Question

If \(A, B, C\) are vertices of a triangle with position vectors \(\vec{a}, \vec{b}, \vec{c}\) respectively, then find the position vector of the point \(D\) where the angle bisector from vertex \(A\) meets \(BC\).

Hint:

To remember the ratio, use the mnemonic: \(BD\) is next to side \(AB\), and \(DC\) is next to side \(AC\). So, \(BD/DC = AB/AC\). This is a very common vector application in MHT-CET!

Answer 1

Step 1: Understanding the Concept:
The internal angle bisector of a triangle divides the opposite side in the ratio of the sides containing the angle.
In \(\triangle ABC\), the angle bisector \(AD\) of \(\angle A\) divides the side \(BC\) in the ratio \(AB : AC\).

Step 2: Key Formula or Approach:
1. Length of side \(AB = c' = |\vec{b} - \vec{a}|\).
2. Length of side \(AC = b' = |\vec{c} - \vec{a}|\).
3. Internal Section Formula: If \(D\) divides \(BC\) in ratio \(m : n\), then \(\vec{d} = \frac{m\vec{c} + n\vec{b}}{m + n}\).

Step 3: Detailed Explanation:
The ratio in which \(D\) divides \(BC\) is \(BD : DC = AB : AC\).
Let \(m = AB = |\vec{b} - \vec{a}|\) and \(n = AC = |\vec{c} - \vec{a}|\).
Wait, the ratio is \(BD : DC = m : n\).
Applying the section formula for internal division:
\[ \vec{d} = \frac{BD \cdot \vec{c} + DC \cdot \vec{b}}{BD + DC} \] Since \(BD/DC = AB/AC\), we can write \(BD = k \cdot AB\) and \(DC = k \cdot AC\).
\[ \vec{d} = \frac{(AB)\vec{c} + (AC)\vec{b}}{AB + AC} \] Substituting the vector magnitudes:
\[ \vec{d} = \frac{|\vec{b} - \vec{a}| \vec{c} + |\vec{c} - \vec{a}| \vec{b}}{|\vec{b} - \vec{a}| + |\vec{c} - \vec{a}|} \] Rearranging for standard order:
\[ \vec{d} = \frac{|\vec{c} - \vec{a}| \vec{b} + |\vec{b} - \vec{a}| \vec{c}}{|\vec{c} - \vec{a}| + |\vec{b} - \vec{a}|} \]
Step 4: Final Answer:
The position vector of \(D\) is \(\vec{d} = \frac{|\vec{c} - \vec{a}| \vec{b} + |\vec{b} - \vec{a}| \vec{c}}{|\vec{c} - \vec{a}| + |\vec{b} - \vec{a}|}\).