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Home » \[ 2a + b = 6 \] - Baxtercollege

\[ 2a + b = 6 \] - Baxtercollege

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 Thus, \( p = 2 \), \( q = 2 \), \( r = 6 \).

📌 \boxed{p = 2, \, q = 2, \, r = 6}

📌 \]Question: A tech entrepreneur in Athens designs a drone navigation system where the drone's path is represented by unit vectors $\mathbf{u}$ and $\mathbf{v}$ with an angle $\theta$ between them. If $\mathbf{u} \cdot \mathbf{v} = \frac{\sqrt{3}}{2}$, find $\theta$.

📌 Solution: The dot product of two unit vectors is $\mathbf{u} \cdot \mathbf{v} = \cos\theta$. Given $\cos\theta = \frac{\sqrt{3}}{2}$, the angle $\theta$ satisfies $\theta = \arccos\left(\frac{\sqrt{3}}{2}\right)$. This corresponds to $\theta = 30^\circ$ or $\frac{\pi}{6}$ radians. However, since cosine is positive in both the first and fourth quadrants, but angles between vectors are typically taken in $[0, \pi]$, the solution is $\boxed{\dfrac{\pi}{6}}$.

📌 Question: A biochemistry technician measures the angle between two molecular bonds modeled as vectors $\mathbf{a} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$. Compute $\cos\theta$ where $\theta$ is the angle between them.

📌 Solution: The cosine of the angle between vectors $\mathbf{a}$ and $\mathbf{b}$ is given by $\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}$. Compute the dot product: $\mathbf{a} \cdot \mathbf{b} = (1)(0) + (0)(1) + (1)(1) = 1$. The magnitudes are $\|\mathbf{a}\| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$ and $\|\mathbf{b}\| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$. Thus, $\cos\theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$. The final answer is $\boxed{\dfrac{1}{2}}$.

📌 Question: A science policy analyst models the efficiency of a renewable energy grid using complex numbers. If $z = \cos\theta + i\sin\theta$ satisfies $z^6 = -1$, find $\theta$ in radians.

📌 Solution: By De Moivre's Theorem, $z^6 = \cos(6\theta) + i\sin(6\theta) = -1 + 0i$. This implies $6\theta = \pi + 2\pi k$ for integer $k$. Solving for $\theta$ gives $\theta = \frac{\pi}{6} + \frac{\pi k}{3}$. The principal solution in $[0, 2\pi)$ is $\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$. The smallest positive $\theta$ is $\boxed{\dfrac{\pi}{6}}$.

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