Key Concepts and Formulas

• Dot Product of Vectors: For two vectors $\vec{U}$ and $\vec{V}$, the dot product is defined as $\vec{U} \cdot \vec{V} = |\vec{U}| |\vec{V}| \cos \theta$, where $\theta$ is the angle between them. This can be rearranged to find the cosine of the angle: $\cos \theta = \frac{\vec{U} \cdot \vec{V}}{|\vec{U}||\vec{V}|}$.
• Properties of Unit Vectors: A unit vector has a magnitude of 1. If $\hat{a}$ and $\hat{b}$ are unit vectors, then $|\hat{a}| = 1$ and $|\hat{b}| = 1$.
• Cross Product of Unit Vectors: For unit vectors $\hat{a}$ and $\hat{b}$ with an angle $\alpha$ between them, the magnitude of their cross product is $|\hat{a} \times \hat{b}| = |\hat{a}| |\hat{b}| \sin \alpha = \sin \alpha$. The direction of $\hat{a} \times \hat{b}$ is perpendicular to both $\hat{a}$ and $\hat{b}$.
• Scalar Triple Product Property: $(\vec{A} \times \vec{B}) \cdot \vec{C} = \vec{A} \cdot (\vec{B} \times \vec{C})$. Also, if $\vec{C}$ is perpendicular to both $\vec{A}$ and $\vec{B}$, then $\vec{A} \cdot \vec{C} = 0$ and $\vec{B} \cdot \vec{C} = 0$.

Step-by-Step Solution

Step 1: Define the vectors and given information. We are given two unit vectors $\hat{a}$ and $\hat{b}$ such that the angle between them is $\frac{\pi}{4}$. This means $|\hat{a}| = 1$, $|\hat{b}| = 1$, and the angle between $\hat{a}$ and $\hat{b}$ is $\alpha = \frac{\pi}{4}$.

Step 2: Identify the two vectors for which we need to find the angle $\theta$. Let $\vec{U} = \hat{a} + \hat{b}$ and $\vec{V} = \hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})$. We need to find the angle $\theta$ between $\vec{U}$ and $\vec{V}$.

Step 3: Calculate the dot product $\vec{U} \cdot \vec{V}$. We expand the dot product: $\vec{U} \cdot \vec{V} = (\hat{a} + \hat{b}) \cdot (\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b}))$ $\vec{U} \cdot \vec{V} = \hat{a} \cdot (\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})) + \hat{b} \cdot (\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b}))$ $\vec{U} \cdot \vec{V} = (\hat{a} \cdot \hat{a}) + 2(\hat{a} \cdot \hat{b}) + 2(\hat{a} \cdot (\hat{a} \times \hat{b})) + (\hat{b} \cdot \hat{a}) + 2(\hat{b} \cdot \hat{b}) + 2(\hat{b} \cdot (\hat{a} \times \hat{b}))$ We know that $\hat{a} \cdot \hat{a} = |\hat{a}|^2 = 1^2 = 1$, $\hat{b} \cdot \hat{b} = |\hat{b}|^2 = 1^2 = 1$, and $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{a} = |\hat{a}||\hat{b}|\cos(\frac{\pi}{4}) = 1 \cdot 1 \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$. The term $\hat{a} \cdot (\hat{a} \times \hat{b})$ represents the scalar triple product of $\hat{a}$, $\hat{a}$, and $\hat{b}$. Since two of the vectors are identical, the scalar triple product is zero. This is because $\hat{a} \times \hat{b}$ is perpendicular to $\hat{a}$, so their dot product is 0. Similarly, $\hat{b} \cdot (\hat{a} \times \hat{b})$ is also zero because $\hat{a} \times \hat{b}$ is perpendicular to $\hat{b}$. Substituting these values: $\vec{U} \cdot \vec{V} = 1 + 2\left(\frac{\sqrt{2}}{2}\right) + 2(0) + \frac{\sqrt{2}}{2} + 2(1) + 2(0)$ $\vec{U} \cdot \vec{V} = 1 + \sqrt{2} + \frac{\sqrt{2}}{2} + 2 = 3 + \frac{3\sqrt{2}}{2}$

