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Merge pull request #5 from lambda-feedback/new_pipeline
New pipeline
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‎src/agents/utils/example_inputs/example_input_1.json

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"question_response_details": "This is the question I am currently working on. I am currently working on Part (a). Below, you'll find its details, including the parts of the question, my submissions for each response area, and feedback on my progress. This information highlights my efforts and progress so far. \n Maths equations are in KaTex format, preserve them the same.\n # Question: Dot Product;\n Guidance to Solve the Question: None;\n Description of Question: $$\n\\vec{a}=\\begin{bmatrix}1 \\\\ 3\\\\ -2\\end{bmatrix} \\quad \\vec{b}=\\begin{bmatrix}0 \\\\ 3\\\\ 1\\end{bmatrix} \\quad \\vec{c}=\\begin{bmatrix} 1 \\\\\\ -1\\\\ -3\\end{bmatrix}\n$$;\n Expected Time to Complete the Question: 1 - 4 min;\n Time Spent on the Question This Session: 72 minutes which is too much time spent on this question.;\n \n # [CURRENTLY WORKING ON] Part (a):\n Part Content: No content;\n \n ## Response Area: 1\n Area task: What is $\\vec{a} \\ \\cdot \\ \\vec{b} \\ =$ ?\n (Secret) Expected Answer: 7;\n My Latest Submission: none made;\n No direct answer\n \n ## Worked Solution 1: \n Assuming the given basis is orthonormal, the dot product between two vectors, $\\vec{a}$ and $\\vec{b}$, can be calculated simply by multiplying their corresponding components and summing them up:\n; ;\n$$\n\\begin{array}{rl}\\vec{a} \\cdot \\vec{b} &=\\begin{bmatrix}1 \\\\ 3\\\\ -2\\end{bmatrix}•\\begin{bmatrix}0 \\\\ 3\\\\ 1\\end{bmatrix} \\\\\\\\ &= (1 \\cdot 0) + (3 \\cdot 3) + (-2 \\cdot 1) \\\\\\\\ &= 0 + 9 + (-2) \\\\\\\\ &= 7\\end{array}\n$$\n; ;\nTherefore, the dot product between vectors $\\vec{a}$ and $\\vec{b}$ is \\$7\\$\n \n \n # Part (b):\n Part Content: No content;\n \n ## Response Area: 1\n Area task: What is $\\left(\\vec{a} - \\vec{b}\\right)\\cdot \\vec{c}\\ =$ ?\n (Secret) Expected Answer: 10;\n My Latest Submission: none made;\n No direct answer\n \n ## Worked Solution 1: \n To calculate $\\left(\\vec{a} - \\vec{b}\\right) \\cdot \\vec{c}$ , we first need to find the vector resulting from the subtraction of $\\vec{b}$ from $\\vec{a}$:\n; ;\n$$\n\\begin{array}{rl}\\vec{a} - \\vec{b} &= \\begin{bmatrix}1 \\\\ 3 \\\\ -2\\end{bmatrix} - \\begin{bmatrix}0 \\\\ 3 \\\\ 1\\end{bmatrix} \\\\\\\\&= \\begin{bmatrix}1 - 0 \\\\ 3 - 3 \\\\ -2 - 1\\end{bmatrix} \\\\\\\\&= \\begin{bmatrix}1 \\\\ 0 \\\\ -3\\end{bmatrix}\\end{array}\n$$\n\nNext, we can compute the dot product of $\\left(\\vec{a} - \\vec{b}\\right)$ and $\\vec{c}$\n; ;\n$$\n\\begin{array}{rl}\\left(\\vec{a} - \\vec{b}\\right) \\cdot \\vec{c} &= \\begin{bmatrix}1 \\\\ 0 \\\\ -3\\end{bmatrix} \\cdot \\begin{bmatrix}1 \\\\ -1 \\\\ -3\\end{bmatrix} \\\\\\\\ &= (1 \\cdot 1) + (0 \\cdot -1) + (-3 \\cdot -3) \\\\\\\\&= 1 + 0 + 9 \\\\\\\\&= 10\\end{array}\n$$\n\nTherefore, $\\left(\\vec{a} - \\vec{b}\\right) \\cdot \\vec{c}$ is equal to \\$10\n \n \n # Part (c):\n Part Content: No content;\n \n ## Response Area: 1\n Area task: What is $\\left( \\ \\vec{a} \\cdot \\vec{c} \\ \\right) \\vec{b}\\ =$ ?