Key Concepts and Formulas

• Scalar Triple Product (STP): For three vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$, the scalar triple product is given by $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$. Geometrically, its absolute value represents the volume of the parallelepiped formed by these vectors.
• STP in terms of Magnitudes and Angle: $[\vec{u} \vec{v} \vec{w}] = |\vec{u}| |\vec{v} \times \vec{w}| \cos \theta$, where $\theta$ is the angle between $\vec{u}$ and $(\vec{v} \times \vec{w})$.
• Minimum/Maximum of STP: The minimum value of the STP occurs when $\cos \theta = -1$ (vectors are antiparallel), and the maximum value occurs when $\cos \theta = 1$ (vectors are parallel).
• Vector Dot Product: $\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z$. Specifically, $\hat{i} \cdot \hat{i} = 1$, $\hat{i} \cdot \hat{j} = 0$, etc.
• Magnitude of a Vector: For a vector $\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$, its magnitude is $|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}$.

Step-by-Step Solution

Step 1: Calculate the Cross Product $\vec{v} \times \vec{w}$ To evaluate the scalar triple product, we first need to compute the cross product of $\vec{v}$ and $\vec{w}$. This vector is perpendicular to both $\vec{v}$ and $\vec{w}$.

Given: $\vec{v} = \alpha \hat{i} + 2 \hat{j} - 3 \hat{k}$ $\vec{w} = 2\alpha \hat{i} + \hat{j} - \hat{k}$

The cross product is calculated using the determinant: $\vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -3 \\ 2\alpha & 1 & -1 \end{vmatrix}$ Expanding the determinant: $\vec{v} \times \vec{w} = \hat{i} ((2)(-1) - (-3)(1)) - \hat{j} ((\alpha)(-1) - (-3)(2\alpha)) + \hat{k} ((\alpha)(1) - (2)(2\alpha))$ $\vec{v} \times \vec{w} = \hat{i} (-2 + 3) - \hat{j} (-\alpha + 6\alpha) + \hat{k} (\alpha - 4\alpha)$ $\vec{v} \times \vec{w} = 1\hat{i} - 5\alpha\hat{j} - 3\alpha\hat{k}$ So, $\vec{v} \times \vec{w} = \hat{i} - 5\alpha\hat{j} - 3\alpha\hat{k}$.

Step 2: Calculate the Magnitude of $\vec{v} \times \vec{w}$ We need the magnitude of the cross product vector for the STP formula. $|\vec{v} \times \vec{w}| = \sqrt{(1)^2 + (-5\alpha)^2 + (-3\alpha)^2}$ $|\vec{v} \times \vec{w}| = \sqrt{1 + 25\alpha^2 + 9\alpha^2}$ $|\vec{v} \times \vec{w}| = \sqrt{1 + 34\alpha^2}$

Step 3: Express the Scalar Triple Product and Use the Minimum Value Condition The scalar triple product is given by $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$. Using the formula with magnitudes and angle: $[\vec{u} \vec{v} \vec{w}] = |\vec{u}| |\vec{v} \times \vec{w}| \cos \theta$ We are given $|\vec{u}| = \alpha$ and we found $|\vec{v} \times \vec{w}| = \sqrt{1 + 34\alpha^2}$. So, the STP is: $[\vec{u} \vec{v} \vec{w}] = \alpha \sqrt{1 + 34\alpha^2} \cos \theta$ We are given that the minimum value of the STP is $-\alpha \sqrt{3401}$. Since $\alpha > 0$ and $\sqrt{1 + 34\alpha^2}$ is always positive, the minimum value of the expression is achieved when $\cos \theta$ is at its minimum, which is $-1$. Setting $\cos \theta = -1$: $[\vec{u} \vec{v} \vec{w}]_{\min} = \alpha \sqrt{1 + 34\alpha^2} (-1) = -\alpha \sqrt{1 + 34\alpha^2}$ Equating this to the given minimum value: $-\alpha \sqrt{1 + 34\alpha^2} = -\alpha \sqrt{3401}$ Since $\alpha > 0$, we can divide both sides by $-\alpha$: $\sqrt{1 + 34\alpha^2} = \sqrt{3401}$ Squaring both sides: $1 + 34\alpha^2 = 3401$ $34\alpha^2 = 3400$ $\alpha^2 = \frac{3400}{34} = 100$ Since $\alpha > 0$, we have $\alpha = 10$.

