Solution: Let $ \mathbf{a} = \langle 1, 2, 3 \rangle $, $ \mathbf{b} = \langle 1, -2, 1 \rangle $. We are to find $ \mathbf{v} $ such that:

["Solution: Finding Vector $ \mathbf{v} $ Such That $ \mathbf{v} \ imes \mathbf{a} = \mathbf{b} $, Given $ \mathbf{a} = \langle 1, 2, 3 \rangle $, $ \mathbf{b} = \langle 1, -2, 1 \rangle $", "When solving for a vector $ \mathbf{v} $ such that\n$$\n\mathbf{v} \ imes \mathbf{a} = \mathbf{b},\n$$\nwhere $ \mathbf{a} = \langle 1, 2, 3 \rangle $ and $ \mathbf{b} = \langle 1, -2, 1 \rangle $, we enter a fundamental operation in vector calculus: computing the cross product in reverse. This problem is common in physics, engineering, and computer graphics—anywhere 3D spatial reasoning is required.", "### Understanding the Cross Product Equation", "The cross product $ \mathbf{v} \ imes \mathbf{a} = \mathbf{b} $ means that $ \mathbf{b} $ is perpendicular to both $ \mathbf{v} $ and $ \mathbf{a} $. For a solution $ \mathbf{v} $ to exist, $ \mathbf{b} $ must lie in the plane perpendicular to $ \mathbf{a} $, i.e., $ \mathbf{b} \cdot \mathbf{a} = 0 $. Let’s verify this condition first.", "Compute the dot product:\n$$\n\mathbf{a} \cdot \mathbf{b} = (1)(1) + (2)(-2) + (3)(1) = 1 - 4 + 3 = 0.\n$$\nSince the dot product is zero, $ \mathbf{b} \perp \mathbf{a} $, confirming feasibility—there may be one or more solutions, depending on vector alignment.", "### Can We Find $ \mathbf{v} $? Solving the System", "Let $ \mathbf{v} = \langle x, y, z \rangle $. Then,\n$$\n\mathbf{v} \ imes \mathbf{a} = \n\begin{vmatrix}\n\mathbf{i} & \mathbf{j} & \mathbf{k} \\nx & y & z \\n1 & 2 & 3 \\n\end{vmatrix}\n= \langle 3y - 2z, -(3x - z), 2x - y \rangle = \langle 3y - 2z, -3x + z, 2x - y \rangle.\n$$\nSet this equal to $ \mathbf{b} = \langle 1, -2, 1 \rangle $, giving the system:\n$$\n\begin{cases}\n3y - 2z = 1 \quad \ ext{(1)}\\n-3x + z = -2 \quad \ ext{(2)}\\n2x - y = 1 \quad \ ext{(3)}\n\end{cases}\n$$", "### Step-by-Step Solution", "From equation (3):\n$$\ny = 2x - 1 \quad \ ext{(a)}\n$$", "From equation (2):\n$$\nz = 3x - 2 \quad \ ext{(b)}\n$$", "Substitute (a) and (b) into equation (1):\n$$\n3(2x - 1) - 2(3x - 2) = 1 \\n6x - 3 - 6x + 4 = 1 \\n1 = 1.\n$$", "This identity confirms consistency—our system has infinitely many solutions, parameterized by $ x $. Thus, the general solution is expressed in terms of a free variable.", "Using (a) and (b), all solutions are:\n$$\n\mathbf{v} = \langle x,, 2x - 1,, 3x - 2 \rangle = x\langle 1, 2, 3 \rangle + \langle 0, -1, -2 \rangle.\n$$", "Note: The vector $ \langle 0, -1, -2 \rangle $ is a particular solution, and $ x\langle 1,2,3 \rangle $ represents the homogeneous solution (since the zero vector cross product is zero, and any scalar multiple is valid).", "### Selecting a Specific Solution", "To illustrate, choose $ x = 0 $. Then:\n$$\n\mathbf{v} = \langle 0, -1, -2 \rangle.\n$$\nVerify:\n$$\n\mathbf{v} \ imes \mathbf{a} = \n\langle 1, 2, 3 \rangle \ imes \langle 0, -1, -2 \rangle = \n\langle (2)(-2) - (3)(-1), -[(1)(-2) - (3)(0)], (1)(-1) - (2)(0) \rangle = \langle -4 + 3, -(-2), -1 \rangle = \langle -1, 2, -1 \rangle.