Question

If $\overrightarrow{AB}=\hat{j}+\hat{k}$ and $\overrightarrow{AC}=3\hat{i}-\hat{j}+4\hat{k}$ represent the two vectors along the sides $AB$ and $AC$ of $\triangle ABC$, prove that the median $\overrightarrow{AD}=\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{2}$ where $D$ is the midpoint of $BC$. Hence, find the length of the median $AD$.

Hint:

In vector geometry, the midpoint vector is always the average of the two endpoint vectors. Hence, the median from one vertex of a triangle is half the sum of the two side vectors starting from that vertex.

Answer 1

Let the position of the point \( A \) be \( \overrightarrow{A} \), the position of point \( B \) be \( \overrightarrow{B} \), and the position of point \( C \) be \( \overrightarrow{C} \). The vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are defined as follows:
\[ \overrightarrow{AB} = \hat{j} + \hat{k} \quad \text{and} \quad \overrightarrow{AC} = 3\hat{i} - \hat{j} + 4\hat{k} \] The median \( \overrightarrow{AD} \) is the vector from point \( A \) to the midpoint \( D \) of side \( BC \). We are asked to prove that:
\[ \overrightarrow{AD} = \frac{\overrightarrow{AB} + \overrightarrow{AC}}{2} \]

Step 1: Midpoint of \( BC \)
The midpoint \( D \) of side \( BC \) can be represented by the position vector of \( D \), which is the average of the position vectors of points \( B \) and \( C \): \[ \overrightarrow{D} = \frac{\overrightarrow{B} + \overrightarrow{C}}{2} \] From vector addition, we know that \( \overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A} \) and \( \overrightarrow{AC} = \overrightarrow{C} - \overrightarrow{A} \). Therefore, we can express the position vectors of \( B \) and \( C \) as:
\[ \overrightarrow{B} = \overrightarrow{A} + \overrightarrow{AB} \quad \text{and} \quad \overrightarrow{C} = \overrightarrow{A} + \overrightarrow{AC} \] Substituting these into the equation for \( \overrightarrow{D} \), we get: \[ \overrightarrow{D} = \frac{(\overrightarrow{A} + \overrightarrow{AB}) + (\overrightarrow{A} + \overrightarrow{AC})}{2} \] Simplifying: \[ \overrightarrow{D} = \frac{2\overrightarrow{A} + \overrightarrow{AB} + \overrightarrow{AC}}{2} \] Now, subtracting \( \overrightarrow{A} \) from both sides to find \( \overrightarrow{AD} \), we get: \[ \overrightarrow{AD} = \frac{\overrightarrow{AB} + \overrightarrow{AC}}{2} \] This proves the given statement.

Step 2: Finding the Length of Median \( AD \)
To find the length of the median \( AD \), we need to calculate the magnitude of \( \overrightarrow{AD} \). We are given: \[ \overrightarrow{AB} = \hat{j} + \hat{k} \quad \text{and} \quad \overrightarrow{AC} = 3\hat{i} - \hat{j} + 4\hat{k} \] Adding these vectors: \[ \overrightarrow{AB} + \overrightarrow{AC} = (\hat{j} + \hat{k}) + (3\hat{i} - \hat{j} + 4\hat{k}) = 3\hat{i} + 5\hat{k} \] Now, divide this by 2: \[ \overrightarrow{AD} = \frac{1}{2} (3\hat{i} + 5\hat{k}) = \frac{3}{2} \hat{i} + \frac{5}{2} \hat{k} \] The magnitude of \( \overrightarrow{AD} \) is: \[ |\overrightarrow{AD}| = \sqrt{\left( \frac{3}{2} \right)^2 + \left( \frac{5}{2} \right)^2} = \sqrt{\frac{9}{4} + \frac{25}{4}} = \sqrt{\frac{34}{4}} = \frac{\sqrt{34}}{2} \] Therefore, the length of the median \( AD \) is \( \boxed{\frac{\sqrt{34}}{2}} \).