Question

If $3\hat{j}$, $4\hat{k}$ and $3\hat{j} + 4\hat{k}$ are the position vectors of the vertices $A, B, C$ respectively of $\Delta ABC$, then the position vector of the point in which the bisector of $\angle A$ meets $BC$ is

• A. $\frac{5}{3}\hat{j} - 4\hat{k}$
• B. $5\hat{j} - 4\hat{k}$
• C. $5\hat{j} + 4\hat{k}$
• D. $\frac{5}{3}\hat{j} + 4\hat{k}$

Hint:

Notice that the $\hat{k}$ component in the final answer can be verified very quickly. The section formula calculation for the $\hat{k}$ coefficients alone is $\frac{5(4) + 4(4)}{9} = \frac{20 + 16}{9} = \frac{36}{9} = 4$. This tells you immediately that the final vector must end with $+4\hat{k}$, leaving options (A) and (D) as the only candidates.

Answer 1

The Correct Option is D

Step 1: List the position vectors.
$\vec{a} = 3\hat{j}$ (vertex A), $\vec{b} = 4\hat{k}$ (vertex B), $\vec{c} = 3\hat{j} + 4\hat{k}$ (vertex C). The bisector of angle A meets BC at a point D.
Step 2: Recall the angle bisector ratio.
The internal bisector from A divides BC in the ratio $BD:DC = AB:AC$, the ratio of the two adjacent sides.
Step 3: Compute AB.
$\vec{AB} = \vec{b} - \vec{a} = -3\hat{j} + 4\hat{k}$, so $AB = \sqrt{9 + 16} = 5$.
Step 4: Compute AC.
$\vec{AC} = \vec{c} - \vec{a} = 4\hat{k}$, so $AC = 4$. Hence D divides BC in the ratio $5:4$ from B to C.
Step 5: Apply the section formula.
With $BD:DC = 5:4$, $\vec{d} = \dfrac{5\vec{c} + 4\vec{b}}{5 + 4} = \dfrac{5(3\hat{j} + 4\hat{k}) + 4(4\hat{k})}{9}$.
Step 6: Simplify.
Numerator $= 15\hat{j} + 20\hat{k} + 16\hat{k} = 15\hat{j} + 36\hat{k}$, so $\vec{d} = \dfrac{15}{9}\hat{j} + \dfrac{36}{9}\hat{k} = \dfrac{5}{3}\hat{j} + 4\hat{k}$.
\[ \boxed{\dfrac{5}{3}\hat{j} + 4\hat{k}} \]