Key Concepts and Formulas

• Position Vectors and Vector Subtraction: The position vector of a point $P$ is denoted by $\vec{P}$. The vector representing the segment from point $A$ to point $B$ is given by $\vec{AB} = \vec{B} - \vec{A}$.
• Magnitude of a Vector: The length (magnitude) of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.
• Angle Bisector Theorem: For a triangle $ABC$, if $AD$ is the angle bisector of $\angle BAC$ where $D$ lies on $BC$, then $\frac{BD}{DC} = \frac{AB}{AC}$.
• Midpoint of a Line Segment: If $D$ is the midpoint of a line segment $BC$, then its position vector is $\vec{D} = \frac{\vec{B} + \vec{C}}{2}$.
• Property of Isosceles Triangles: In an isosceles triangle, the angle bisector of the vertex angle is also the median to the base.

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Step-by-Step Solution

1. Identify the Coordinates of the Vertices

The given position vectors of the vertices A, B, and C are: $\vec{A} = 2 \hat{i}-3 \hat{j}+3 \hat{k} \implies A(2, -3, 3)$ $\vec{B} = 2 \hat{i}+2 \hat{j}+3 \hat{k} \implies B(2, 2, 3)$ $\vec{C} = -\hat{i}+\hat{j}+3 \hat{k} \implies C(-1, 1, 3)$

2. Calculate the Lengths of Sides AB and AC

To determine the nature of the triangle and apply the angle bisector theorem or its special case, we calculate the lengths of the sides AB and AC.

• Vector $\vec{AB}$: $\vec{AB} = \vec{B} - \vec{A} = (2\hat{i}+2\hat{j}+3\hat{k}) - (2\hat{i}-3\hat{j}+3\hat{k}) = (2-2)\hat{i} + (2-(-3))\hat{j} + (3-3)\hat{k} = 0\hat{i} + 5\hat{j} + 0\hat{k} = 5\hat{j}$

• Length of AB: $AB = |\vec{AB}| = |5\hat{j}| = \sqrt{0^2 + 5^2 + 0^2} = \sqrt{25} = 5 \text{ units}$

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

• Length of AC: $AC = |\vec{AC}| = |-3\hat{i} + 4\hat{j}| = \sqrt{(-3)^2 + 4^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ units}$

3. Analyze the Triangle and Locate Point D

Since $AB = AC = 5$, triangle $ABC$ is an isosceles triangle with $AB$ and $AC$ as the equal sides. In an isosceles triangle, the angle bisector of the vertex angle (here, $\angle BAC$) is also the median to the base. Therefore, point $D$, which lies on $BC$ and is the foot of the angle bisector from $A$, must be the midpoint of the line segment $BC$.

4. Determine the Position Vector of Point D

As $D$ is the midpoint of $BC$, we can find its position vector using the midpoint formula: $\vec{D} = \frac{\vec{B} + \vec{C}}{2}$ $\vec{D} = \frac{(2\hat{i}+2\hat{j}+3\hat{k}) + (-\hat{i}+\hat{j}+3\hat{k})}{2}$ $\vec{D} = \frac{(2-1)\hat{i} + (2+1)\hat{j} + (3+3)\hat{k}}{2}$ $\vec{D} = \frac{\hat{i} + 3\hat{j} + 6\hat{k}}{2} = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} + 3\hat{k}$ The coordinates of point D are $(\frac{1}{2}, \frac{3}{2}, 3)$.

5. Calculate the Length of the Angle Bisector AD ($l$)

The length $l$ of the angle bisector $AD$ is the distance between points $A(2, -3, 3)$ and $D(\frac{1}{2}, \frac{3}{2}, 3)$. We find the vector $\vec{AD}$: $\vec{AD} = \vec{D} - \vec{A} = \left(\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} + 3\hat{k}\right) - (2\hat{i}-3\hat{j}+3\hat{k})$ $\vec{AD} = \left(\frac{1}{2}-2\right)\hat{i} + \left(\frac{3}{2}-(-3)\right)\hat{j} + (3-3)\hat{k}$ $\vec{AD} = \left(\frac{1-4}{2}\right)\hat{i} + \left(\frac{3+6}{2}\right)\hat{j} + 0\hat{k}$ $\vec{AD} = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j}$ The length $l$ is the magnitude of $\vec{AD}$: $l = |\vec{AD}| = \sqrt{\left(-\frac{3}{2}\right)^2 + \left(\frac{9}{2}\right)^2 + 0^2}$ $l = \sqrt{\frac{9}{4} + \frac{81}{4}} = \sqrt{\frac{90}{4}} = \sqrt{\frac{45}{2}}$

6. Compute the Value of $2l^2$

The problem asks for the value of $2l^2$. $2l^2 = 2 \times \left(\sqrt{\frac{45}{2}}\right)^2$ $2l^2 = 2 \times \frac{45}{2}$ $2l^2 = 45$

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Common Mistakes & Tips

• Forgetting the isosceles property: If one does not notice that $AB=AC$, they might proceed with the general angle bisector formula, which is more complex and prone to errors. Always check for symmetry or equality of side lengths first.
• Calculation errors with fractions: Pay close attention to arithmetic operations, especially when dealing with fractions and negative signs, as seen in step 5.
• Confusing position vectors with displacement vectors: Ensure that when calculating lengths or directions between points, you are using displacement vectors (e.g., $\vec{AB} = \vec{B} - \vec{A}$) and not position vectors directly.

Summary

We were given the position vectors of the vertices of a triangle ABC. By calculating the lengths of sides AB and AC, we identified that the triangle is isosceles ($AB=AC$). This crucial observation allowed us to deduce that the angle bisector AD is also the median to BC, meaning D is the midpoint of BC. We then found the position vector of D and subsequently calculated the length $l$ of the angle bisector AD. Finally, we computed $2l^2$, which resulted in 45.

The final answer is $\boxed{\text{45}}$.