Question

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

• A. \( \dfrac{\vec b + \vec c}{2} \)
• B. \( \dfrac{|\vec{AC}|\,\vec b + |\vec{AB}|\,\vec c}{|\vec{AB}| + |\vec{AC}|} \)
• C. \( \dfrac{|\vec{AB}|\,\vec b + |\vec{AC}|\,\vec c}{|\vec{AB}| + |\vec{AC}|} \)
• D. \( \dfrac{\vec a + \vec b + \vec c}{3} \)

Hint:

The internal angle bisector divides the opposite side in the ratio of the adjacent sides. When using vectors, combine this ratio with the section formula to obtain the required position vector.

Answer 1

The Correct Option is C

Concept: The Angle Bisector Theorem states that the internal angle bisector of a triangle divides the opposite side in the ratio of the adjacent sides. If the angle bisector from \(A\) meets \(BC\) at \(D\), then \[ \frac{BD}{DC} = \frac{AB}{AC} \] Thus, point \(D\) divides the line segment \(BC\) internally in the ratio \(AB : AC\). If a point divides the line joining vectors \(\vec b\) and \(\vec c\) in the ratio \(m:n\), then its position vector is \[ \frac{m\vec c + n\vec b}{m+n} \]
Step 1: {Apply the Angle Bisector Theorem.} Since \[ \frac{BD}{DC} = \frac{AB}{AC} \] point \(D\) divides \(BC\) internally in the ratio \[ AB : AC \]
Step 2: {Use the section formula in vector form.} If a point divides the segment joining \(\vec b\) and \(\vec c\) in the ratio \(AB:AC\), its position vector is \[ \vec{OD} = \frac{AB\,\vec c + AC\,\vec b}{AB + AC} \] Rearranging, \[ \vec{OD} = \frac{|\vec{AB}|\,\vec b + |\vec{AC}|\,\vec c}{|\vec{AB}| + |\vec{AC}|} \]
Step 3: {Write the final position vector.} Thus the position vector of \(D\) is \[ \boxed{ \vec{OD} = \frac{|\vec{AB}|\,\vec b + |\vec{AC}|\,\vec c}{|\vec{AB}| + |\vec{AC}|} } \]