Key Concepts and Formulas

1. Collinearity of Vectors: Three points A, B, C are collinear if the vectors formed by them, such as $\overrightarrow{AB}$ and $\overrightarrow{AC}$, are parallel. Two vectors are parallel if one is a scalar multiple of the other, or equivalently, if their corresponding components are proportional.
2. Position Vectors and Vector Subtraction: The vector from point A to point B is given by $\overrightarrow{AB} = \overrightarrow{b} - \overrightarrow{a}$, where $\overrightarrow{a}$ and $\overrightarrow{b}$ are the position vectors of A and B, respectively.
3. Midpoint Formula: The position vector of the midpoint M of a line segment BC is given by $\overrightarrow{m} = \frac{\overrightarrow{b} + \overrightarrow{c}}{2}$.
4. Distance Formula: The distance between two points with position vectors $\overrightarrow{p}$ and $\overrightarrow{q}$ is the magnitude of the vector $\overrightarrow{pq} = \overrightarrow{q} - \overrightarrow{p}$, i.e., $|\overrightarrow{q} - \overrightarrow{p}|$. If $\overrightarrow{p} = x_1\widehat i + y_1\widehat j + z_1\widehat k$ and $\overrightarrow{q} = x_2\widehat i + y_2\widehat j + z_2\widehat k$, the distance is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

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Step-by-Step Solution

Step 1: Find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ and establish the condition for collinearity. We are given the position vectors: $\overrightarrow a = \widehat i + 4\widehat j + 3\widehat k$ $\overrightarrow b = 2\widehat i + \alpha \widehat j + 4\widehat k$ $\overrightarrow c = 3\widehat i - 2\widehat j + 5\widehat k$

The vector $\overrightarrow{AB}$ is given by: $\overrightarrow{AB} = \overrightarrow b - \overrightarrow a = (2\widehat i + \alpha \widehat j + 4\widehat k) - (\widehat i + 4\widehat j + 3\widehat k)$ $\overrightarrow{AB} = (2-1)\widehat i + (\alpha-4)\widehat j + (4-3)\widehat k = \widehat i + (\alpha-4)\widehat j + \widehat k$

The vector $\overrightarrow{AC}$ is given by: $\overrightarrow{AC} = \overrightarrow c - \overrightarrow a = (3\widehat i - 2\widehat j + 5\widehat k) - (\widehat i + 4\widehat j + 3\widehat k)$ $\overrightarrow{AC} = (3-1)\widehat i + (-2-4)\widehat j + (5-3)\widehat k = 2\widehat i - 6\widehat j + 2\widehat k$

For points A, B, and C to be collinear, the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ must be parallel. This means their corresponding components must be proportional: $\frac{1}{2} = \frac{\alpha-4}{-6} = \frac{1}{2}$

Step 2: Solve for $\alpha$ that makes the points collinear. From the proportionality condition, we can equate the first two ratios: $\frac{1}{2} = \frac{\alpha-4}{-6}$ Multiplying both sides by $-6$: $-3 = \alpha-4$ Adding 4 to both sides: $\alpha = 1$ Thus, if $\alpha = 1$, the points A, B, and C are collinear.

Step 3: Determine the smallest positive integer value of $\alpha$ for which the points are non-collinear. The problem requires $\alpha$ to be the smallest positive integer for which A, B, and C are non-collinear. Positive integers are $1, 2, 3, \dots$. Since $\alpha = 1$ leads to collinearity, the smallest positive integer value for $\alpha$ that ensures non-collinearity is $\alpha = 2$.

Step 4: Find the position vector of the midpoint M of the side BC. With $\alpha = 2$, the position vector $\overrightarrow b$ becomes: $\overrightarrow b = 2\widehat i + 2\widehat j + 4\widehat k$ The position vector of point C is: $\overrightarrow c = 3\widehat i - 2\widehat j + 5\widehat k$ Let M be the midpoint of BC. Its position vector $\overrightarrow m$ is given by the midpoint formula: $\overrightarrow m = \frac{\overrightarrow b + \overrightarrow c}{2} = \frac{(2\widehat i + 2\widehat j + 4\widehat k) + (3\widehat i - 2\widehat j + 5\widehat k)}{2}$ $\overrightarrow m = \frac{(2+3)\widehat i + (2-2)\widehat j + (4+5)\widehat k}{2} = \frac{5\widehat i + 0\widehat j + 9\widehat k}{2}$ $\overrightarrow m = \frac{5}{2}\widehat i + 0\widehat j + \frac{9}{2}\widehat k$

Step 5: Calculate the length of the median AM. The median through A is the line segment AM, where M is the midpoint of BC. We need to find the length of AM. The position vector of A is $\overrightarrow a = \widehat i + 4\widehat j + 3\widehat k$. The position vector of M is $\overrightarrow m = \frac{5}{2}\widehat i + \frac{9}{2}\widehat k$.

The vector $\overrightarrow{AM}$ is: $\overrightarrow{AM} = \overrightarrow m - \overrightarrow a = \left(\frac{5}{2}\widehat i + \frac{9}{2}\widehat k\right) - (\widehat i + 4\widehat j + 3\widehat k)$ $\overrightarrow{AM} = \left(\frac{5}{2}-1\right)\widehat i + (0-4)\widehat j + \left(\frac{9}{2}-3\right)\widehat k$ $\overrightarrow{AM} = \left(\frac{5-2}{2}\right)\widehat i - 4\widehat j + \left(\frac{9-6}{2}\right)\widehat k$ $\overrightarrow{AM} = \frac{3}{2}\widehat i - 4\widehat j + \frac{3}{2}\widehat k$

The length of the median AM is the magnitude of the vector $\overrightarrow{AM}$: $AM = |\overrightarrow{AM}| = \sqrt{\left(\frac{3}{2}\right)^2 + (-4)^2 + \left(\frac{3}{2}\right)^2}$ $AM = \sqrt{\frac{9}{4} + 16 + \frac{9}{4}}$ $AM = \sqrt{\frac{18}{4} + 16} = \sqrt{\frac{9}{2} + 16}$ To add the terms, find a common denominator: $AM = \sqrt{\frac{9}{2} + \frac{32}{2}} = \sqrt{\frac{41}{2}}$ To express this in the desired format, we can write it as: $AM = \sqrt{\frac{41 \times 2}{2 \times 2}} = \sqrt{\frac{82}{4}} = \frac{\sqrt{82}}{\sqrt{4}} = \frac{\sqrt{82}}{2}$

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Common Mistakes & Tips

• Collinearity vs. Coplanarity: Ensure you are applying the correct condition. Collinearity of three points means the vectors formed between them are parallel.
• Smallest Positive Integer: Carefully read the question to identify the exact condition for $\alpha$. "Smallest positive integer" implies $\alpha \in \{1, 2, 3, \dots\}$, and we must exclude values that cause collinearity.
• Arithmetic with Fractions: Pay close attention to calculations involving fractions, especially when squaring or adding them, to avoid errors in the final distance calculation.

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Summary

The problem first requires determining the value of $\alpha$ that leads to collinearity by checking the proportionality of the components of vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. We found that $\alpha=1$ causes collinearity. The question asks for the smallest positive integer $\alpha$ for non-collinearity, which is $\alpha=2$. With this value of $\alpha$, we then found the midpoint M of BC and calculated the length of the median AM using the distance formula.

The final answer is $\boxed{\frac{\sqrt {82} }{2}}$.