1. Key Concepts and Formulas

• Centroid of a Triangle: The centroid of a triangle is the point of intersection of its medians. A fundamental property is that the centroid divides each median in the ratio 2:1 from the vertex to the midpoint of the opposite side. If the vertices of a triangle are $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$, then the coordinates of its centroid $G$ are given by the average of the coordinates: $G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)$
• Angle between Two Vectors using Dot Product: If $\theta$ is the angle between two non-zero vectors $\vec{a}$ and $\vec{b}$, their dot product is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$. From this, the cosine of the angle can be found as: $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$
• Magnitude of a Vector: For a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$, its magnitude (or length) is: $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$
• Position Vector: The position vector of a point $P(x, y, z)$ from the origin $O(0, 0, 0)$ is $\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}$.

2. Step-by-Step Solution

Step 1: Identify point G as the centroid of $\triangle ABC$.

• Why this step is taken: The problem states that M is the midpoint of AC, making BM a median of $\triangle ABC$. G divides BM in the ratio 2:1 from B to M. This is the defining property of a centroid. Recognizing this allows us to use the centroid formula directly, simplifying the calculation of G's coordinates.
• Explanation: Since M is the midpoint of AC, BM is a median. G divides the median BM in the ratio 2:1 (from vertex B to midpoint M). By definition, the centroid of a triangle is the point that divides each median in a 2:1 ratio from the vertex. Therefore, G is the centroid of $\triangle ABC$.

Step 2: Calculate the coordinates of the centroid G.

• Why this step is taken: To find the vector $\vec{OG}$ and its magnitude, we first need the coordinates of G.
• Explanation: Using the centroid formula with the given vertices A(3, 0, –1), B(2, 10, 6), and C(1, 2, 1): $G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}, \frac{z_A + z_B + z_C}{3} \right)$ $G = \left( \frac{3 + 2 + 1}{3}, \frac{0 + 10 + 2}{3}, \frac{-1 + 6 + 1}{3} \right)$ $G = \left( \frac{6}{3}, \frac{12}{3}, \frac{6}{3} \right)$ $G = (2, 4, 2)$

Step 3: Determine the position vectors $\vec{OA}$ and $\vec{OG}$.

• Why this step is taken: The angle $\angle GOA$ is formed by the vectors originating from the origin O and pointing to points G and A. We need these vectors to apply the dot product formula.
• Explanation: The position vector of a point $P(x, y, z)$ from the origin $O(0,0,0)$ is given by $\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}$. For point A(3, 0, –1): $\vec{OA} = 3\hat{i} + 0\hat{j} - 1\hat{k} = 3\hat{i} - \hat{k}$ For point G(2, 4, 2): $\vec{OG} = 2\hat{i} + 4\hat{j} + 2\hat{k}$

Step 4: Calculate the dot product $\vec{OA} \cdot \vec{OG}$.

• Why this step is taken: The dot product is the numerator in the formula for the cosine of the angle between two vectors.
• Explanation: The dot product of two vectors $\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$ and $\vec{b} = b_x\hat{i} + b_y\hat{j} + b_z\hat{k}$ is $a_x b_x + a_y b_y + a_z b_z$. $\vec{OA} \cdot \vec{OG} = (3)(2) + (0)(4) + (-1)(2)$ $\vec{OA} \cdot \vec{OG} = 6 + 0 - 2$ $\vec{OA} \cdot \vec{OG} = 4$

Step 5: Calculate the magnitudes $|\vec{OA}|$ and $|\vec{OG}|$.

• Why this step is taken: The magnitudes of the vectors form the denominator in the formula for the cosine of the angle.
• Explanation: The magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $\sqrt{x^2 + y^2 + z^2}$. For $\vec{OA} = 3\hat{i} - \hat{k}$: $|\vec{OA}| = \sqrt{(3)^2 + (0)^2 + (-1)^2} = \sqrt{9 + 0 + 1} = \sqrt{10}$ For $\vec{OG} = 2\hat{i} + 4\hat{j} + 2\hat{k}$: $|\vec{OG}| = \sqrt{(2)^2 + (4)^2 + (2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24}$ We can simplify $\sqrt{24}$ as $\sqrt{4 \times 6} = 2\sqrt{6}$.

Step 6: Compute cos($\angle$GOA) using the dot product formula.

• Why this step is taken: This is the final calculation to determine the cosine of the angle between the vectors $\vec{OA}$ and $\vec{OG}$, which directly answers the question.
• Explanation: Using the formula $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$, where $\theta = \angle GOA$: $\cos(\angle GOA) = \frac{\vec{OA} \cdot \vec{OG}}{|\vec{OA}| |\vec{OG}|}$ $\cos(\angle GOA) = \frac{4}{\sqrt{10} \cdot 2\sqrt{6}}$ $\cos(\angle GOA) = \frac{4}{2\sqrt{10 \times 6}}$ $\cos(\angle GOA) = \frac{4}{2\sqrt{60}}$ Simplify the expression: $\cos(\angle GOA) = \frac{2}{\sqrt{60}}$ Now, simplify the radical $\sqrt{60}$: $\sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$. $\cos(\angle GOA) = \frac{2}{2\sqrt{15}}$ $\cos(\angle GOA) = \frac{1}{\sqrt{15}}$

3. Common Mistakes & Tips

• Centroid Identification: The most common shortcut is to immediately recognize G as the centroid. If you don't, you might first calculate M's coordinates and then use the section formula for G, which is more prone to errors.
• Vector Direction: Ensure the vectors used for the angle calculation ($\vec{OA}$ and $\vec{OG}$) both originate from the same point, in this case, the origin O, as implied by $\angle GOA$.
• Radical Simplification: Always simplify square roots (e.g., $\sqrt{24}$ to $2\sqrt{6}$, $\sqrt{60}$ to $2\sqrt{15}$) to match the format of the given options.

4. Summary

The problem involves finding the cosine of the angle between two position vectors, $\vec{OA}$ and $\vec{OG}$. The key insight is recognizing that point G, which divides the median BM in a 2:1 ratio, is the centroid of $\triangle ABC$. By calculating the coordinates of G using the centroid formula, we can then determine the position vectors $\vec{OA}$ and $\vec{OG}$. Applying the dot product formula for the angle between vectors, we compute the cosine of $\angle GOA$ by finding the dot product and magnitudes of these vectors.

The final answer is $\boxed{\frac{1}{\sqrt{15}}}$, which corresponds to option (A).