Question

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

Hint:

The parallelogram law of vector addition can be visualised here: if you complete a parallelogram with sides \(\vec{AB}\) and \(\vec{AC}\), the diagonal is exactly \(\vec{AB} + \vec{AC}\). Since the diagonals bisect each other, the vector to the midpoint of the base is half of the main diagonal vector!

Answer 1

Concept: Let us apply vector algebra rules to triangle geometries. Let the position vectors of vertices \(A, B,\) and \(C\) relative to an arbitrary origin be \(\vec{a}, \vec{b},\) and \(\vec{c}\) respectively.
• The side vectors are expressed as: \(\vec{AB} = \vec{b} - \vec{a}\) and \(\vec{AC} = \vec{c} - \vec{a}\).
• Since \(D\) is the midpoint of side \(BC\), its position vector \(\vec{d}\) is given by the section formula: \(\vec{d} = \frac{\vec{b} + \vec{c}}{2}\).
• The median vector is \(\vec{AD} = \vec{d} - \vec{a}\).

Step 1: Vector proof for the median relation.
Let the position vectors of the vertices \(A, B, C\) be \(\vec{a}, \vec{b}, \vec{c}\). Then, we can write the given vectors as differences of position vectors: \[ \vec{AB} = \vec{b} - \vec{a} \quad \cdots (1) \] \[ \vec{AC} = \vec{c} - \vec{a} \quad \cdots (2) \] Adding equation (1) and equation (2) together: \[ \vec{AB} + \vec{AC} = (\vec{b} - \vec{a}) + (\vec{c} - \vec{a}) = \vec{b} + \vec{c} - 2\vec{a} \quad \cdots (3) \] Now, because \(D\) is explicitly specified as the midpoint of the line segment \(BC\), its position vector is: \[ \vec{d} = \frac{\vec{b} + \vec{c}}{2} \] The position vector of the median line vector \(\vec{AD}\) is: \[ \vec{AD} = \vec{d} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2} \] Multiplying both sides by 2 gives: \[ 2\vec{AD} = \vec{b} + \vec{c} - 2\vec{a} \quad \cdots (4) \] Comparing equation (3) and equation (4), we find that their right-hand sides are completely identical. Therefore, we have proven: \[ \vec{AB} + \vec{AC} = 2\vec{AD} \]

Step 2: Compute the component form of the vector \(\vec{AD}\).
We are given: \[ \vec{AB} = 0\hat{i} + 1\hat{j} + 1\hat{k} \] \[ \vec{AC} = 3\hat{i} - 1\hat{j} + 4\hat{k} \] Let us calculate the vector sum \(\vec{AB} + \vec{AC}\): \[ \vec{AB} + \vec{AC} = (0+3)\hat{i} + (1 - 1)\hat{j} + (1 + 4)\hat{k} = 3\hat{i} + 0\hat{j} + 5\hat{k} \] Using our proven relation \(2\vec{AD} = \vec{AB} + \vec{AC}\): \[ 2\vec{AD} = 3\hat{i} + 5\hat{k} \quad \Rightarrow \quad \vec{AD} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k} \]

Step 3: Calculate the magnitude length of the median vector \(\vec{AD}\).
The length of the median is simply the magnitude of vector \(\vec{AD}\): \[ |\vec{AD}| = \sqrt{\left(\frac{3}{2}\right)^2 + (0)^2 + \left(\frac{5}{2}\right)^2} \] \[ |\vec{AD}| = \sqrt{\frac{9}{4} + 0 + \frac{25}{4}} = \sqrt{\frac{34}{4}} = \frac{\sqrt{34}}{2} \] Let us re-verify the terms. The sum vector components are \(3\hat{i} + 5\hat{k}\). Square of components is \(3^2 + 5^2 = 9 + 25 = 34\). Divided by 4 under root becomes \(\frac{\sqrt{34}}{2}\).