Step 4: Calculate the magnitude of vector $\vec{U}$. $|\vec{U}| = |\hat{a} + \hat{b}|$ We use the property $|\vec{X}|^2 = \vec{X} \cdot \vec{X}$: $|\vec{U}|^2 = (\hat{a} + \hat{b}) \cdot (\hat{a} + \hat{b}) = \hat{a} \cdot \hat{a} + 2(\hat{a} \cdot \hat{b}) + \hat{b} \cdot \hat{b}$ $|\vec{U}|^2 = 1 + 2\left(\frac{\sqrt{2}}{2}\right) + 1 = 2 + \sqrt{2}$ So, $|\vec{U}| = \sqrt{2 + \sqrt{2}}$

Step 5: Calculate the magnitude of vector $\vec{V}$. $|\vec{V}|^2 = |\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})|^2$ Let $\vec{W} = \hat{a} \times \hat{b}$. Then $\vec{V} = \hat{a} + 2\hat{b} + 2\vec{W}$. We know that $\vec{W}$ is perpendicular to both $\hat{a}$ and $\hat{b}$. $|\vec{V}|^2 = (\hat{a} + 2\hat{b} + 2\vec{W}) \cdot (\hat{a} + 2\hat{b} + 2\vec{W})$ $|\vec{V}|^2 = \hat{a} \cdot \hat{a} + \hat{a} \cdot 2\hat{b} + \hat{a} \cdot 2\vec{W} + 2\hat{b} \cdot \hat{a} + 2\hat{b} \cdot 2\hat{b} + 2\hat{b} \cdot 2\vec{W} + 2\vec{W} \cdot \hat{a} + 2\vec{W} \cdot 2\hat{b} + 2\vec{W} \cdot 2\vec{W}$ $|\vec{V}|^2 = |\hat{a}|^2 + 2(\hat{a} \cdot \hat{b}) + 2(\hat{a} \cdot \vec{W}) + 2(\hat{b} \cdot \hat{a}) + 4|\hat{b}|^2 + 4(\hat{b} \cdot \vec{W}) + 2(\vec{W} \cdot \hat{a}) + 4(\vec{W} \cdot \hat{b}) + 4|\vec{W}|^2$ Since $\vec{W} = \hat{a} \times \hat{b}$, we have: $\hat{a} \cdot \vec{W} = 0$ $\hat{b} \cdot \vec{W} = 0$ $\vec{W} \cdot \hat{a} = 0$ $\vec{W} \cdot \hat{b} = 0$ Also, $|\vec{W}| = |\hat{a} \times \hat{b}| = |\hat{a}||\hat{b}|\sin(\frac{\pi}{4}) = 1 \cdot 1 \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$. So, $|\vec{W}|^2 = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2}$ Substituting the known values: $|\vec{V}|^2 = 1 + 2\left(\frac{\sqrt{2}}{2}\right) + 0 + 2\left(\frac{\sqrt{2}}{2}\right) + 4(1) + 0 + 0 + 0 + 4\left(\frac{1}{2}\right)$ $|\vec{V}|^2 = 1 + \sqrt{2} + \sqrt{2} + 4 + 2 = 7 + 2\sqrt{2}$ So, $|\vec{V}| = \sqrt{7 + 2\sqrt{2}}$

Step 6: Calculate $\cos \theta$. $\cos \theta = \frac{\vec{U} \cdot \vec{V}}{|\vec{U}||\vec{V}|} = \frac{3 + \frac{3\sqrt{2}}{2}}{\sqrt{2 + \sqrt{2}} \sqrt{7 + 2\sqrt{2}}}$ $\cos \theta = \frac{\frac{6 + 3\sqrt{2}}{2}}{\sqrt{(2 + \sqrt{2})(7 + 2\sqrt{2})}} = \frac{3(2 + \sqrt{2})}{2\sqrt{14 + 4\sqrt{2} + 7\sqrt{2} + 4}}$ $\cos \theta = \frac{3(2 + \sqrt{2})}{2\sqrt{18 + 11\sqrt{2}}}$ This approach seems to lead to a complicated expression. Let's re-examine the problem and potentially find a simplification or an alternative calculation for $\cos^2 \theta$.

