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: Look at where B and C sit.
Both $B(0,0,4)$ and $C(0,5,4)$ have an x value of 0, so the whole segment BC lies in the plane where $x = 0$.

Step 2: Use that on the point D.
Any point on BC must also have $x = 0$, which means its position vector has no $\hat i$ part.

Step 3: Test the options.
Three options carry an $\hat i$ term, so they cannot lie on BC. Only $\dfrac{5\hat j + 12\hat k}{3}$ has no $\hat i$, so it is the point.
\[ \boxed{\dfrac{5\hat j + 12\hat k}{3}} \]