Question

A(1, -1, -3), B(2, 1, -2) & C(-5, 2, -6) are the position vectors of the vertices of a triangle ABC, length of the bisector of its internal angle at A is :

• A. $\sqrt{10}/4$
• B. $3\sqrt{10}/4$
• C. $\sqrt{10}$
• D. none

Answer 1

Answer: (B)

$3\sqrt{10}/4$

Answer 2

The problem asks for the length of the internal angle bisector at vertex A of triangle ABC, given the position vectors of its vertices.

Let the position vectors of A, B, and C be $\vec{a}$, $\vec{b}$, and $\vec{c}$ respectively.

Given: $\vec{a} = (1, -1, -3)$ $\vec{b} = (2, 1, -2)$ $\vec{c} = (-5, 2, -6)$

Let AD be the angle bisector of angle A, where D is a point on side BC. According to the Angle Bisector Theorem, the point D divides the side BC in the ratio of the lengths of the adjacent sides AB and AC. So, BD : DC = AB : AC.

First, calculate the lengths of the sides AB and AC:

Length of side AB: $\vec{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (1-(-1))\hat{j} + (-2-(-3))\hat{k} = \hat{i} + 2\hat{j} + \hat{k}$ $|\vec{AB}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

Length of side AC: $\vec{AC} = \vec{c} - \vec{a} = (-5-1)\hat{i} + (2-(-1))\hat{j} + (-6-(-3))\hat{k} = -6\hat{i} + 3\hat{j} - 3\hat{k}$ $|\vec{AC}| = \sqrt{(-6)^2 + 3^2 + (-3)^2} = \sqrt{36 + 9 + 9} = \sqrt{54} = \sqrt{9 \times 6} = 3\sqrt{6}$

Now, find the ratio BD : DC: BD : DC = AB : AC = $\sqrt{6} : 3\sqrt{6} = 1 : 3$.

Next, find the position vector of point D using the section formula. D divides BC in the ratio 1:3. $\vec{d} = \frac{3\vec{b} + 1\vec{c}}{1+3} = \frac{3\vec{b} + \vec{c}}{4}$

Substitute the position vectors of B and C: $3\vec{b} = 3(2\hat{i} + \hat{j} - 2\hat{k}) = 6\hat{i} + 3\hat{j} - 6\hat{k}$ $\vec{c} = -5\hat{i} + 2\hat{j} - 6\hat{k}$ $\vec{d} = \frac{(6\hat{i} + 3\hat{j} - 6\hat{k}) + (-5\hat{i} + 2\hat{j} - 6\hat{k})}{4} = \frac{(6-5)\hat{i} + (3+2)\hat{j} + (-6-6)\hat{k}}{4}$ $\vec{d} = \frac{\hat{i} + 5\hat{j} - 12\hat{k}}{4} = \frac{1}{4}\hat{i} + \frac{5}{4}\hat{j} - 3\hat{k}$

Finally, calculate the length of the angle bisector AD: $\vec{AD} = \vec{d} - \vec{a} = (\frac{1}{4}\hat{i} + \frac{5}{4}\hat{j} - 3\hat{k}) - (\hat{i} - \hat{j} - 3\hat{k})$ $\vec{AD} = (\frac{1}{4} - 1)\hat{i} + (\frac{5}{4} - (-1))\hat{j} + (-3 - (-3))\hat{k}$ $\vec{AD} = (\frac{1-4}{4})\hat{i} + (\frac{5+4}{4})\hat{j} + (-3+3)\hat{k}$ $\vec{AD} = -\frac{3}{4}\hat{i} + \frac{9}{4}\hat{j} + 0\hat{k}$

Length of AD: $|\vec{AD}| = \sqrt{(-\frac{3}{4})^2 + (\frac{9}{4})^2 + 0^2}$ $|\vec{AD}| = \sqrt{\frac{9}{16} + \frac{81}{16}}$ $|\vec{AD}| = \sqrt{\frac{90}{16}}$ $|\vec{AD}| = \frac{\sqrt{90}}{\sqrt{16}} = \frac{\sqrt{9 \times 10}}{4} = \frac{3\sqrt{10}}{4}$

The final answer is $3\sqrt{10}/4$.