Question

The vertices of triangle ABC are $A \equiv (3,0,0) ; B \equiv (0,0,4) ; C \equiv (0,5,4)$. Find the position vector of the point in which the bisector of angle A meets BC is

• A. $5\hat{i} + 12\hat{j}$
• B. $\frac{5\hat{j} + 12\hat{k}}{3}$
• C. $\frac{5\hat{i} + 12\hat{j}}{13}$
• D. $\frac{5\hat{i} - 12\hat{j}}{3}$

Hint:

Always scan the coordinates for zeroes before diving into heavy section formula calculations! Identifying that a line segment lies entirely within a specific 2D plane (like $x=0$) allows you to instantly eliminate impossible vector options.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the 3D coordinates of the vertices of a triangle. We must find the position vector of a specific point $D$ on the line segment $BC$, where the internal angle bisector of vertex $A$ intersects the segment $BC$.

Step 2: Key Formula or Approach:
The Angle Bisector Theorem states that the angle bisector of an angle of a triangle divides the opposite side into segments that are proportional to the lengths of the other two adjacent sides.
Therefore, point $D$ must divide the segment $BC$ in the ratio $AB : AC$.
Once the ratio is known, the position vector of $D$ can theoretically be found using the section formula: $\vec{D} = \frac{m\vec{C} + n\vec{B}}{m + n}$.

Step 3: Detailed Explanation:
Let's analyze the coordinates of the points. There is a powerful visual shortcut here that saves us from performing complex distance calculations!
Point $B(0,0,4)$ lies strictly on the z-axis, which is contained entirely within the yz-plane (since its x-coordinate is 0).
Point $C(0,5,4)$ also lies strictly in the yz-plane (since its x-coordinate is 0).
Because both vertices $B$ and $C$ lie flat in the yz-plane, the entire straight line segment connecting them ($BC$) must also reside exclusively in the yz-plane.
Therefore, any point $D$ that lies anywhere on the segment $BC$ must inherently have an x-coordinate of 0. This means its position vector can only contain $\hat{j}$ and $\hat{k}$ components, and it cannot contain an $\hat{i}$ component.
Let's evaluate the four given options:
(A) $5\hat{i} + 12\hat{j}$ has an $\hat{i}$ component.
(C) $\frac{5\hat{i} + 12\hat{j}}{13}$ has an $\hat{i}$ component.
(D) $\frac{5\hat{i} - 12\hat{j}}{3}$ has an $\hat{i}$ component.
Option (B) is the only vector provided that lacks an $\hat{i}$ component, making it mathematically the only possible point that could lie on the line segment $BC$.
(Note: A full calculation of distances $AB=5$ and $AC=5\sqrt{2}$ yields a ratio of $1:\sqrt{2}$, which suggests there is likely a minor typographical error in the provided exam coordinates for C. A ratio of $1:2$ is required to perfectly compute option B using the section formula. However, the logical elimination method holds true regardless of the typo!)

Step 4: Final Answer:
The position vector of the point is $\frac{5\hat{j} + 12\hat{k}}{3}$, which corresponds to option (B).