Step 4: Determine the Vector $\vec{u}$ The minimum value of the STP occurs when $\cos \theta = -1$, which means $\vec{u}$ is in the opposite direction to $\vec{v} \times \vec{w}$. So, $\vec{u}$ is a unit vector in the direction of $-(\vec{v} \times \vec{w})$, scaled by $|\vec{u}| = \alpha$. $\vec{u} = |\vec{u}| \left( - \frac{\vec{v} \times \vec{w}}{|\vec{v} \times \vec{w}|} \right)$ We found $\alpha = 10$, $|\vec{u}| = 10$, and $|\vec{v} \times \vec{w}| = \sqrt{3401}$. From Step 1, $\vec{v} \times \vec{w} = \hat{i} - 5\alpha\hat{j} - 3\alpha\hat{k}$. Substituting $\alpha = 10$: $\vec{v} \times \vec{w} = \hat{i} - 5(10)\hat{j} - 3(10)\hat{k} = \hat{i} - 50\hat{j} - 30\hat{k}$. Now, substituting these into the expression for $\vec{u}$: $\vec{u} = 10 \left( - \frac{\hat{i} - 50\hat{j} - 30\hat{k}}{\sqrt{3401}} \right)$ $\vec{u} = \frac{-10}{\sqrt{3401}} (\hat{i} - 50\hat{j} - 30\hat{k})$

Step 5: Calculate $|\vec{u} \cdot \hat{i}|^2$ and Find $m+n$ We need to compute $|\vec{u} \cdot \hat{i}|^2$. First, find the dot product $\vec{u} \cdot \hat{i}$: $\vec{u} \cdot \hat{i} = \left( \frac{-10}{\sqrt{3401}} (\hat{i} - 50\hat{j} - 30\hat{k}) \right) \cdot \hat{i}$ $\vec{u} \cdot \hat{i} = \frac{-10}{\sqrt{3401}} ((\hat{i} \cdot \hat{i}) - 50(\hat{j} \cdot \hat{i}) - 30(\hat{k} \cdot \hat{i}))$ Since $\hat{i} \cdot \hat{i} = 1$ and $\hat{j} \cdot \hat{i} = \hat{k} \cdot \hat{i} = 0$: $\vec{u} \cdot \hat{i} = \frac{-10}{\sqrt{3401}} (1 - 0 - 0) = \frac{-10}{\sqrt{3401}}$ Now, we square this value: $|\vec{u} \cdot \hat{i}|^2 = \left| \frac{-10}{\sqrt{3401}} \right|^2 = \frac{(-10)^2}{(\sqrt{3401})^2} = \frac{100}{3401}$ We are given that $|\vec{u} \cdot \hat{i}|^2 = \frac{m}{n}$, where $m$ and $n$ are coprime natural numbers. So, $m = 100$ and $n = 3401$. We need to confirm that $m$ and $n$ are coprime. $100 = 2^2 \times 5^2$. The prime factors of 100 are 2 and 5. For $n = 3401$, it is not divisible by 2 (it's odd) or 5 (doesn't end in 0 or 5). Thus, 100 and 3401 share no common prime factors and are coprime. Finally, we calculate $m+n$: $m+n = 100 + 3401 = 3501$

Common Mistakes & Tips

• Sign Errors in Cross Product: Carefully compute the determinant for the cross product, paying attention to the signs of the unit vectors and the terms within the determinant.
• Minimum Value of STP: Remember that the minimum value of $\vec{u} \cdot \vec{x}$ is $-|\vec{u}||\vec{x}|$, which occurs when $\vec{u}$ is antiparallel to $\vec{x}$. This is key for problems involving minimum or maximum values of the scalar triple product.
• Coprimality Check: Ensure that the numerator and denominator of the fraction $\frac{m}{n}$ have no common factors before calculating $m+n$.

Summary

The problem involves finding the minimum value of a scalar triple product and then using that information to determine the value of $\alpha$. We first calculated the cross product $\vec{v} \times \vec{w}$ and its magnitude. By relating the minimum value of the scalar triple product to the condition $\cos \theta = -1$, we solved for $\alpha$. Subsequently, we determined the vector $\vec{u}$ based on the direction dictated by the minimum STP. Finally, we computed $|\vec{u} \cdot \hat{i}|^2$, identified $m$ and $n$, verified their coprimality, and calculated their sum.

The final answer is $\boxed{3501}$.