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top 10 players of all time nba - RTA

📅 May 10, 2026 👤 scraface

May 10, 2026

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📌 Now solve the first and new second equation:

📌 z + 4y = -1

📌 \Rightarrow \text{Subtract: } 0 = 6 \Rightarrow \text{Contradiction}

📌 This suggests no solution exists unless the given cross product is inconsistent with the vector.

📌 However, cross product outputs are orthogonal to $\mathbf{a}$. Check:

📌 \begin{pmatrix} 5 \\ -6 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = 10 + 6 - 4 = 12 \ne 0

📌 So the given vector is not orthogonal to $\mathbf{a}$, which violates the cross product property.

📌 Therefore, no such vector $\mathbf{v}$ exists.

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