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what number is travis kelce - Network4

📅 May 18, 2026 👤 scraface

May 18, 2026

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📌 Given \(\mathbf{v} \times \mathbf{b} = \begin{pmatrix} 0 \\ 3 \\ -4 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}\), let \(\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}\).

📌 The cross product \(\mathbf{v} \times \mathbf{b}\) is:

📌 \mathbf{v} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ v_1 & v_2 & v_3 \\ 1 & 0 & 2 \end{vmatrix} = \begin{pmatrix} (v_2 \cdot 2 - v_3 \cdot 0) \\ (v_3 \cdot 1 - v_1 \cdot 2) \\ (v_1 \cdot 0 - v_2 \cdot 1) \end{pmatrix} = \begin{pmatrix} 2v_2 \\ v_3 - 2v_1 \\ -v_2 \end{pmatrix}

📌 Set this equal to \(\begin{pmatrix} 0 \\ 3 \\ -4 \end{pmatrix}\):

📌 2v_2 = 0 \\

📌 v_3 - 2v_1 = 3 \\

📌 From the first equation, \(2v_2 = 0 \implies v_2 = 0\).

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