Question

If $4i + 7j + 8k, 2i + 3j + 4k$ and $2i + 5j + 7k$ are the position vectors of the vertices A, B and C respectively of triangle ABC, then the position vector of the point where the internal bisector of angle A meets BC is

• A. $2\hat{i} + \frac{13}{3}\hat{j} + 6\hat{k}$
• B. $2\hat{i} + 5\hat{j} + 7\hat{k}$
• C. $4\hat{i} + 7\hat{j} + 8\hat{k}$
• D. $2\hat{i} + 3\hat{j} + 4\hat{k}$

Answer 1

Answer: (A)

$2\hat{i} + \frac{13}{3}\hat{j} + 6\hat{k}$

Answer 2

The position vector of the point where the internal bisector of angle A meets BC can be found using the Angle Bisector Theorem and the section formula.

Let $\vec{a}$, $\vec{b}$, and $\vec{c}$ be the position vectors of vertices A, B, and C, respectively. Given: $\vec{a} = 4\hat{i} + 7\hat{j} + 8\hat{k}$ $\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ $\vec{c} = 2\hat{i} + 5\hat{j} + 7\hat{k}$

Let D be the point on BC where the internal bisector of angle A meets BC. According to the Angle Bisector Theorem, D divides BC in the ratio $AB:AC$.

First, calculate the lengths of sides AB and AC: $\vec{AB} = \vec{b} - \vec{a} = (2-4)\hat{i} + (3-7)\hat{j} + (4-8)\hat{k} = -2\hat{i} - 4\hat{j} - 4\hat{k}$ $AB = |\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.

$\vec{AC} = \vec{c} - \vec{a} = (2-4)\hat{i} + (5-7)\hat{j} + (7-8)\hat{k} = -2\hat{i} - 2\hat{j} - \hat{k}$ $AC = |\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.

The ratio $BD:DC = AB:AC = 6:3 = 2:1$. Using the section formula for internal division, the position vector of D ($\vec{d}$) is: $\vec{d} = \frac{1 \cdot \vec{b} + 2 \cdot \vec{c}}{1+2}$ $\vec{d} = \frac{1 \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) + 2 \cdot (2\hat{i} + 5\hat{j} + 7\hat{k})}{3}$ $\vec{d} = \frac{(2\hat{i} + 3\hat{j} + 4\hat{k}) + (4\hat{i} + 10\hat{j} + 14\hat{k})}{3}$ $\vec{d} = \frac{(2+4)\hat{i} + (3+10)\hat{j} + (4+14)\hat{k}}{3}$ $\vec{d} = \frac{6\hat{i} + 13\hat{j} + 18\hat{k}}{3}$ $\vec{d} = 2\hat{i} + \frac{13}{3}\hat{j} + 6\hat{k}$