Key Concepts and Formulas

1. Cross Product: The cross product of two vectors $\vec{u}$ and $\vec{v}$ yields a vector perpendicular to both $\vec{u}$ and $\vec{v}$. Its magnitude is $|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin\theta$, and its direction is given by the right-hand rule.
2. Unit Vector: A unit vector $\hat{a}$ has a magnitude of 1 ($|\hat{a}|=1$). If a vector $\vec{A}$ is perpendicular to two other vectors, the unit vector in that direction is $\hat{a} = \pm \frac{\vec{A}}{|\vec{A}|}$.
3. Dot Product and Angle between Vectors: The dot product of two vectors $\vec{A}$ and $\vec{B}$ is given by $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta$, where $\theta$ is the angle between them. For unit vectors, $\hat{a} \cdot \hat{b} = \cos \theta$.

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Step-by-Step Solution

Step 1: Find a vector perpendicular to $\vec{b}$ and $\vec{c}$

We are given that $\hat{a}$ is a unit vector perpendicular to $\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{c} = 2\hat{i} + 3\hat{j} - \hat{k}$. A vector perpendicular to both $\vec{b}$ and $\vec{c}$ can be found using their cross product, $\vec{b} \times \vec{c}$.

Calculating the cross product: $\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix}$ $\vec{b} \times \vec{c} = \hat{i}((-2)(-1) - (3)(3)) - \hat{j}((1)(-1) - (3)(2)) + \hat{k}((1)(3) - (-2)(2))$ $\vec{b} \times \vec{c} = \hat{i}(2 - 9) - \hat{j}(-1 - 6) + \hat{k}(3 + 4)$ $\vec{b} \times \vec{c} = -7\hat{i} + 7\hat{j} + 7\hat{k}$ We can factor out 7 for simplicity: $\vec{b} \times \vec{c} = 7(-\hat{i} + \hat{j} + \hat{k})$

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Step 2: Determine the possible unit vectors for $\hat{a}$

Since $\hat{a}$ is a unit vector perpendicular to $\vec{b}$ and $\vec{c}$, it must be parallel to $\vec{b} \times \vec{c}$. There are two possible directions for $\hat{a}$, given by the normalized cross product and its negative.

First, we find the magnitude of $\vec{b} \times \vec{c}$: $|\vec{b} \times \vec{c}| = |7(-\hat{i} + \hat{j} + \hat{k})| = 7 \sqrt{(-1)^2 + 1^2 + 1^2} = 7 \sqrt{1 + 1 + 1} = 7\sqrt{3}$

The two possible unit vectors for $\hat{a}$ are: $\hat{a}_1 = \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|} = \frac{7(-\hat{i} + \hat{j} + \hat{k})}{7\sqrt{3}} = \frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$ $\hat{a}_2 = -\frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|} = -\left(\frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}\right) = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$

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Step 3: Use the first angle condition to select the correct $\hat{a}$

We are given that $\hat{a}$ makes an angle of $\cos^{-1}\left(-\frac{1}{3}\right)$ with the vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$. This means $\cos \theta_1 = -\frac{1}{3}$, where $\theta_1$ is the angle between $\hat{a}$ and $\vec{v}$. Using the dot product formula for unit vectors: $\cos \theta_1 = \hat{a} \cdot \vec{v}$. The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.

Let's test $\hat{a}_1$: $\hat{a}_1 \cdot \vec{v} = \left(\frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}\right) \cdot (\hat{i} + \hat{j} + \hat{k}) = \frac{(-1)(1) + (1)(1) + (1)(1)}{\sqrt{3}} = \frac{-1 + 1 + 1}{\sqrt{3}} = \frac{1}{\sqrt{3}}$ So, $\cos \theta_1 = \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} = \frac{1}{3}$. This does not match $-\frac{1}{3}$.

