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But this diverges. - Databee Business Systems

📅 February 26, 2026 👤 scraface

Feb 26, 2026

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Solution: Let $\mathbf{v} = \langle v_1, v_2, v_3 \rangle$. The cross...
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
v_1 & v_2 & v_3 \\
& 2 & 3 \\
\end{vmatrix} = \langle 2v_3 - 3v_2, 3v_1 - v_3, v_2 - 2v_1 \rangle =...
This yields the system:
2v_3 - 3v_2 = 4, \quad 3v_1 - v_3 = 5, \quad v_2 - 2v_1 = 6.
From the third equation, $v_2 = 6 + 2v_1$. Substituting into the first...
2v_3 - 3(6 + 2v_1) = 4 \implies 2v_3 - 18 - 6v_1 = 4 \implies 2v_3 = 2...

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