\n$$\nWait—this does not match $ \mathbf{b} = \langle 1, -2, 1 \rangle $. There’s a sign error in sign convention.", "Let’s recheck. Our general solution form is correct: $ \mathbf{v} = x\mathbf{a} + \mathbf{v}_0 $, where $ \mathbf{v}_0 $ is a particular solution.", "We found $ \mathbf{v}_0 = \langle 0, -1, -2 \rangle $, but verification fails. Re-solve the system carefully.", "Go back to equations:\n1. $ 3y - 2z = 1 $\n2. $ -3x + z = -2 \Rightarrow z = 3x - 2 $\n3. $ 2x - y = 1 \Rightarrow y = 2x - 1 $", "Plug into (1):\n$$\n3(2x - 1) - 2(3x - 2) = 6x - 3 - 6x + 4 = 1 \quad \ ext{✓}\n$$\nSo algebra is correct, and the identity $ 1 = 1 $ confirms the entire line of solutions works.", "Now recompute cross product with general $ \mathbf{v} = \langle x, 2x-1, 3x-2 \rangle $, $ \mathbf{a} = \langle 1,2,3 \rangle $:", "$$\n\mathbf{v} \ imes \mathbf{a} = \langle (2y - 3z), -(3x - z), 2x - y \rangle\n$$", "Compute components using $ y = 2x - 1 $, $ z = 3x - 2 $:", "- $ 3y - 2z = 3(2x - 1) - 2(3x - 2) = 6x - 3 - 6x + 4 = 1 $ ✅\n- $ -3x + z = -3x + 3x - 2 = -2 $ ✅\n- $ 2x - y = 2x - (2x - 1) = 1 $ ✅", "Thus, the cross product is $ \langle 1, -2, 1 \rangle $, exactly $ \mathbf{b} $. Confirmation complete.", "### Final Answer", "All vectors $ \mathbf{v} $ satisfying $ \mathbf{v} \ imes \mathbf{a} = \mathbf{b} $ are of the form:\n$$\n\mathbf{v} = \langle x,, 2x - 1,, 3x - 2 \rangle, \quad x \in \mathbb{R}.\n$$\nThis represents a line of solutions in 3D space—spanning infinitely many vectors that produce $ \mathbf{b} $ when crossed with $ \mathbf{a} $. A particular solution occurs when $ x = 0 $:\n$$\n\mathbf{v} = \langle 0,, -1,, -2 \rangle.\n$$\nVerification earlier appears flawed due to miscalculation. Re-evaluation confirms its correctness.", "### Why This Matters", "Understanding vector equations like $ \mathbf{v} \ imes \mathbf{a} = \mathbf{b} $ enables solutions in:", "- Physics: Calculating angular momentum $ \mathbf{L} = \mathbf{r} \ imes \mathbf{p} $, where $ \mathbf{v} $ may represent velocity in constraint systems.\n- Computer Graphics: Determining normal vectors from cross products for rendering and lighting.\n- Robotics: Computing required joint forces via spatial vector relations.", "Though the solution set is infinite, the structure reveals deep symmetry in 3D vector spaces—highlighting the interplay between perpendicularity, linear independence, and cross product invertibility.", "---", "Key Takeaways:\n- Cross product equations may have no solution or infinitely many depending on $ \mathbf{b} \cdot \mathbf{a} = 0 $.\n- When consistent, solutions form a line through a particular vector.\n- Particular solutions can be found via inspection (e.g., $ x = 0 $ gives $ \langle 0, -1, -2 \rangle $).\n- The general solution includes all vectors differing by scalar multiples of $ \mathbf{a} $, reflecting the homogeneous solution space.", "This framework empowers precise vector modeling in engineering, science, and computational design."]