}}{2}$

$\vec{r} = \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2}$

Step 3: Scalar Multiplication

Divide each component by the scalar $2$:

$\vec{r} = \left(\frac{6}{2}\right)\hat{i} + \left(\frac{4}{2}\right)\hat{j} + \left(\frac{2}{2}\right)\hat{k}$

$\vec{r} = 3\hat{i} + 2\hat{j} + 1\hat{k}$

Visual Representation

P(2, 3, 4) R(3, 2, 1) Q(4, 1, -2)

Final Solution: The position vector of the midpoint is $3\hat{i} + 2\hat{j} + \hat{k}$.


More Questions from Class 12 Mathematics Vector Algebra EXERCISE 10.2


CBSE Solutions for Class 12 Mathematics Vector Algebra


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