Solution: Let $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}$. The cross product $\mathbf{v} \times \mathbf{w}$ is $\begin{pmatrix} 3v_2 + v_3 \\ 2v_3 - 3v_1 \\ -v_1 - 2v_2 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ -2 \end{pmatrix}$. This gives the system:

["Understanding the Cross Product Equation: Solving for $\mathbf{v}$ Given the Result", "The cross product is a fundamental operation in vector algebra, widely used in physics and 3D geometry to find perpendicular vectors, calculate torque, and define areas in space. Given a vector $\mathbf{v} = \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix}$ and another vector $\mathbf{w}$, their cross product $\mathbf{v} \ imes \mathbf{w} = \begin{pmatrix} 3 \ 0 \ -2 \end{pmatrix}$ provides a powerful way to derive constraints on $\mathbf{v}$. Here’s how to interpret and solve such a system.", "---", "### The Cross Product Definition", "Let’s first recall the formula for the cross product of $\mathbf{v} = \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix}$ and an unknown $\mathbf{w} = \begin{pmatrix} w_1 \ w_2 \ w_3 \end{pmatrix}$:", "$$\n\mathbf{v} \ imes \mathbf{w} = \n\begin{pmatrix}\nv_2 w_3 - v_3 w_2 \\nv_3 w_1 - v_1 w_3 \\nv_1 w_2 - v_2 w_1\n\end{pmatrix}\n= \begin{pmatrix}\n3 \\n0 \\n-2\n\end{pmatrix}\n$$", "This yields the following system of equations:", "1. $ v_2 w_3 - v_3 w_2 = 3 $  (1)\n2. $ v_3 w_1 - v_1 w_3 = 0 $    (2)\n3. $ v_1 w_2 - v_2 w_1 = -2 $   (3)", "However, $\mathbf{w}$ is not specified—it’s unknown. But notice: this is a linear system in variables $w_1, w_2, w_3$, and we’re solving for $\mathbf{v}$. To isolate $\mathbf{v}$, consider $\mathbf{v}$ as unknown and $\mathbf{w}$ as a parameter. But since $\mathbf{w}$ is unknown, a smarter approach is to analyze the structure and constraints imposed by the equations.", "---", "### Rewriting the system in terms of $\mathbf{v}$", "Let $\mathbf{v} = \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix}$ and treat the system as a linear relation involving components of $\mathbf{v}$ multiplied by components of $\mathbf{w}$. But without $\mathbf{w}$, we seek values of $\mathbf{v}$ such that the cross product equals $\begin{pmatrix} 3 \ 0 \ -2 \end{pmatrix}$ for some $\mathbf{w}$. This turns into a feasibility problem for $\mathbf{v}$.", "Instead, suppose $\mathbf{w}$ is an arbitrary but nonzero vector producing the given cross product. We can eliminate $\mathbf{w}$ by expressing equations in terms of $\mathbf{v}$ and consistency conditions.", "---", "### Step 1: Analyze Equation (2): $ v_3 w_1 = v_1 w_3 $", "Assuming $w_3 <br/>\ne 0$, solve for $w_1$:", "$$\nw_1 = \frac{v_1}{v_3} w_3 \quad \ ext{if } v_3 <br/>\ne 0\n$$", "If $v_3 = 0$, equation (2) becomes $0 = 0$, which gives no info on $w_1$, so $w_1$ can be arbitrary. But check equation (3):\nIf $v_3 = 0$, equation (3): $v_1 w_2 - v_2 w_1 = -2$. This may restrict relations, but proceed with $v_3 <br/>\ne 0$ for generality.", "Similarly, examine equation (1) and (3) for consistency across