Question: Find the vector $ \mathbf{v} $ such that $ \mathbf{v} \times \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} -4 \\ 6 \\ 2 \end{pmatrix} $.

["Title: Solving for $ \mathbf{v} $: A Step-by-Step Guide to Find the Vector Cross Product Question", "Meta Description:\nStruggling to find the vector $ \mathbf{v} $ such that $ \mathbf{v} \ imes \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} = \begin{pmatrix} -4 \ 6 \ 2 \end{pmatrix} $? This detailed article guides you through the cross product solution step by step using vector algebra.", "---", "### Introduction: Solving Cross Product Equations", "In vector mathematics, finding a vector $ \mathbf{v} $ that satisfies a cross product equation is a fundamental problem with wide applications in physics and engineering. Today, we focus on a specific equation:\n$$\n\mathbf{v} \ imes \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} = \begin{pmatrix} -4 \ 6 \ 2 \end{pmatrix}\n$$\nLet $ \mathbf{v} = \begin{pmatrix} x \ y \ z \end{pmatrix} $, $ \mathbf{a} = \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} $, and the known result $ \mathbf{b} = \begin{pmatrix} -4 \ 6 \ 2 \end{pmatrix} $. We seek $ \mathbf{v} $ such that $ \mathbf{v} \ imes \mathbf{a} = \mathbf{b} $.", "This article explains how to solve for $ x $, $ y $, and $ z $ using the standard method for cross product equations.", "---", "### Step 1: Write the Cross Product Explicitly", "The cross product $ \mathbf{v} \ imes \mathbf{a} $ for vectors in $ \mathbb{R}^3 $ is given by the determinant:\n$$\n\mathbf{v} \ imes \mathbf{a} = \n\begin{vmatrix}\n\mathbf{i} & \mathbf{j} & \mathbf{k} \\nx & y & z \\n1 & 0 & 2\n\end{vmatrix}\n= \mathbf{i}(y \cdot 2 - z \cdot 0) - \mathbf{j}(x \cdot 2 - z \cdot 1) + \mathbf{k}(x \cdot 0 - y \cdot 1)\n$$\nSimplifying:\n$$\n\mathbf{v} \ imes \mathbf{a} = \begin{pmatrix} 2y \ -(2x - z) \ -y \end{pmatrix} = \begin{pmatrix} 2y \ -2x + z \ -y \end{pmatrix}\n$$", "---", "### Step 2: Set Up the System of Equations", "Equate this result to $ \mathbf{b} = \begin{pmatrix} -4 \ 6 \ 2 \end{pmatrix} $:\n$$\n\begin{cases}\n2y = -4 \\n-2x + z = 6 \\n-y = 2\n\end{cases}\n$$", "---", "### Step 3: Solve Each Component Equation", "From the first component:\n$$\n2y = -4 \implies y = -2\n$$", "From the third component:\n$$\n-y = 2 \implies y = -2 \quad \ ext{(consistent with above)}\n$$", "Substitute $ y = -2 $ into the second equation:\n$$\n-2x + z = 6 \implies z = 2x + 6\n$$", "---", "### Step 4: Express General Solution", "We now have:\n- $ y = -2 $\n- $ z = 2x + 6 $", "Since no constraints fix $ x $, it remains a free variable. Thus, the general solution is:\n$$\n\mathbf{v} = \begin{pmatrix} x \ -2 \ 2x + 6 \end{pmatrix}, \quad x \in \mathbb{R}\n$$\nThis means the vector $ \mathbf{v} $ is not unique — it forms a line of solutions in 3D space, all lying parallel to a fixed direction vector.", "---", "### Step 5: Find a Particular Solution (Optional)", "For visualization or application, choosing $ x = 0 $ gives a simple particular solution:\n$$\n\mathbf{v}_0 = \begin{pmatrix} 0 \ -2 \ 6 \end{pmatrix}\n$$\nCheck:\n$$\n\begin{pmatrix} 0 \ -2 \ 6 \end{pmatrix} \ imes \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} = \begin{pmatrix} (-2)(2) - (6)(0) \ -(0)(2) - (6)(1) \ (0)(0) - (-2)(1) \end{pmatrix} = \begin{pmatrix} -4 \ -6 \ 2 \end{pmatrix}\n$$\nOops! The $ y $-component is $ -6 $, not $ 6 $. Wait — correction:\nThe cross product calculation confirms:\n$$\n\mathbf{v} = \begin{pmatrix} 0 \ -2 \ 6 \end{pmatrix} \Rightarrow \mathbf{v} \ imes \mathbf{a} = \begin{pmatrix} 2(-2) \ -(0 - 6) \ -(-2) \end{pmatrix} = \begin{pmatrix} -4 \ 6 \ 2 \end{pmatrix}\n$$\n✅ Correct! So our solution is verified.", "---", "### Conclusion: The Vector $ \mathbf{v} $ Satisfies the Cross Product Equation", "The vector $ \mathbf{v} $ such that:\n$$\n\mathbf{v} \ imes \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} = \begin{pmatrix} -4 \ 6 \ 2 \end{pmatrix}\n$$\nis any vector of the form:\n$$\n\mathbf{v} = \begin{pmatrix} x \ -2 \ 2x + 6 \end{pmatrix}, \quad x \in \mathbb{R}\n$$\nThis illustrates how cross product equations often yield a general solution involving one free parameter — common when solving vector problems in physics and linear algebra.", "---", "### SEO Keywords:\nvector cross product solution, solve v × (1,0,2) = (-4,6,2), find vector v cross product,\nfind vector v such that v × i = -4j + 6k, cross product equation solution,\nvector algebra, solve v × a = b, cross product inverse problem", "---", "Ready to apply cross product techniques? Use this method anytime you need to find a vector given its cross product with another vector!"]