Alternative Approach for $\cos^2 \theta$

We need to calculate $164 \cos^2 \theta$. Let's focus on calculating $\cos^2 \theta$ directly. $\cos^2 \theta = \frac{(\vec{U} \cdot \vec{V})^2}{|\vec{U}|^2 |\vec{V}|^2}$ From Step 3, $\vec{U} \cdot \vec{V} = 3 + \frac{3\sqrt{2}}{2} = \frac{6 + 3\sqrt{2}}{2}$ $(\vec{U} \cdot \vec{V})^2 = \left(\frac{6 + 3\sqrt{2}}{2}\right)^2 = \frac{36 + 36\sqrt{2} + 18}{4} = \frac{54 + 36\sqrt{2}}{4} = \frac{27 + 18\sqrt{2}}{2}$ From Step 4, $|\vec{U}|^2 = 2 + \sqrt{2}$ From Step 5, $|\vec{V}|^2 = 7 + 2\sqrt{2}$ Now, multiply the denominators: $|\vec{U}|^2 |\vec{V}|^2 = (2 + \sqrt{2})(7 + 2\sqrt{2}) = 14 + 4\sqrt{2} + 7\sqrt{2} + 4 = 18 + 11\sqrt{2}$ So, $\cos^2 \theta = \frac{\frac{27 + 18\sqrt{2}}{2}}{18 + 11\sqrt{2}} = \frac{27 + 18\sqrt{2}}{2(18 + 11\sqrt{2})} = \frac{9(3 + 2\sqrt{2})}{2(18 + 11\sqrt{2})}$ This still looks complicated. Let's check the calculations.

Let's re-evaluate the dot product $\vec{U} \cdot \vec{V}$ and magnitudes. $\vec{U} = \hat{a} + \hat{b}$ $\vec{V} = \hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})$

Dot product: $\vec{U} \cdot \vec{V} = (\hat{a} + \hat{b}) \cdot (\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b}))$ $= \hat{a}\cdot\hat{a} + 2\hat{a}\cdot\hat{b} + 2\hat{a}\cdot(\hat{a} \times \hat{b}) + \hat{b}\cdot\hat{a} + 2\hat{b}\cdot\hat{b} + 2\hat{b}\cdot(\hat{a} \times \hat{b})$ $= 1 + 2(\frac{\sqrt{2}}{2}) + 0 + \frac{\sqrt{2}}{2} + 2(1) + 0$ $= 1 + \sqrt{2} + \frac{\sqrt{2}}{2} + 2 = 3 + \frac{3\sqrt{2}}{2}$. This is correct.

Magnitude of $\vec{U}$: $|\vec{U}|^2 = (\hat{a} + \hat{b}) \cdot (\hat{a} + \hat{b}) = |\hat{a}|^2 + 2\hat{a}\cdot\hat{b} + |\hat{b}|^2 = 1 + 2(\frac{\sqrt{2}}{2}) + 1 = 2 + \sqrt{2}$. This is correct.

Magnitude of $\vec{V}$: Let $\vec{W} = \hat{a} \times \hat{b}$. $|\vec{W}| = \frac{\sqrt{2}}{2}$. $|\vec{V}|^2 = |\hat{a} + 2\hat{b} + 2\vec{W}|^2$ $= (\hat{a} + 2\hat{b} + 2\vec{W}) \cdot (\hat{a} + 2\hat{b} + 2\vec{W})$ $= |\hat{a}|^2 + |2\hat{b}|^2 + |2\vec{W}|^2 + 2(\hat{a} \cdot 2\hat{b}) + 2(\hat{a} \cdot 2\vec{W}) + 2(2\hat{b} \cdot 2\vec{W})$ $= 1 + 4(1) + 4(\frac{1}{2}) + 4\hat{a}\cdot\hat{b} + 4\hat{a}\cdot\vec{W} + 8\hat{b}\cdot\vec{W}$ $= 1 + 4 + 2 + 4(\frac{\sqrt{2}}{2}) + 0 + 0$ $= 7 + 2\sqrt{2}$. This is also correct.

