Key Concepts and Formulas

1. Angle Bisector Theorem & Section Formula: In $\triangle ABC$, if $AD$ is the angle bisector of $\angle BAC$ meeting $BC$ at $D$, then $D$ divides the side $BC$ in the ratio $AB:AC$. If a point $D$ divides the line segment joining $B(x_B, y_B, z_B)$ and $C(x_C, y_C, z_C)$ in the ratio $m:n$, its coordinates are $D = \left(\frac{nx_B+mx_C}{m+n}, \frac{ny_B+my_C}{m+n}, \frac{nz_B+mz_C}{m+n}\right)$.
2. Vector Operations and Magnitude: For points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$, the vector $\overrightarrow{PQ}$ is given by $(x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$. The magnitude of a vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ is $|\vec{v}| = \sqrt{a^2+b^2+c^2}$.
3. Scalar Projection: The length of the scalar projection of a vector $\vec{u}$ onto another vector $\vec{v}$ is given by the formula $\text{Proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$. The dot product of $\vec{u} = u_x\hat{i} + u_y\hat{j} + u_z\hat{k}$ and $\vec{v} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}$ is $\vec{u} \cdot \vec{v} = u_xv_x + u_yv_y + u_zv_z$.

Step-by-Step Solution

We are given the vertices of $\triangle ABC$ as $A(1,3,2)$, $B(-2,8,0)$, and $C(3,6,7)$. We need to find the length of the projection of $\overrightarrow{AD}$ on $\overrightarrow{AC}$.

Step 1: Calculate the lengths of sides AB and AC to determine the ratio for point D. According to the Angle Bisector Theorem, point $D$ divides the side $BC$ in the ratio $BD:DC = AB:AC$. Therefore, we first need to find the magnitudes of vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

• Calculate $\overrightarrow{AB}$ and its magnitude $AB$: $\overrightarrow{AB} = B - A = (-2-1)\hat{i} + (8-3)\hat{j} + (0-2)\hat{k} = -3\hat{i} + 5\hat{j} - 2\hat{k}$ The length $AB$ is the magnitude of $\overrightarrow{AB}$: $AB = |\overrightarrow{AB}| = \sqrt{(-3)^2 + (5)^2 + (-2)^2} = \sqrt{9 + 25 + 4} = \sqrt{38}$

• Calculate $\overrightarrow{AC}$ and its magnitude $AC$: $\overrightarrow{AC} = C - A = (3-1)\hat{i} + (6-3)\hat{j} + (7-2)\hat{k} = 2\hat{i} + 3\hat{j} + 5\hat{k}$ The length $AC$ is the magnitude of $\overrightarrow{AC}$: $AC = |\overrightarrow{AC}| = \sqrt{(2)^2 + (3)^2 + (5)^2} = \sqrt{4 + 9 + 25} = \sqrt{38}$

• Determine the ratio $BD:DC$: Since $AB = \sqrt{38}$ and $AC = \sqrt{38}$, we have $AB:AC = \sqrt{38}:\sqrt{38} = 1:1$. This implies that $D$ is the midpoint of the line segment $BC$.

Step 2: Determine the coordinates of point D. Since $D$ is the midpoint of $BC$, we can use the midpoint formula. Given $B(-2,8,0)$ and $C(3,6,7)$: $D = \left(\frac{x_B+x_C}{2}, \frac{y_B+y_C}{2}, \frac{z_B+z_C}{2}\right)$ $D = \left(\frac{-2+3}{2}, \frac{8+6}{2}, \frac{0+7}{2}\right) = \left(\frac{1}{2}, \frac{14}{2}, \frac{7}{2}\right) = \left(\frac{1}{2}, 7, \frac{7}{2}\right)$ So, the coordinates of point $D$ are $(\frac{1}{2}, 7, \frac{7}{2})$.

Step 3: Formulate the vectors $\overrightarrow{AD}$ and $\overrightarrow{AC}$. We need these vectors for the projection formula. We already have $\overrightarrow{AC}$ from Step 1.

• Calculate $\overrightarrow{AD}$: Using $A(1,3,2)$ and $D(\frac{1}{2}, 7, \frac{7}{2})$: $\overrightarrow{AD} = D - A = \left(\frac{1}{2}-1\right)\hat{i} + (7-3)\hat{j} + \left(\frac{7}{2}-2\right)\hat{k}$ $\overrightarrow{AD} = \left(-\frac{1}{2}\right)\hat{i} + 4\hat{j} + \left(\frac{7-4}{2}\right)\hat{k} = -\frac{1}{2}\hat{i} + 4\hat{j} + \frac{3}{2}\hat{k}$

• Recall $\overrightarrow{AC}$: From Step 1, we have: $\overrightarrow{AC} = 2\hat{i} + 3\hat{j} + 5\hat{k}$

Step 4: Calculate the length of the projection of $\overrightarrow{AD}$ on $\overrightarrow{AC}$. Using the scalar projection formula $\text{Proj}_{\overrightarrow{AC}} \overrightarrow{AD} = \frac{\overrightarrow{AD} \cdot \overrightarrow{AC}}{|\overrightarrow{AC}|}$.

• Calculate the dot product $\overrightarrow{AD} \cdot \overrightarrow{AC}$: $\overrightarrow{AD} \cdot \overrightarrow{AC} = \left(-\frac{1}{2}\right)(2) + (4)(3) + \left(\frac{3}{2}\right)(5)$ $= -1 + 12 + \frac{15}{2}$ $= 11 + \frac{15}{2} = \frac{22}{2} + \frac{15}{2} = \frac{37}{2}$

• Recall the magnitude $|\overrightarrow{AC}|$: From Step 1, we found $|\overrightarrow{AC}| = \sqrt{38}$.

• Apply the projection formula: $\text{Length of Projection} = \frac{\frac{37}{2}}{\sqrt{38}} = \frac{37}{2\sqrt{38}}$

Common Mistakes & Tips

• Arithmetic Precision: Be extremely careful with fraction arithmetic and sign conventions, especially when calculating vector components and dot products.
• Angle Bisector Theorem: Ensure you correctly apply the theorem. A common error is to mix up the ratio $AB:AC$ with $AC:AB$. Recognizing $AB=AC$ is a significant simplification.
• Projection Formula: Memorize the scalar projection formula $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$ correctly. It's the magnitude of the projection, hence always non-negative.

Summary

This problem required a systematic application of 3D geometry and vector algebra principles. We began by using the Angle Bisector Theorem to determine the ratio in which point $D$ divides $BC$. The key observation that $AB=AC$ simplified $D$ to being the midpoint of $BC$. We then calculated the coordinates of $D$, formed the vectors $\overrightarrow{AD}$ and $\overrightarrow{AC}$, and finally used the dot product and magnitude to compute the length of the scalar projection of $\overrightarrow{AD}$ onto $\overrightarrow{AC}$. The calculated projection length is $\frac{37}{2\sqrt{38}}$.

The final answer is $\boxed{\text{(A)}}$.