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 Question:
In a triangle \( ABC \), the internal angle bisector of \( \angle A \) divides the opposite side \( BC \) into segments \( BD \) and \( DC \) such that the ratio of these segments is equal to the ratio of the sides containing the angle.
Step 2: Key Formula or Approach:
1. Angle Bisector Theorem: \( \frac{BD}{DC} = \frac{AB}{AC} = \frac{c}{b} \), where \( c = |\vec{AB}| \) and \( b = |\vec{AC}| \).
2. Section Formula: If point \( P \) divides the line segment joining \( \vec{r_1} \) and \( \vec{r_2} \) in the ratio \( m:n \), then \( \vec{r} = \frac{m\vec{r_2} + n\vec{r_1}}{m+n} \).
Step 3: Detailed Explanation:
Let the lengths of the sides opposite to vertices \( A, B, C \) be \( a, b, c \) respectively.
The angle bisector at \( A \) meets \( BC \) at \( D \).
According to the property of the internal angle bisector, \( D \) divides \( BC \) internally in the ratio \( AB : AC \).
Thus, the ratio is \( c : b \).
The position vectors of \( B \) and \( C \) are \( \vec{b} \) and \( \vec{c} \) respectively.
Applying the internal section formula for the ratio \( c:b \):
\[ \vec{d} = \frac{c\vec{c} + b\vec{b}}{c + b} \]
Rearranging the terms, we get:
\[ \vec{d} = \frac{b\vec{b} + c\vec{c}}{b+c} \]
Step 4: Final Answer:
The position vector of point \( D \) is \( \frac{b\vec{b} + c\vec{c}}{b+c} \).