#### 75000Question: Given vectors $\mathbf{c} = egin{pmatrix} 2 \ -1 \ 3 \end{pmatrix}$ and $\mathbf{d} = egin{pmatrix} -4 \ 5 \ 1 \end{pmatrix}$, find the vector $\mathbf{v}$ such that $\mathbf{v} imes \mathbf{c} = \mathbf{d}$.
["# How to Solve for Vector $\mathbf{v}$ in $\mathbf{v} \ imes \mathbf{c} = \mathbf{d}$: A Step-by-Step Guide", "When faced with the vector equation $\mathbf{v} \ imes \mathbf{c} = \mathbf{d}$, solving for $\mathbf{v}$ may seem challenging at first—especially since the cross product does not have a direct inverse like matrix equations. However, given specific vectors $\mathbf{c} = \begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix}$ and $\mathbf{d} = \begin{pmatrix} -4 \ 5 \ 1 \end{pmatrix}$, we can systematically determine $\mathbf{v}$ using fundamental principles of vector algebra.", "## Understanding the Cross Product Equation", "The cross product $\mathbf{v} \ imes \mathbf{c} = \mathbf{d}$ implies that the resulting vector $\mathbf{d}$ must be perpendicular to both $\mathbf{v}$ and $\mathbf{c}$. A crucial property is that $\mathbf{c} \cdot \mathbf{d} = 0$ — if this condition is not satisfied, no solution exists.", "Check the dot product:", "$$\n\mathbf{c} \cdot \mathbf{d} = (2)(-4) + (-1)(5) + (3)(1) = -8 - 5 + 3 = -10 <br/>\neq 0\n$$", "Wait—this is not zero! Since $\mathbf{c} \cdot \mathbf{d} <br/>\ne 0$, the equation $\mathbf{v} \ imes \mathbf{c} = \mathbf{d}$ has no solution, because the vector $\mathbf{d}$ is not perpendicular to $\mathbf{c}$, violating a geometric constraint of cross products.", "However, if this dot product were zero, we could proceed to solve for $\mathbf{v}$. In many applied contexts—such as robotics, physics, or computer graphics—this condition arises naturally, so understanding when a solution exists is just as important as finding one.", "## When a Solution Exists: Using Matrix Determinants", "Suppose we were given vectors $\mathbf{c}$ and $\mathbf{d}$ such that $\mathbf{c} \cdot \mathbf{d} = 0$. Then $\mathbf{v}$ can be found using a method involving a helper vector and a determinant-based formula.", "Let $\mathbf{v} = \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix}$. The cross product $\mathbf{v} \ imes \mathbf{c}$ is:", "$$\n\mathbf{v} \ imes \mathbf{c} = \begin{vmatrix}\n\mathbf{i} & \mathbf{j} & \mathbf{k} \\nv_1 & v_2 & v_3 \\n2 & -1 & 3\n\end{vmatrix}\n= \mathbf{i}(v_2 \cdot 3 - v_3 \cdot (-1)) - \mathbf{j}(v_1 \cdot 3 - v_3 \cdot 2) + \mathbf{k}(v_1 \cdot (-1) - v_2 \cdot 2)\n$$\n$$\n= \begin{pmatrix}\n3v_2 + v_3 \\n-3v_1 + 2v_3 \\n-v_1 - 2v_2\n\end{pmatrix}\n$$", "Set this equal to $\mathbf{d} = \begin{pmatrix} -4 \ 5 \ 1 \end{pmatrix}$:", "$$\n\begin{cases}\n3v_2 + v_3 = -4 \\n-3v_1 + 2v_3 = 5 \\n-v_1 - 2v_2 = 1\n\end{cases}\n$$", "This system can be solved using substitution or matrix methods. However, recall: since $\mathbf{c} \cdot \mathbf{d} = -10 <br/>\ne 0$, no solution exists—the system is inconsistent.", "## Practical Insight and Next Steps", "If you encounter $\mathbf{v} \ imes \mathbf{c} = \mathbf{d}$ in practice and check $\mathbf{c} \cdot \mathbf{d} <br/>\ne 0$, there is no such vector $\mathbf{v}$. This fact prevents unnecessary computation and signals the need to revisit assumptions in physics models, error inputs, or coordinate system alignments.", "For learning purposes, however, mastering vector cross products and their properties strengthens your ability in linear algebra, electromagnetism, and kinematics.", "## Final Summary", "- The equation $\mathbf{v} \ imes \mathbf{c} = \mathbf{d}$ has a solution if and only if $\mathbf{c} \cdot \mathbf{d} = 0$.\n- With $\mathbf{c} = \begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix}$, $\mathbf{d} = \begin{pmatrix} -4 \ 5 \ 1 \end{pmatrix}$, compute:\n $$\n \mathbf{c} \cdot \mathbf{d} = -8 - 5 + 3 = -10 <br/>\ne 0\n $$\n- Hence, no solution exists.\n- Always verify orthogonality of $\mathbf{c}$ and $\mathbf{d}$ before attempting to solve for $\mathbf{v}$.", "Understanding these principles equips you to handle vector equations confidently in science and engineering applications.", "---", "Keywords: vector cross product solution, solve $\mathbf{v} \ imes \mathbf{c} = \mathbf{d}$, when does $\mathbf{v}$ exist, $\mathbf{c} \cdot \mathbf{d} = 0 condition, vector algebra, physics applications."]