\n (Secret) Expected Answer: 0,12,4;\n My Latest Submission: none made;\n No direct answer\n \n ## Worked Solution 1: \n To calculate $\\left( \\ \\vec{a} \\cdot \\vec{c} \\ \\right) \\vec{b}$ , we first need to find the dot product of vectors $\\vec{a}$ and $\\vec{c}$\n; ;\n$$\n\\begin{array}{rl}\\vec{a} \\cdot \\vec{c} &= \\begin{bmatrix}1 \\\\ 3 \\\\ -2\\end{bmatrix} \\cdot \\begin{bmatrix}1 \\\\ -1 \\\\ -3\\end{bmatrix} \\\\\\\\ &= (1 \\cdot 1) + (3 \\cdot -1) + (-2 \\cdot -3) \\\\\\\\&= 1 - 3 + 6 \\\\\\\\&= 4\\end{array}\n$$\n; ;\nNext, we can scale $\\vec{b}$ by $\\vec{a} \\cdot \\vec{c}$\n; ;\n$$\n\\begin{array}{rl}\\left( \\ \\vec{a} \\cdot \\vec{c} \\ \\right) \\vec{b} &= 4\\begin{bmatrix}0 \\\\ 3 \\\\ 1\\end{bmatrix} \\\\\\\\ &= \\begin{bmatrix}4\\cdot0 \\\\ 4\\cdot3 \\\\ 4\\cdot1\\end{bmatrix} \\\\\\\\&= \\begin{bmatrix}0 \\\\ 12 \\\\ 4\\end{bmatrix} \\end{array}\n$$\n; ;\n \n ",
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"questionContent": "$$\n\\vec{a}=\\begin{bmatrix}1 \\\\ 3\\\\ -2\\end{bmatrix} \\quad \\vec{b}=\\begin{bmatrix}0 \\\\ 3\\\\ 1\\end{bmatrix} \\quad \\vec{c}=\\begin{bmatrix} 1 \\\\\\ -1\\\\ -3\\end{bmatrix}\n$$",
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"content": "Assuming the given basis is orthonormal, the dot product between two vectors, $\\vec{a}$ and $\\vec{b}$, can be calculated simply by multiplying their corresponding components and summing them up:\n\n   \n\n$$\n\\begin{array}{rl}\\vec{a} \\cdot \\vec{b} &=\\begin{bmatrix}1 \\\\ 3\\\\ -2\\end{bmatrix}•\\begin{bmatrix}0 \\\\ 3\\\\ 1\\end{bmatrix} \\\\\\\\ &= (1 \\cdot 0) + (3 \\cdot 3) + (-2 \\cdot 1) \\\\\\\\ &= 0 + 9 + (-2) \\\\\\\\ &= 7\\end{array}\n$$\n\n   \n\nTherefore, the dot product between vectors $\\vec{a}$ and $\\vec{b}$ is \\$7\\$"
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"content": "To calculate $\\left(\\vec{a} - \\vec{b}\\right) \\cdot \\vec{c}$ , we first need to find the vector resulting from the subtraction of $\\vec{b}$ from $\\vec{a}$:\n\n   \n\n$$\n\\begin{array}{rl}\\vec{a} - \\vec{b} &= \\begin{bmatrix}1 \\\\ 3 \\\\ -2\\end{bmatrix} - \\begin{bmatrix}0 \\\\ 3 \\\\ 1\\end{bmatrix} \\\\\\\\&= \\begin{bmatrix}1 - 0 \\\\ 3 - 3 \\\\ -2 - 1\\end{bmatrix} \\\\\\\\&= \\begin{bmatrix}1 \\\\ 0 \\\\ -3\\end{bmatrix}\\end{array}\n$$\n\n\n\nNext, we can compute the dot product of $\\left(\\vec{a} - \\vec{b}\\right)$ and $\\vec{c}$\n\n   \n\n$$\n\\begin{array}{rl}\\left(\\vec{a} - \\vec{b}\\right) \\cdot \\vec{c} &= \\begin{bmatrix}1 \\\\ 0 \\\\ -3\\end{bmatrix} \\cdot \\begin{bmatrix}1 \\\\ -1 \\\\ -3\\end{bmatrix} \\\\\\\\ &= (1 \\cdot 1) + (0 \\cdot -1) + (-3 \\cdot -3) \\\\\\\\&= 1 + 0 + 9 \\\\\\\\&= 10\\end{array}\n$$\n\n\n\nTherefore, $\\left(\\vec{a} - \\vec{b}\\right) \\cdot \\vec{c}$ is equal to \\$10"
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"content": "To calculate $\\left( \\ \\vec{a} \\cdot \\vec{c} \\ \\right) \\vec{b}$ , we first need to find the dot product of vectors $\\vec{a}$ and $\\vec{c}$\n\n   \n\n$$\n\\begin{array}{rl}\\vec{a} \\cdot \\vec{c} &= \\begin{bmatrix}1 \\\\ 3 \\\\ -2\\end{bmatrix} \\cdot \\begin{bmatrix}1 \\\\ -1 \\\\ -3\\end{bmatrix} \\\\\\\\ &= (1 \\cdot 1) + (3 \\cdot -1) + (-2 \\cdot -3) \\\\\\\\&= 1 - 3 + 6 \\\\\\\\&= 4\\end{array}\n$$\n\n   \n\nNext, we can scale $\\vec{b}$ by $\\vec{a} \\cdot \\vec{c}$\n\n   \n\n$$\n\\begin{array}{rl}\\left( \\ \\vec{a} \\cdot \\vec{c} \\ \\right) \\vec{b} &= 4\\begin{bmatrix}0 \\\\ 3 \\\\ 1\\end{bmatrix} \\\\\\\\ &= \\begin{bmatrix}4\\cdot0 \\\\ 4\\cdot3 \\\\ 4\\cdot1\\end{bmatrix} \\\\\\\\&= \\begin{bmatrix}0 \\\\ 12 \\\\ 4\\end{bmatrix} \\end{array}\n$$\n\n   \n"
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