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so you think you think you can dance - Malaeb

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} \times \begin{pmatrix} 4 \\ 0 \\ -1 \end{pmatrix} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & 3 \\ 4 & 0 & -1 \end{vmatrix}

📌 = \mathbf{i}((-2)(-1) - (3)(0)) - \mathbf{j}((1)(-1) - (3)(4)) + \mathbf{k}((1)(0) - (-2)(4))

📌 = \mathbf{i}(2) - \mathbf{j}(-1 - 12) + \mathbf{k}(0 + 8) = 2\mathbf{i} + 13\mathbf{j} + 8\mathbf{k}

📌 Thus, the vector is $\boxed{\begin{pmatrix} 2 \\ 13 \\ 8 \end{pmatrix}}$.

📌 Question:** A linguist modeling word embeddings in 3D space defines two unit vectors $\mathbf{u}$ and $\mathbf{v}$ with an angle of $60^\circ$ between them. What is the maximum possible value of $|\mathbf{u} \cdot \mathbf{v}|$?

📌 Solution:** The dot product of two unit vectors is given by $\mathbf{u} \cdot \mathbf{v} = \cos\theta$, where $\theta$ is the angle between them. Since $\theta = 60^\circ$, we compute:

📌 |\mathbf{u} \cdot \mathbf{v}| = |\cos 60^\circ| = \left|\frac{1}{2}\right| = \boxed{\frac{1}{2}}

📌 Question:** A cloud computing system routes data along vectors in $\mathbb{R}^3$. Given $\mathbf{v} \times \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 5 \\ -6 \\ -1 \end{pmatrix}$, find $\mathbf{v}$.

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