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Home » "This Surprise Birthday Gift Will Have You Regretting Not Giving It Earlier! - GetMeFoodie

"This Surprise Birthday Gift Will Have You Regretting Not Giving It Earlier! - GetMeFoodie

📅 May 2, 2026 👤 scraface

May 02, 2026

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📌 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

📌 So no such \(\mathbf{v}\) exists because the given vector is not perpendicular to \(\mathbf{a}\), violating a fundamental identity: \(\mathbf{v} imes \mathbf{a} \perp \mathbf{a}\).

📌 However, if we assume a typo and the right-hand side is perpendicular, suppose the intended result is \(egin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\), but it's given as \(egin{pmatrix} 0 \ 0 \ 5 \end{pmatrix}\). Since no vector satisfies the equation, we must conclude the system is inconsistent.

📌 But suppose we seek a \(\mathbf{v}\) such that the cross product equals a vector parallel to \(\mathbf{a}^\perp\). Alternatively, perhaps the problem meant a different result.

📌 Wait â reconsider: **There is always a unique solution up to addition of a multiple of \(\mathbf{a}\)** to the cross product, but the equation must be consistent with \(\mathbf{v} imes \mathbf{a} \perp \mathbf{a}\).

📌 Since \(egin{pmatrix} 0 \ 0 \ 5 \end{pmatrix}

📌 ot\perp egin{pmatrix} 1 \ 2 \ 3 \end{pmatrix}\), **no such vector \(\mathbf{v}\) exists**.

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