\boxed{43}**Question: A vector \(\mathbf{v}\) satisfies \(\mathbf{v} \times \mathbf{b} = \begin{pmatrix} 0 \\ 3 \\ -4 \end{pmatrix}\) for \(\mathbf{b} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}\). Find \(\mathbf{v}\).

["Understanding the Vector Cross Product Equation: Solving for (\mathbf{v}) Given (\mathbf{v} \ imes \mathbf{b} = \begin{pmatrix} 0 \ 3 \ -4 \end{pmatrix})", "When given a vector equation involving a cross product, such as\n[\n\mathbf{v} \ imes \mathbf{b} = \begin{pmatrix} 0 \ 3 \ -4 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix},\n]\nfinding the unknown vector (\mathbf{v}) involves solving a system derived from the properties of the cross product. This article explains step-by-step how to determine (\mathbf{v}).", "---", "### The Cross Product Basics", "Let\n[\n\mathbf{v} = \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix}.\n]\nThe cross product\n[\n\mathbf{v} \ imes \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ v_1 & v_2 & v_3 \ 1 & 0 & 2 \end{vmatrix}\n]\nis computed as:\n[\n\mathbf{v} \ imes \mathbf{b} = \mathbf{i}(v_2 \cdot 2 - v_3 \cdot 0) - \mathbf{j}(v_1 \cdot 2 - v_3 \cdot 1) + \mathbf{k}(v_1 \cdot 0 - v_2 \cdot 1)\n]\n[\n= \begin{pmatrix} 2v_2 \ - (2v_1 - v_3) \ -v_2 \end{pmatrix} = \begin{pmatrix} 2v_2 \ v_3 - 2v_1 \ -v_2 \end{pmatrix}.\n]", "---", "### Setting Up the Equation", "We now equate this result to the given vector:\n[\n\begin{pmatrix} 2v_2 \ v_3 - 2v_1 \ -v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 3 \ -4 \end{pmatrix}.\n]", "This gives three component equations:", "1. ( 2v_2 = 0 )\n2. ( v_3 - 2v_1 = 3 )\n3. ( -v_2 = -4 )", "---", "### Solving the System", "From equation (3):\n[\n-v_2 = -4 \implies v_2 = 4.\n]\nBut equation (1) gives:\n[\n2v_2 = 0 \implies v_2 = 0.\n]\nThe contradiction (v_2 = 0) and (v_2 = 4) suggests an issue — but wait: double-checking the cross product expansion.", "---", "### Rechecking the Cross Product Computation", "Expanding again:\n[\n\mathbf{v} \ imes \mathbf{b} = \begin{pmatrix} v_2 \cdot 2 - v_3 \cdot 0 \ -(v_1 \cdot 2 - v_3 \cdot 1) \ v_1 \cdot 0 - v_2 \cdot 1 \end{pmatrix} = \begin{pmatrix} 2v_2 \ -2v_1 + v_3 \ -v_2 \end{pmatrix}.\n]\nThe earlier result was correct. The contradiction implies we must accept that the system must be consistent.", "But here:\nFrom (1): (2v_2 = 0 \Rightarrow v_2 = 0)\nFrom (3): (-v_2 = -4 \Rightarrow v_2 = 4) — inconsistent.", "This contradiction suggests no solution unless the given vector is not compatible with (\mathbf{b}). But cross products yield vectors orthogonal to both (\mathbf{v}) and (\mathbf{b}), so (\mathbf{v} \ imes \mathbf{b}) must be orthogonal to (\mathbf{b}).", "---", "### Orthogonality Condition: Essential Check", "Check if the right-hand side vector (\begin{pmatrix} 0 \ 3 \ -4 \end{pmatrix}) is orthogonal to (\mathbf{b} = \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix}):\nDot product:\n[\n\begin{pmatrix} 0 \ 3 \ -4 \end{pmatrix} \cdot \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} = 0 \cdot 1 + 3 \cdot 0 + (-4) \cdot 2 = -8 <br/>\ne 0.\n]", "Since the cross product (\mathbf{v} \ imes \mathbf{b}) must be orthogonal to (\mathbf{b}), and (-8 <br/>\ne 0), no such vector (\mathbf{v}) exists that satisfies the equation.", "---", "### Conclusion: No Solution Exists", "Given the incompatibility — the vector (\begin{pmatrix} 0 \ 3 \ -4 \end{pmatrix}) is not orthogonal to (\mathbf{b} = \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix}) — the equation\n[\n\mathbf{v} \ imes \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} = \begin{pmatrix} 0 \ 3 \ -4 \end{pmatrix}\n]\nhas no solution.", "This highlights an essential property in vector algebra: the cross product (\mathbf{v} \ imes \mathbf{b}) is always perpendicular to (\mathbf{b}). Since the given result violates this orthogonality, no such (\mathbf{v}) exists.", "---", "### Final Takeaways", "- Always verify orthogonality: (\mathbf{v} \ imes \mathbf{b} \perp \mathbf{b}).\n- The contradiction arises here due to the dot product not being zero.\n- Solutions to (\mathbf{v} \ imes \mathbf{b} = \mathbf{c}) exist only if (\mathbf{c} \cdot \mathbf{b} = 0).\n- If no solution exists, analyze for errors or adjust the problem accordingly.", "For further practice, solving such equations requires respecting geometric constraints — orthogonality being fundamental.", "---", "Keywords: vector cross product, solve for vector, (\mathbf{v} \ imes \mathbf{b} = \begin{pmatrix}0\3\-4\end{pmatrix}), orthogonality condition, vector algebra, rejection of non-orthogonal cross products."]