Let's test $\hat{a}_2$: $\hat{a}_2 \cdot \vec{v} = \left(\frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}\right) \cdot (\hat{i} + \hat{j} + \hat{k}) = \frac{(1)(1) + (-1)(1) + (-1)(1)}{\sqrt{3}} = \frac{1 - 1 - 1}{\sqrt{3}} = \frac{-1}{\sqrt{3}}$ So, $\cos \theta_1 = \frac{-1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} = -\frac{1}{3}$. This matches the given condition. Therefore, the correct unit vector is $\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$.

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Step 4: Use the second angle condition to find $\alpha$

We are given that $\hat{a}$ makes an angle of $\frac{\pi}{3}$ with the vector $\vec{w} = \hat{i} + \alpha\hat{j} + \hat{k}$. This means $\cos \frac{\pi}{3} = \frac{1}{2}$. Using the dot product formula: $\cos \frac{\pi}{3} = \hat{a} \cdot \vec{w}$, since $|\hat{a}|=1$. The magnitude of $\vec{w}$ is $|\vec{w}| = \sqrt{1^2 + \alpha^2 + 1^2} = \sqrt{\alpha^2 + 2}$.

Calculate the dot product $\hat{a} \cdot \vec{w}$: $\hat{a} \cdot \vec{w} = \left(\frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}\right) \cdot (\hat{i} + \alpha\hat{j} + \hat{k}) = \frac{(1)(1) + (-1)(\alpha) + (-1)(1)}{\sqrt{3}} = \frac{1 - \alpha - 1}{\sqrt{3}} = \frac{-\alpha}{\sqrt{3}}$

Now, substitute these into the dot product formula: $\cos \frac{\pi}{3} = \frac{\hat{a} \cdot \vec{w}}{|\vec{w}|}$ $\frac{1}{2} = \frac{\frac{-\alpha}{\sqrt{3}}}{\sqrt{\alpha^2 + 2}}$ $\frac{1}{2} = \frac{-\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}}$

To solve for $\alpha$, we first observe that the left side is positive ($1/2$), and the denominator on the right side ($\sqrt{3}\sqrt{\alpha^2+2}$) is also positive. Therefore, the numerator, $-\alpha$, must be positive, which implies $\alpha < 0$.

Rearrange the equation: $\sqrt{3}\sqrt{\alpha^2 + 2} = -2\alpha$

Square both sides: $(\sqrt{3}\sqrt{\alpha^2 + 2})^2 = (-2\alpha)^2$ $3(\alpha^2 + 2) = 4\alpha^2$ $3\alpha^2 + 6 = 4\alpha^2$ $6 = 4\alpha^2 - 3\alpha^2$ $\alpha^2 = 6$ $\alpha = \pm\sqrt{6}$

Considering our earlier observation that $\alpha < 0$, we choose the negative value. Thus, $\alpha = -\sqrt{6}$.

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Common Mistakes & Tips

• Sign Ambiguity in Cross Product: Always remember that there are two unit vectors perpendicular to a plane defined by two vectors (one in each direction). Failing to consider both possibilities can lead to incorrect solutions.
• Squaring Equations: When squaring an equation involving square roots, be mindful of the signs of the expressions before squaring. An extraneous solution might be introduced if the signs are not handled correctly. In this case, $\alpha$ had to be negative for the equality to hold before squaring.
• Dot Product Formula: Ensure correct application of the dot product formula, especially when dealing with unit vectors and known angles.

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Summary

We began by finding a vector perpendicular to the given vectors $\vec{b}$ and $\vec{c}$ using the cross product. This vector was then normalized to find the two possible unit vectors for $\hat{a}$. The first angle condition, relating $\hat{a}$ to $\hat{i} + \hat{j} + \hat{k}$, allowed us to uniquely identify the correct $\hat{a}$. Finally, the second angle condition, relating $\hat{a}$ to $\hat{i} + \alpha\hat{j} + \hat{k}$, was used to set up an equation for $\alpha$. By carefully solving this equation, considering the sign constraint on $\alpha$ derived from the angle condition, we found the value of $\alpha$.

The final answer is $\boxed{-\sqrt{6}}$.