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: Understanding the Question:
We are given the position vectors of three vertices of a triangle $ABC$. An internal angle bisector drawn from vertex $A$ intersects the opposite side $BC$ at a point $D$. We need to compute the position vector of this intersection point $D$.

Step 2: Key Formula or Approach:
1. According to the Angle Bisector Theorem, the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the lengths of the remaining two sides: $$BD : DC = AB : AC$$ 2. Calculate the lengths of the vector sides: $AB = |\vec{b} - \vec{a}|$ and $AC = |\vec{c} - \vec{a}|$.
3. Use the vector section formula to find the position vector of point $D$ dividing side $BC$ in the ratio $m : n$: $$\vec{d} = \frac{m\vec{c} + n\vec{b}}{m + n}$$

Step 3: Detailed Explanation:
Let the position vectors of the vertices be: $$\vec{a} = 3\hat{j}, \quad \vec{b} = 4\hat{k}, \quad \vec{c} = 3\hat{j} + 4\hat{k}$$ 4. Compute the side vector $\vec{AB}$ and its absolute magnitude: $$\vec{AB} = \vec{b} - \vec{a} = -3\hat{j} + 4\hat{k}$$ $$AB = |\vec{AB}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$ 5. Compute the side vector $\vec{AC}$ and its absolute magnitude: $$\vec{AC} = \vec{c} - \vec{a} = (3\hat{j} + 4\hat{k}) - 3\hat{j} = 4\hat{k}$$ $$AC = |\vec{AC}| = \sqrt{0^2 + 0^2 + 4^2} = 4$$ 6. Therefore, the point $D$ cuts the baseline segment $BC$ internally in the ratio $m : n = 5 : 4$.
Apply the internal section formula to find $\vec{d}$: $$\vec{d} = \frac{5\vec{c} + 4\vec{b}}{5 + 4}$$ $$\vec{d} = \frac{5(3\hat{j} + 4\hat{k}) + 4(4\hat{k})}{9}$$ $$\vec{d} = \frac{15\hat{j} + 20\hat{k} + 16\hat{k}}{9} = \frac{15\hat{j} + 36\hat{k}}{9}$$ 7. Divide each component by the denominator 9: $$\vec{d} = \left(\frac{15}{9}\right)\hat{j} + \left(\frac{36}{9}\right)\hat{k} = \frac{5}{3}\hat{j} + 4\hat{k}$$

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