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📅 March 5, 2026 👤 scraface

Mar 05, 2026

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From the third equation: $x = 1 - 2y$.
2z - 4(1 - 2y) = -6 \Rightarrow 2z - 4 + 8y = -6 \Rightarrow 2z + 8y =...
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 inc...
However, cross product outputs are orthogonal to $\mathbf{a}$. Check:
\begin{pmatrix} 5 \\ -6 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\...
So the given vector is not orthogonal to $\mathbf{a}$, which violates...
Therefore, no such vector $\mathbf{v}$ exists.

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