components.", "---", "### Step 2: Express system as linear equations in $w_1, w_2, w_3$", "From (1): $ v_2 w_3 - v_3 w_2 = 3 $\nFrom (2): $ v_3 w_1 - v_1 w_3 = 0 \Rightarrow v_3 w_1 = v_1 w_3 $\nFrom (3): $ v_1 w_2 - v_2 w_1 = -2 $", "This forms a homogeneous and inhomogeneous system in $w_1, w_2, w_3$. For a solution $w$ to exist, the system must be consistent. But since $\mathbf{w}$ exists (as per problem setup), the system must have a nontrivial solution.", "We seek conditions on $\mathbf{v}$ such that this 3×3 system has a solution. But our goal is to solve for $\mathbf{v}$ — not $\mathbf{w}$ — so reverse intuition: Given the output cross product, find $\mathbf{v}$.", "---", "### Step 3: Use cross product properties and guess a plausible form", "Let us suppose $\mathbf{w}$ has a known structure. Alternatively, eliminate $\mathbf{w}$ by expressing ratios.", "From (2): $ \frac{w_1}{w_3} = \frac{v_1}{v_3} $, if $v_3 <br/>\ne 0 $", "Let $ a = \frac{v_1}{v_3} $, $ b = \frac{v_2}{v_3} $, so $\mathbf{v} = v_3 \begin{pmatrix} a \ b \ 1 \end{pmatrix}$", "Let $\mathbf{w} = w_3 \begin{pmatrix} \frac{v_1}{v_3} \ \frac{v_2}{v_3} \ 1 \end{pmatrix} = v_3 \begin{pmatrix} a \ b \ 1 \end{pmatrix}$ — but wait, this assumes the same scaling, risky unless $\mathbf{w}$ is parallel.", "Better: let $\mathbf{w} = \begin{pmatrix} w_1 \ w_2 \ w_3 \end{pmatrix}$ satisfy the 3 equations. Substitute ratios.", "From (1): $ v_2 w_3 - v_3 w_2 = 3 \Rightarrow v_3 b w_3 - v_3 w_2 = 3 \Rightarrow v_3 (b w_3 - w_2) = 3 $", "From (2): $ v_3 w_1 = v_1 w_3 = v_3 a w_3 \Rightarrow w_1 = a w_3 $", "From (3): $ v_1 w_2 - v_2 w_1 = v_3 a w_2 - v_3 b v_3 w_1 = v_3 (a w_2 - b v_3 w_1) = -2 $", "But $ w_1 = a w_3 $, so:", "$$\nv_3 (a w_2 - b v_3 a w_3) = -2 \Rightarrow v_3 a (w_2 - b v_3 w_3) = -2\n$$", "Now use $ w_2 = b w_3 + \frac{3}{v_3} $ from (1)", "Substitute:", "$$\nv_3 a \left( (b w_3 + \frac{3}{v_3}) - b v_3 w_3 \right)\n= v_3 a \left( b w_3 + \frac{3}{v_3} - b v_3 w_3 \right)\n= v_3 a \left( \frac{3}{v_3} + b w_3 (1 - v_3^2) \right)\n$$", "Simplify:", "$$\n= a \left( 3 + v_3^2 b w_3 (1 - v_3^2) \right)\n= -2\n$$", "So:", "$$\na \left( 3 + b v_3^2 (1 - v_3^2) \right) w_3 = -2\n$$", "This equation links $a, b, v_3, w_3$, but $w_3$ appears freely — unless coefficients balance independently.", "To make this solvable for some $w_3 <br/>\ne 0$, the term in parentheses must allow a real $w_3$. But to solve for $\mathbf{v}$, suppose we assume a simple form.", "---", "### Step 4: Try a specific solution by inspection", "Assume a solution where components of $\mathbf{v}$ are integers or simple fractions.", "Try hypothesis: suppose $v_3 <br/>\ne 0$, and suppose $v_1 = 1$, $v_2 = 1$, then from (3):", "$$\nw_1 - w_2 = -2 \Rightarrow w_1 = w_2 - 2\n$$", "From (2): $v_3 w_1 = v_1 w_3 = w_3 \Rightarrow w_1 = \frac{w_3}{v_3}$", "From (1): $v_2 w_3 - v_3 w_2 = w_3 - v_3 w_2 = 3$", "Try $v_3 = 1$: then equation (1): $w_3 - w_2 = 3 \Rightarrow w_3 = w_2 + 3$", "From (2): $w_1 = w_3 = w_2 + 3$", "From (3): $w_1 = w_2 - 2$ ⇒ $w_2 + 3 = w_2 - 2$ → contradiction.", "Try $v_3 = 2$: now (1): $w_3 - 2w_2 = 3 \Rightarrow w_3 = 2w_2 + 3$\n(2): $2w_1 = w_3 \Rightarrow w_1 = \frac{w_3}{2} = w_2 + \frac{3}{2}$\n(3): $v_1 w_2 - v_2 w_1 = 1\cdot w_2 - 1\cdot w_1 = w_2 - w_1 = -2$ ⇒ $w_2 - w_1 = -2$", "Substitute $w_1 = w_2 + \frac{3}{2}$:", "$$\nw_2 - (w_2 + \frac{3}{2}) = -\frac{3}{2} <br/>\ne -2\n$$", "Close, but not exact.", "Try $v_1 = 2$, $v_2 = 1$, $v_3 = 1$", "Then (3): $2w_2 - 1\cdot w_1 = -2 \Rightarrow 2w_2 - w_1 = -2 \Rightarrow w_1 = 2w_2 + 2$", "(2): $1\cdot w_1 - 2 w_3 = 0 \Rightarrow w_1 = 2w_3$", "(1): $1\cdot w_3 - 1\cdot w_2 = 3 \Rightarrow w_3 - w_2 = 3 \Rightarrow w_2 = w_3 - 3$", "Now substitute: $w_1 = 2w_3$, $w_2 = w_3 - 3$ → check (3): $2(w_3 - 3) - (w_3 + 2) = 2w_3 - 6 - w_3 - 2 = w_3 - 8 = -2 \Rightarrow w_3 = 6$", "Then $w_2 = 3$, $w_1 = 12$", "Now verify all:", "- (1): $v_2 w_3 - v_3 w_2 = 1\cdot6 - 1\cdot3 = 3$ ✅\n- (2): $v_3 w_1 - v_1 w_3 = 1\cdot12 - 2\cdot6 = 12 - 12 = 0$ ✅\n- (3): $2\cdot3 - 1\cdot12 = 6 - 12 = -6 <br/>\ne -2$ ❌", "Too low. Try $v_1 = 1$, $v_2 = 2$", "Then (3): $1\cdot w_2 - 2 w_1 = -2$", "(2): $v_3 w_1 - w_3 = 0 \Rightarrow w_3 = v_3 w_1$", "(1): $2 w_3 - v_3 w_2 = 3 \Rightarrow 2(v_3 w_1) - v_3 w_2 = 3 \Rightarrow v_3 (2w_1 - w_2) = 3$", "Let $v_3 = 1$: then $2w_1 - w_2 = 3$", "From (3): $w_2 - 2w_1 = -2$", "Add: $ (2w_1 - w_2) + (w_2 - 2w_1) = 3 - 2 \Rightarrow 0 = 1 $ ❌", "Try $v_3 = 3$: then (1): $2 w_3 - 3 w_2 = 3$, (2): $3w_1 = w_3$ ⇒ $w_3 = 3w_1$, so $2(3w_1) - 3w_2 = 3 \Rightarrow 6w_1 - 3w_2 = 3 \Rightarrow 2w_1 - w_2 = 1$", "(3): $1\cdot w_2 - 2 w_1 = w_2 - 2w_1 = -2$", "Now solve:\n$2w_1 - w_2 = 1$\n$-2w_1 + w_2 = -2$\nAdd: $0 = -1$ ❌", "Try $v_1 = 3$, $v_2 = 1$, $v_3 = 1$", "(3): $3w_2 - w_1 = -2 \Rightarrow w_1 = 3w_2 + 2$", "(2): $1\cdot w_1 - 3 w_3 = 0 \Rightarrow w_1 = 3w_3$", "(1): $1\cdot w_3 - 1\cdot w_2 = 3 \Rightarrow w_3 - w_2 = 3 \Rightarrow w_2 = w_3 - 3$", "Now: $3w_2 + 2 = w_1 = 3w_3$\nSo $3(w_3 - 3) + 2 = 3w_3 \Rightarrow 3w_3 - 9 + 2 = 3w_3 \Rightarrow -7 = 0$ ❌", "---", "### Insight: Use vector algebra identity", "Recall: For any two vectors $\mathbf{v}, \mathbf{w}$,\n$$\n|\mathbf{v} \ imes \mathbf{w}|^2 = |\mathbf{v}|^2 |\mathbf{w}|^2 - (\mathbf{v} \cdot \mathbf{w})^2\n$$", "But this gives magnitude, not components.", "Instead, observe that the system:", "$$\n\mathbf{v} \ imes \mathbf{w} = \mathbf{u}, \quad |\mathbf{u}| = \begin{pmatrix} 3 \ 0 \ -2 \end{pmatrix}, |\mathbf{u}|^2 = 9 + 4 = 13\n$$", "But without $\mathbf{w}$, hard to exploit.", "---", "### Key Realization: The system is underdetermined — infinitely many $\mathbf{v}$ may satisfy existence of $\mathbf{w}$", "But the problem implies a unique or specific solution — so likely assumes $\mathbf{w}$ is known or relates globally.", "Wait: re-read — “The cross product $\mathbf{v} \ imes \mathbf{w}$ is” — and gives a vector. But $\mathbf{w}$ is unknown. So the equation is:", "$$\n\mathbf{v} \ imes \mathbf{w} = \begin{pmatrix} 3 \ 0 \ -2 \end{pmatrix}\n$$", "But without $\mathbf{w}$, we cannot solve for $\mathbf{v}$ uniquely — many $\mathbf{v}$ pairs can produce the same cross product for some $\mathbf{w}$.", "Unless $\mathbf{w}$ is implied to be, say, unit or aligned in a way — but not stated.", "Alternative interpretation: perhaps $\mathbf{w}$ is a standard basis vector? But not specified.", "---", "### Correct Mathematical Approach: Consistency condition", "The equation $\mathbf{v} \ imes \mathbf{w} = \mathbf{u}$ has solution $\mathbf{w}$ if and only if $\mathbf{u} \cdot \mathbf{v} = 0$. Why?", "Because $\mathbf{v} \ imes \mathbf{w} \perp \mathbf{v}$, so $\mathbf{u} \perp \mathbf{v} \Rightarrow \mathbf{u} \cdot \mathbf{v} = 0$", "But here, $\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$, and since $\mathbf{u} = \mathbf{v} \ imes \mathbf{w}$, by vector identity:\n$$\n\mathbf{v} \cdot (\mathbf{v} \ imes \mathbf{w}) = 0\n$$\nAlways true. So every $\mathbf{v}$ admits some $\mathbf{w}$ such that $\mathbf{v} \ imes \mathbf{w} = \mathbf{u}$ — but the setup assumes such $\mathbf{w}$ exists, and we solve for $\mathbf{v}$ given the full cross product output.", "But the system fixes the cross product, so $\mathbf{v}$ is constrained.", "In fact, the solution set for $\mathbf{v}$ such that $\mathbf{v} \ imes \mathbf{w} = \mathbf{u}$ for some $\mathbf{w}$ is all vectors $\mathbf{v}$, because for any $\mathbf{v} <br/>\ne 0$, pick $\mathbf{w} = \frac{1}{v_3} \mathbf{v} \ imes \mathbf{u}$ (if $v_3 <br/>\ne 0$) — then $\mathbf{v} \ imes \mathbf{w} = \mathbf{u}$.", "Thus, the equation gives no constraint on $\mathbf{v}$ — every $\mathbf{v}$ works with suitable $\mathbf{w}$.", "But the problem asks to “solve” — implying a specific $\mathbf{v}$. So likely, $\mathbf{w}$ is given or inferred.", "Wait — perhaps the problem is misstated. More plausible: suppose $\mathbf{w}$ is known, say $\mathbf{w} = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}$, standard basis.", "Try that.", "---", "### Assume $\mathbf{w} = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}$", "Then:\n$$\n\mathbf{v} \ imes \mathbf{w} = \n\begin{pmatrix}\nv_2 \cdot 0 - v_3 \cdot 1 \\nv_3 \cdot 0 - v_1 \cdot 0 \\nv_1 \cdot 1 - v_2 \cdot 0\n\end{pmatrix}\n= \begin{pmatrix} -v_3 \ 0 \ v_1 \end{pmatrix}\n$$", "Set equal to $\begin{pmatrix} 3 \ 0 \ -2 \end{pmatrix}$:", "$$\n-v_3 = 3 \Rightarrow v_3 = -3 \\nv_1 = -2\n$$", "$w_2 = 1$, arbitrary in cross product, but $\mathbf{w}$ fixed.", "So $v_1 = -2$, $v_3 = -3$, $v_2$ arbitrary? But (3) gives $v_1 = -2$, (1) gives $-v_3 = 3 \Rightarrow v_3 = -3$, and (2): $0 = 0$, no constraint.", "But cross product has third component $v_1 = -2$, matches.", "So solution: $\mathbf{v} = \begin{pmatrix} -2 \ t \ -3 \end{pmatrix}$ for any $t$.", "But not unique.", "To get unique $\mathbf{v}$, need more constraints.", "---", "### The only way to get a unique solution is if the system, combined with minimality or orthogonality, pins down $\mathbf{v}$.", "But recall: the only vector that commutes with all cross products is zero, but here we seek $\mathbf{v}$ such that $\mathbf{v} \ imes \mathbf{w} = \mathbf{u}$ for a given $\mathbf{u}$."]