Now, let's calculate $164 \cos^2 \theta$: $164 \cos^2 \theta = 164 \frac{(\vec{U} \cdot \vec{V})^2}{|\vec{U}|^2 |\vec{V}|^2}$ $164 \cos^2 \theta = 164 \frac{\left(3 + \frac{3\sqrt{2}}{2}\right)^2}{(2 + \sqrt{2})(7 + 2\sqrt{2})}$ $164 \cos^2 \theta = 164 \frac{\left(\frac{6 + 3\sqrt{2}}{2}\right)^2}{18 + 11\sqrt{2}}$ $164 \cos^2 \theta = 164 \frac{\frac{36 + 36\sqrt{2} + 18}{4}}{18 + 11\sqrt{2}} = 164 \frac{\frac{54 + 36\sqrt{2}}{4}}{18 + 11\sqrt{2}}$ $164 \cos^2 \theta = 164 \frac{54 + 36\sqrt{2}}{4(18 + 11\sqrt{2})} = 41 \frac{54 + 36\sqrt{2}}{18 + 11\sqrt{2}}$ $164 \cos^2 \theta = 41 \frac{18(3 + 2\sqrt{2})}{18 + 11\sqrt{2}}$ Let's try to rationalize the denominator $18 + 11\sqrt{2}$. Multiply by the conjugate $18 - 11\sqrt{2}$: $\frac{18(3 + 2\sqrt{2})}{18 + 11\sqrt{2}} \times \frac{18 - 11\sqrt{2}}{18 - 11\sqrt{2}} = \frac{18( (3)(18) - (3)(11\sqrt{2}) + (2\sqrt{2})(18) - (2\sqrt{2})(11\sqrt{2}) )}{18^2 - (11\sqrt{2})^2}$ $= \frac{18(54 - 33\sqrt{2} + 36\sqrt{2} - 44)}{324 - 121(2)} = \frac{18(10 + 3\sqrt{2})}{324 - 242} = \frac{18(10 + 3\sqrt{2})}{82}$ $= \frac{9(10 + 3\sqrt{2})}{41}$ Now, substitute this back into the expression for $164 \cos^2 \theta$: $164 \cos^2 \theta = 41 \times \frac{9(10 + 3\sqrt{2})}{41} = 9(10 + 3\sqrt{2}) = 90 + 27\sqrt{2}$

Step 7: Verify the result with the options. The calculated value is $90 + 27\sqrt{2}$. This matches option (A).

Common Mistakes & Tips

• Scalar Triple Product: Remember that the scalar triple product of three vectors where two are identical (e.g., $\vec{a} \cdot (\vec{a} \times \vec{b})$) is always zero.
• Magnitude Calculation: When calculating the square of the magnitude of a sum of vectors, expand it fully and use the properties of dot products. For example, $|\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}$.
• Cross Product of Unit Vectors: Be careful with the magnitude of the cross product of unit vectors: $|\hat{a} \times \hat{b}| = \sin \alpha$, where $\alpha$ is the angle between them.

Summary

The problem required finding the angle $\theta$ between two complex vectors $\vec{U} = \hat{a} + \hat{b}$ and $\vec{V} = \hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})$, given that $\hat{a}$ and $\hat{b}$ are unit vectors with an angle of $\frac{\pi}{4}$ between them. We used the dot product formula $\cos \theta = \frac{\vec{U} \cdot \vec{V}}{|\vec{U}||\vec{V}|}$ and calculated $\cos^2 \theta = \frac{(\vec{U} \cdot \vec{V})^2}{|\vec{U}|^2 |\vec{V}|^2}$. By carefully expanding the dot products and magnitudes, and utilizing the properties of unit vectors and the cross product, we found the value of $164 \cos^2 \theta$ to be $90 + 27\sqrt{2}$.

The final answer is \boxed{90+27 \sqrt{2}}.