\mathbf{v} \times \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} b(2) - c(0) \\ c(1) - a(2) \\ a(0) - b(1) \end{pmatrix} = \begin{pmatrix} 2b \\ c - 2a \\ -b \end{pmatrix} = \begin{pmatrix} -4 \\ 6 \\ 2 \end{pmatrix}.

["Solving the Cross Product Equation: Step-by-Step Breakdown of\n  \ extbf{v} × \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} = \begin{pmatrix} -4 \ 6 \ 2 \end{pmatrix}", "When faced with a vector cross product equation like\n  \ extbf{v} × \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} = \begin{pmatrix} -4 \ 6 \ 2 \end{pmatrix},\nthe solution involves translating the geometric operation into algebraic equations and solving a system derived from the cross product formula. In this article, we break down how vector cross products work, apply the formula rigorously, and solve:\n  \ extbf{v} × \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} = \begin{pmatrix} -4 \ 6 \ 2 \end{pmatrix},\nto find the components (a), (b), and (c) of vector v = \begin{pmatrix} a \ b \ c \end{pmatrix}.", "---", "### Understanding the Cross Product Formula", "For two 3D vectors:\n  \ extbf{u} = \begin{pmatrix} a \ b \ c \end{pmatrix},  \ extbf{w} = \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix},\nthe cross product is defined as:\n  \ extbf{u} × \ extbf{w} =\n  \begin{pmatrix}\n  b \cdot 2 - c \cdot 0 \\n  c \cdot 1 - a \cdot 2 \\n  a \cdot 0 - b \cdot 1\n \end{pmatrix}\n =\n  \begin{pmatrix}\n  2b \\n  c - 2a \\n  - b\n \end{pmatrix}", "This establishes the vector result:\n  \ extbf{v} × \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} = \begin{pmatrix} 2b \ c - 2a \ -b \end{pmatrix}", "---", "### Setting Equal Components", "Given the equality\n  \begin{pmatrix} 2b \ c - 2a \ -b \end{pmatrix} = \begin{pmatrix} -4 \ 6 \ 2 \end{pmatrix},\nwe equate each component to form a system of linear equations:\n1)  2b = -4\n2)  c - 2a = 6\n3)  -b = 2", "---", "### Step-by-Step Solution", "Step 1: Solve for (b) using the third equation\nFrom  - b = 2 →           b = -2", "Step 2: Substitute (b = -2) into the first equation\n2b = -4 → 2(-2) = -4 ✓ (consistent)", "Step 3: Use (b = -2) in the third equation for verification\n-b = 2 → -(-2) = 2 ✓ (correct)", "Step 4: Solve for (a) and (c) using equation 2\nc − 2a = 6\nThis is a single equation with two unknowns, so we express one variable in terms of the other:\n  c = 6 + 2a", "At this stage, (a) is a free parameter — meaning there are infinitely many solutions, differing by values of (a), with (c) determined accordingly.", "---", "### General Solution", "The system yields:\n  a = a  (arbitrary real number)\n  b = -2\n  c = 6 + 2a", "So vector v is:\n  \ extbf{v} = \begin{pmatrix} a \ -2 \ 6 + 2a \end{pmatrix},  a ∈ ℝ", "---", "### Unique Particular Solution", "To present a clean, specific answer, choose any convenient value for the free variable (a). For simplicity, take (a = 0):\n  \ extbf{v} =\n  \begin{pmatrix} 0 \ -2 \ 6 \end{pmatrix}", "Check:\n  0 × \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} = \begin{pmatrix} 0 \ 6 \ 0 \end{pmatrix} ≠ correct\nWait — we must inspect the cross product result again.", "Wait — earlier derivation was correct, but verification shows inconsistency?", "No — correction: from earlier:\nThe cross product with v = (a, b, c) gives (\begin{pmatrix} 2b \ c - 2a \ -b \end{pmatrix})\nWith b = -2:\n  First component: 2(-2) = -4 ✓\n  Third component: -(-2) = 2 ✓\nSo only b is constrained; a and c remain linked.", "Thus, (c − 2a = 6) ⇒ c = 6 + 2a, b = -2, a free.", "Checking the full cross product:\n  (−4, c − 2a = 6, 2) matches exact RHS", "Thus v = (a, -2, 6 + 2a) is the solution set.", "---", "### Final Interpretation", "The equation defines a line of solutions in 3D space, parameterized by (a). This occurs because the cross product mapping v → v × w is a linear transformation with a 2-dimensional kernel (only vectors parallel to w map to zero, excluding the zero vector case here), implying a 1-dimensional solution space.", "---", "### Practical Takeaways", "- Cross product results are determined entirely by vector components via a fixed algebraic rule.\n- Equating components generates a system to solve.\n- When inconsistency, no solution — but here system is consistent and underdetermined.\n- The solution is typically expressed in parametric form.", "---", "Conclusion", "By applying the cross product definition rigorously and solving the resulting system, we find that:\n  \ extbf{v} = \begin{pmatrix} a \ -2 \ 6 + 2a \end{pmatrix},  a ∈ ℝ\nis the complete solution to\n  \ extbf{v} × \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} = \begin{pmatrix} -4 \ 6 \ 2 \end{pmatrix}.", "For a standard solution, choosing (a = 0) yields:\n  \ extbf{v} = \begin{pmatrix} 0 \ -2 \ 6 \end{pmatrix}", "Always verify by plugging back into the original cross product to ensure correctness.", "---", "Keywords: vector cross product, v × ⟨1,0,2⟩ = ⟨-4,6,2⟩, solving vector equations, parametric solution, cross product components, linear algebra application."]