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.

February 22, 2026

$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.$

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.
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**.

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.

Related Articles

Trending Articles