Question

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

Hint:

For any triangle, the median vector is the arithmetic mean of the vectors forming the adjacent sides originating from the same vertex.
You can simplify \( \frac{\sqrt{34}}{2} \) to \( \sqrt{\frac{34}{4}} = \sqrt{8.5} \), but keeping the fraction is standard.

Answer 1

Concept:
• Parallelogram law of vector addition: The sum of two vectors representing adjacent sides of a parallelogram is the diagonal.
• Midpoint property: In \( \Delta ABC \), if \( D \) is the midpoint of \( BC \), then \( \vec{AD} \) is half the vector sum of the sides \( AB \) and \( AC \).
• Magnitude of a vector \( x\hat{i} + y\hat{j} + z\hat{k} = \sqrt{x^2 + y^2 + z^2} \).

Step 1: Prove the vector relation for the median
Let the position vectors of vertices \( A, B, \) and \( C \) be \( \vec{a}, \vec{b}, \) and \( \vec{c} \) respectively.
Then \( \vec{AB} = \vec{b} - \vec{a} \) and \( \vec{AC} = \vec{c} - \vec{a} \).
Since \( D \) is the midpoint of \( BC \), its position vector \( \vec{d} \) is given by:
\[ \vec{d} = \frac{\vec{b} + \vec{c}}{2} \]
The median vector \( \vec{AD} \) is given by \( \vec{d} - \vec{a} \):
\[ \vec{AD} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2} \]
\[ \vec{AD} = \frac{(\vec{b} - \vec{a}) + (\vec{c} - \vec{a})}{2} \]
\[ \vec{AD} = \frac{\vec{AB} + \vec{AC}}{2} \] (Proved).

Step 2: Calculate the median vector \( \vec{AD} \)
Substitute the given vectors \( \vec{AB} = \hat{j} + \hat{k} \) and \( \vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k} \):
\[ \vec{AD} = \frac{(\hat{j} + \hat{k}) + (3\hat{i} - \hat{j} + 4\hat{k})}{2} \]
\[ \vec{AD} = \frac{3\hat{i} + 0\hat{j} + 5\hat{k}}{2} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k} \]

Step 3: Find the length of the median
The length of the median is the magnitude of vector \( \vec{AD} \):
\[ |\vec{AD}| = \sqrt{\left(\frac{3}{2}\right)^2 + 0^2 + \left(\frac{5}{2}\right)^2} \]
\[ |\vec{AD}| = \sqrt{\frac{9}{4} + \frac{25}{4}} = \sqrt{\frac{34}{4}} \]
\[ |\vec{AD}| = \frac{\sqrt{34}}{2} \text{ units} \]