السؤال: أوجد المتجه $\mathbf{v}$ بحيث أن $\mathbf{v} \times \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix}$.

["Title: Determine the Vector $\mathbf{v}$ Such That $\mathbf{v} \ imes \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 3 \ -2 \end{pmatrix}$ — A Step-by-Step Solution", "---", "Introduction", "In vector calculus, the cross product is a fundamental operation that yields a vector orthogonal to the input vectors. One common problem is finding a vector $\mathbf{v}$ given the result of a cross product with a known basis vector. This article explains how to find $\mathbf{v}$ such that:", "$$\n\mathbf{v} \ imes \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 3 \ -2 \end{pmatrix}.\n$$", "We will solve this step-by-step using properties of the cross product and algebra.", "---", "Understanding the Cross Product", "Let $\mathbf{v} = \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix}$ and $\mathbf{a} = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}$. The cross product $\mathbf{v} \ imes \mathbf{a}$ is computed using the determinant:", "$$\n\mathbf{v} \ imes \mathbf{a} = \begin{vmatrix}\n\mathbf{i} & \mathbf{j} & \mathbf{k} \\nv_1 & v_2 & v_3 \\n1 & 0 & 0 \\n\end{vmatrix} = \begin{pmatrix}\n0 \cdot v_3 - 0 \cdot v_2 \\n-(v_1 \cdot 0 - v_3 \cdot 1) \\nv_1 \cdot 0 - v_2 \cdot 1\n\end{pmatrix} = \begin{pmatrix} 0 \ v_3 \ -v_2 \end{pmatrix}.\n$$", "---", "Applying the Given Condition", "We equate this result to the given vector:", "$$\n\begin{pmatrix} 0 \ v_3 \ -v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 3 \ -2 \end{pmatrix}.\n$$", "By matching components:", "1. First component: $0 = 0$ — always true (no constraint).\n2. Second component: $v_3 = 3$.\n3. Third component: $-v_2 = -2 \Rightarrow v_2 = 2$.", "Note that $v_1$ is not determined — it does not affect the cross product with $\begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}$.", "---", "General Solution", "Thus, any vector $\mathbf{v}$ satisfying the equation must be of the form:", "$$\n\mathbf{v} = \begin{pmatrix} v_1 \ 2 \ 3 \end{pmatrix}, \quad \ ext{where } v_1 \in \mathbb{R}.\n$$", "This means the solution is a one-parameter family of vectors lying in the plane where $v_2 = 2$, $v_3 = 3$, and $v_1$ is arbitrary.", "---", "Why $v_1$ is Free", "The cross product $\mathbf{v} \ imes \mathbf{a}$ depends only on the components $v_2$ and $v_3$, since:", "- The $i$-component is $v_2 v_3' - v_3 v_2' = v_2 \cdot 0 - v_3 \cdot 0 = 0$,\n- The $j$-component involves $v_1$ and $v_3$, but in our case it simplifies to just $v_3$,\n- The $k$-component involves $v_1 v_2 - v_2 v_1 = 0$.", "Thus, $v_1$ does not influence the result and is free.", "---", "APPLICATION IN PHYSICS AND ENGINEERING", "Cross products model rotational phenomena, such as torque and angular velocity. Knowing a resultant vector and one input vector allows determination of possible force or motion vectors—important for designing mechanical systems or analyzing particle interactions.", "---", "Conclusion", "The vector $\mathbf{v}$ satisfying\n$$\n\mathbf{v} \ imes \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 3 \ -2 \end{pmatrix}\n$$\nexists and is given by:", "$$\n\mathbf{v} = \begin{pmatrix} t \ 2 \ 3 \end{pmatrix}, \quad t \in \mathbb{R}.\n$$", "This infinite family represents all possible solutions, with $v_1$ fully unbounded.", "---", "Keywords:\nvector cross product, solve for vector v, $\mathbf{v} \ imes \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 3 \ -2 \end{pmatrix}$, solution formula, $v_1$ free, physics applications", "---", "Meta Description:", "Find the vector $\mathbf{v}$ such that $\mathbf{v} \ imes \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 3 \ -2 \end{pmatrix}$. Learn how $v_2 = 2$, $v_3 = 3$, and $v_1$ is arbitrary via step-by-step vector algebra.", "---", "Further Reading:", "- Cross product properties and geometric interpretation\n- Solving vector equations using determinants\n- Applications of cross products in physics", "---", "Author’s Note:\nAlways verify solutions by substitution—plugging $\mathbf{v} = \begin{pmatrix} 0 \ 2 \ 3 \end{pmatrix}$ confirms:\n$$\n\begin{pmatrix} 0 \ 2 \ 3 \end{pmatrix} \ imes \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 3 \ -2 \end{pmatrix}.\n$$\nThis validates our solution."]