Question

The position vectors of the vertices $A, B$ and $C$ of a triangle are $2\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}, \quad 2\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \quad \text{and} \quad -\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ respectively. Let $l$ denote the length of the angle bisector $AD$ of $\angle BAC$ where $D$ is on the line segment $BC$. Then $2l^2$ equals:

• A. 49
• B. 42
• C. 50
• D. 45

Answer 1

Answer: (D)

45

Answer 2

First, find the lengths of $AB$ and $AC$:
$\vec{AB} = \vec{B} - \vec{A} = (2 - 2)\hat{i} + (2 + 3)\hat{j} + (3 - 3)\hat{k} = 0\hat{i} + 5\hat{j} + 0\hat{k}.$

$|\vec{AB}| = \sqrt{0^2 + 5^2 + 0^2} = 5.$

$\vec{AC} = \vec{C} - \vec{A} = (-1 - 2)\hat{i} + (1 + 3)\hat{j} + (3 - 3)\hat{k} = -3\hat{i} + 4\hat{j} + 0\hat{k}.$

$|\vec{AC}| = \sqrt{(-3)^2 + 4^2 + 0^2} = 5.$

Since $AB = AC$, triangle $ABC$ is isosceles. The midpoint $D$ of $BC$ is given by:

$\vec{D} = \frac{\vec{B} + \vec{C}}{2} = \frac{(2\hat{i} + 2\hat{j} + 3\hat{k}) + (-\hat{i} + 3\hat{j} + 3\hat{k})}{2} = \frac{\hat{i} + 5\hat{j} + 6\hat{k}}{2} = \frac{1}{2}\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}.$

The length of the angle bisector $\ell$ is given by:

$\ell = |\vec{A} - \vec{D}| = \left|2\hat{i} - 3\hat{j} - 3\hat{k} - \left(\frac{1}{2}\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}\right)\right|.$

$\ell = \left|\frac{3}{2}\hat{i} - \frac{9}{2}\hat{j} - \frac{9}{2}\hat{k}\right| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(-\frac{9}{2}\right)^2 + \left(-\frac{9}{2}\right)^2}.$

$\ell = \sqrt{\frac{9}{4} + \frac{81}{4} + \frac{81}{4}} = \sqrt{\frac{171}{4}} = \frac{\sqrt{45}}{2}.$

Calculating $2\ell^2$:
$2\ell^2 = 2 \times \left(\frac{\sqrt{45}}{2}\right)^2 = 45.$

The Correct answer is: 45