Solution: To solve for $\mathbf{v}$, we use the cross product equation $\mathbf{v} \times \mathbf{w} = \mathbf{p}$. Let $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}$. The cross product $\mathbf{v} \times \mathbf{w}$ is calculated as $\begin{pmatrix} v_2 \cdot 3 - v_3 \cdot (-1) \\ v_3 \cdot 2 - v_1 \cdot 3 \\ v_1 \cdot (-1) - v_2 \cdot 2 \end{pmatrix} = \begin{pmatrix} 3v_2 + v_3 \\ 2v_3 - 3v_1 \\ -v_1 - 2v_2 \end{pmatrix}$. Setting this equal to $\mathbf{p} = \begin{pmatrix} 5

["Solution to $\mathbf{v} \ imes \mathbf{w} = \mathbf{p}$: Finding $\mathbf{v}$ Using Vector Cross Products", "When solving vector equations like $\mathbf{v} \ imes \mathbf{w} = \mathbf{p}$ for an unknown vector $\mathbf{v}$, one powerful approach lies in leveraging the properties of the cross product. Here, we solve for $\mathbf{v} = \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix}$, given $\mathbf{w} = \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix}$ and $\mathbf{p} = \begin{pmatrix} 5 \ 0 \ -2 \end{pmatrix}$. The equation becomes:", "$$\n\mathbf{v} \ imes \mathbf{w} = \begin{pmatrix} v_2 \cdot 2 - v_3 \cdot (-1) \ v_3 \cdot 3 - v_1 \cdot 2 \ v_1 \cdot (-1) - v_2 \cdot 3 \end{pmatrix} = \begin{pmatrix} 2v_2 + v_3 \ 3v_3 - 2v_1 \ -v_1 - 3v_2 \end{pmatrix} = \begin{pmatrix} 5 \ 0 \ -2 \end{pmatrix}\n$$", "This leads to the system of linear equations:", "$$\n\begin{cases}\n2v_2 + v_3 = 5 \quad \ ext{(1)}\\n3v_3 - 2v_1 = 0 \quad \ ext{(2)}\\n-v_1 - 3v_2 = -2 \quad \ ext{(3)}\n\end{cases}\n$$", "---", "Step 1: Simplify equation (2) for $v_1$\nFrom (2):\n$$\n3v_3 - 2v_1 = 0 \Rightarrow v_1 = \frac{3}{2}v_3\n$$", "---", "Step 2: Substitute $v_1$ into equation (3)\nPlug $v_1 = \frac{3}{2}v_3$ into (3):\n$$\n-\left(\frac{3}{2}v_3\right) - 3v_2 = -2 \Rightarrow -\frac{3}{2}v_3 - 3v_2 = -2\n$$\nMultiply through by 2 to eliminate fractions:\n$$\n-3v_3 - 6v_2 = -4 \Rightarrow 3v_3 + 6v_2 = 4 \quad \ ext{(4)}\n$$", "---", "Step 3: Solve equations (1) and (4) together\nFrom (1):\n$$\nv_3 = 5 - 2v_2 \quad \ ext{(5)}\n$$\nSubstitute (5) into (4):\n$$\n3(5 - 2v_2) + 6v_2 = 4 \Rightarrow 15 - 6v_2 + 6v_2 = 4 \Rightarrow 15 = 4\n$$", "Wait — this yields the contradiction $15 = 4$, indicating no solution exists for $\mathbf{v}$ satisfying $\mathbf{v} \ imes \mathbf{w} = \mathbf{p}$ unless $\mathbf{p}$ is orthogonal to $\mathbf{w}$. This is a key insight.", "---", "Why Orthogonality Matters in Cross Product Equations\nThe cross product $\mathbf{v} \ imes \mathbf{w}$ is always perpendicular to both $\mathbf{v}$ and $\mathbf{w}$. Therefore, for the equation $\mathbf{v} \ imes \mathbf{w} = \mathbf{p}$ to have solutions, $\mathbf{p}$ must lie in the plane perpendicular to $\mathbf{w}$, i.e., $\mathbf{p} \cdot \mathbf{w} = 0$.", "Check this condition:\n$$\n\mathbf{p} \cdot \mathbf{w} = \begin{pmatrix} 5 \ 0 \ -2 \end{pmatrix} \cdot \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix} = 5 \cdot 3 + 0 \cdot (-1) + (-2) \cdot 2 = 15 - 4 = 11 <br/>\neq 0\n$$\nSince $\mathbf{p} \cdot \mathbf{w} <br/>\ne 0$, the vector equation has no solution.", "---", "When Solutions Exist: A Contrast Example\nTo illustrate, suppose the given $\mathbf{p} = \begin{pmatrix} 5 \ 0 \ -11 \end{pmatrix}$ (now orthogonal to $\mathbf{w}$). Then $ \mathbf{p} \cdot \mathbf{w} = 0 $, satisfying necessary condition. Substituting $v_3 = 5 - 2v_2$ into (4) now yields consistency, and the system yields infinitely many solutions depending on a free parameter, typical in cross product setups.", "---", "Conclusion\nSolving $\mathbf{v} \ imes \mathbf{w} = \mathbf{p}$ reduces to analyzing orthogonality: $\mathbf{p}$ must be perpendicular to $\mathbf{w}$ for a solution $\mathbf{v}$ to exist. When this condition fails—as with $\mathbf{p} = \begin{pmatrix} 5 \ 0 \ -2 \end{pmatrix}$—no vector $\mathbf{v}$ satisfies the equation. This principle is vital in physics, engineering, and computer graphics where vector cross products model rotational dynamics and orientation.", "For future vector equation solutions, always verify $\mathbf{p} \cdot \mathbf{w} = 0$ first—this simple check saves time searching where no solution exists.", "Keywords: cross product equation, vector solution $\mathbf{v}$, $\mathbf{v} \ imes \mathbf{w} = \mathbf{p}$, orthogonality condition, linear system from cross products, no solution vector algebra."]