Contradiction! So no solution exists unless the cross product is orthogonal to $\mathbf{a}$, which \(egin{pmatrix} 0 \ 0 \ 5 \end{pmatrix} \cdot egin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} = 15

February 22, 2026

$Contradiction! So no solution exists unless the cross product is orthogonal to $\mathbf{a}$, which \(egin{pmatrix} 0 \ 0 \ 5 \end{pmatrix} \cdot egin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} = 15$

From first: $3y = 2z = 2(3x) = 6x \Rightarrow y = 2x$
Substitute into third:
2x - (2x) = 5 \Rightarrow 0 = 5
e 0\), so inherently inconsistent.
But wait: the cross product must be perpendicular to $\mathbf{a}$. Check:
egin{pmatrix} 0 \ 0 \ 5 \end{pmatrix} \cdot egin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} = 15

Contradiction! So no solution exists unless the cross product is orthogonal to \(\mathbf{a}\), which \(egin{pmatrix} 0 \ 0 \ 5 \end{pmatrix} \cdot egin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} = 15

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