Key Concepts and Formulas

• Dot Product: For two vectors $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$, their dot product is $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$.
• Geometric Interpretation of Dot Product: $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta$, where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$.
• Unit Vector Property: A unit vector $\hat{u}$ has a magnitude of 1, i.e., $|\hat{u}| = 1$. Consequently, $|\hat{u}|^2 = x^2 + y^2 + z^2 = 1$.
• Squared Magnitude of a Vector: For a vector $\vec{A}$, $|\vec{A}|^2 = A_x^2 + A_y^2 + A_z^2$.
• Magnitude of Difference: $|\hat{u} - \vec{v}|^2 = (\hat{u} - \vec{v}) \cdot (\hat{u} - \vec{v})$.

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Step-by-Step Solution

Step 1: Define the given vectors and their properties. We are given a unit vector $\hat{u} = x \hat{i} + y \hat{j} + z \hat{k}$. Since it's a unit vector, $|\hat{u}| = 1$. We are also given three vectors: $\vec{a} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k}$ $\vec{b} = \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$ $\vec{c} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}$ Let's calculate their magnitudes: $|\vec{a}| = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + 0^2 + \left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$ $|\vec{b}| = \sqrt{0^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$ $|\vec{c}| = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + 0^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$ All three vectors $\vec{a}, \vec{b}, \vec{c}$ are also unit vectors.

Step 2: Use the dot product and given angles to form equations. The angle between $\hat{u}$ and a vector $\vec{v}$ is given by $\cos \theta = \frac{\hat{u} \cdot \vec{v}}{|\hat{u}| |\vec{v}|}$. Since $|\hat{u}| = 1$ and $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$, the formula simplifies to $\cos \theta = \hat{u} \cdot \vec{v}$.

• Angle between $\hat{u}$ and $\vec{a}$ is $\frac{\pi}{2}$: $\hat{u} \cdot \vec{a} = \cos\left(\frac{\pi}{2}\right) = 0$ $(x \hat{i} + y \hat{j} + z \hat{k}) \cdot \left(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k}\right) = 0$ $\frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0 \implies x + z = 0 \quad (\text{Equation 1})$

• Angle between $\hat{u}$ and $\vec{b}$ is $\frac{\pi}{3}$: $\hat{u} \cdot \vec{b} = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ $(x \hat{i} + y \hat{j} + z \hat{k}) \cdot \left(\frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}\right) = \frac{1}{2}$ $\frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2} \implies y + z = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \quad (\text{Equation 2})$

• Angle between $\hat{u}$ and $\vec{c}$ is $\frac{2\pi}{3}$: $\hat{u} \cdot \vec{c} = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$ $(x \hat{i} + y \hat{j} + z \hat{k}) \cdot \left(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}\right) = -\frac{1}{2}$ $\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2} \implies x + y = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}} \quad (\text{Equation 3})$

Step 3: Solve the system of linear equations for $x, y, z$. We have the system:

1. $x + z = 0$
2. $y + z = \frac{1}{\sqrt{2}}$
3. $x + y = -\frac{1}{\sqrt{2}}$

From Equation 1, $z = -x$. Substitute $z = -x$ into Equation 2: $y + (-x) = \frac{1}{\sqrt{2}} \implies y - x = \frac{1}{\sqrt{2}} \quad (\text{Equation 4})$

Now we have a system of two equations with $x$ and $y$: 3. $x + y = -\frac{1}{\sqrt{2}}$ 4. $-x + y = \frac{1}{\sqrt{2}}$

Adding Equation 3 and Equation 4: $(x + y) + (-x + y) = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}$ $2y = 0 \implies y = 0$

Substitute $y=0$ into Equation 3: $x + 0 = -\frac{1}{\sqrt{2}} \implies x = -\frac{1}{\sqrt{2}}$

Substitute $x = -\frac{1}{\sqrt{2}}$ into Equation 1: $-\frac{1}{\sqrt{2}} + z = 0 \implies z = \frac{1}{\sqrt{2}}$

Thus, the unit vector is $\hat{u} = -\frac{1}{\sqrt{2}} \hat{i} + 0 \hat{j} + \frac{1}{\sqrt{2}} \hat{k} = -\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k}$.

Step 4: Calculate $|\hat{u} - \vec{v}|^2$. We are given $\vec{v} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$. First, find the difference vector $\hat{u} - \vec{v}$: $\hat{u} - \vec{v} = \left(-\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k}\right) - \left(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}\right)$ $\hat{u} - \vec{v} = \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) \hat{i} + \left(0 - \frac{1}{\sqrt{2}}\right) \hat{j} + \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) \hat{k}$ $\hat{u} - \vec{v} = -\frac{2}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} + 0 \hat{k}$ $\hat{u} - \vec{v} = -\sqrt{2} \hat{i} - \frac{1}{\sqrt{2}} \hat{j}$ Now, calculate the squared magnitude: $|\hat{u} - \vec{v}|^2 = (-\sqrt{2})^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + (0)^2$ $|\hat{u} - \vec{v}|^2 = 2 + \frac{1}{2} + 0$ $|\hat{u} - \vec{v}|^2 = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}$

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Common Mistakes & Tips

• Algebraic Errors: Be meticulous when solving the system of linear equations. A small mistake can lead to an incorrect value for $x, y, z$.
• Angle Interpretation: Ensure you correctly identify the cosine values for the given angles, especially for angles greater than $\frac{\pi}{2}$. For instance, $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$.
• Unit Vector Check: Always verify that your derived $\hat{u}$ is indeed a unit vector by checking if $x^2 + y^2 + z^2 = 1$.

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Summary

The problem involves finding the components of a unit vector $\hat{u}$ given the angles it makes with three other vectors. By utilizing the dot product formula and the geometric interpretation of the angle between vectors, we set up a system of linear equations for the components of $\hat{u}$. Solving this system yields the specific components of $\hat{u}$. Finally, we compute the difference vector between $\hat{u}$ and the given vector $\vec{v}$ and then find its squared magnitude.

The final answer is $\boxed{\frac{5}{2}}$.