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: Note where the points sit.
We have $A(3,0,0)$, $B(0,0,4)$, $C(0,5,4)$. The bisector from $A$ meets $BC$ at some point $D$, and we want the position vector of $D$.
Step 2: Use a smart observation about $B$ and $C$.
Both $B$ and $C$ have their $x$-coordinate equal to $0$. So the whole segment $BC$ lies in the plane $x=0$.
Step 3: What this means for $D$.
Since $D$ lies on $BC$, its $x$-coordinate must also be $0$. So the answer can have only $\hat{j}$ and $\hat{k}$ parts, no $\hat{i}$ part.
Step 4: Check the options against this.
Options (1), (3) and (4) all carry an $\hat{i}$ term, so they cannot lie on $BC$ and are ruled out.
Step 5: The surviving option.
Only option (2), $\dfrac{5\hat{j}+12\hat{k}}{3}$, has no $\hat{i}$ part, so it is the only point that can lie on $BC$.
Step 6: Conclude.
Hence the bisector meets $BC$ at the point given by option (2). \[ \boxed{\bar{D}=\frac{5\hat{j}+12\hat{k}}{3}} \]