+ "question_response_details": "This is the question I am currently working on. I am currently working on Part (c). 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 ( Hard );\n Guidance to Solve the Question: None;\n Description of Question: ;\n Expected Time to Complete the Question: 3 - 7 min;\n Time Spent on the Question This Session: 4 minutes which is a good amount of time spent on this question.;\n \n # Part (a):\n Part Content: True or False: The formula for the dot product is always equal to multiplying the corresponding components of two vectors and summing them up;\n \n ## Response Area: 1\n \n (Secret) Expected Answer: false;\n My Latest Submission: none made;\n No direct answer\n No worked solutions for part (a);\n \n # Part (b):\n Part Content: Given that $\\vec{a}=\\begin{bmatrix}4 \\\\ -2 \\\\ -5 \\\\ 2\\end{bmatrix}$, what is its length?;\n \n ## Response Area: 1\n \n (Secret) Expected Answer: 7;\n My Latest Submission: none made;\n No direct answer\n \n ## Worked Solution 1: \n We can find the length of vector $\\vec{a}$ using the dot product\n; ;\n$$\n|\\vec{a}| = \\sqrt{\\vec{a} \\cdot \\vec{a}}\n$$\n; ;\nFirst, let's calculate the dot product of $\\vec{a}$ with itself:\n; ;\n$$\n\\begin{array}{rl}\\vec{a} \\cdot \\vec{a} &= \\begin{bmatrix}4 \\\\ -2 \\\\ -5 \\\\ 2\\end{bmatrix} \\cdot \\begin{bmatrix}4 \\\\ -2 \\\\ -5 \\\\ 2\\end{bmatrix} \\\\\\\\&= 4^2+(-2)^2+(-5)^2+2^2 \\\\\\\\ &= 49\\end{array}\n$$\n; ;\nNow, we can calculate the length of $\\vec{a}$:\n; ;\n$$\n\\begin{array}{rl}|\\vec{a}|&=\\sqrt{\\vec{a} \\cdot \\vec{a}} \\\\\\\\ &=\\sqrt{49} \\\\\\\\ &= 7\\end{array} \n$$\n; ;\nTherefore, the length of vector $\\vec{a}$ is approximately 7 arbitrary units\n \n \n # [CURRENTLY WORKING ON] Part (c):\n Part Content: Given that $\\vec{N}=\\begin{bmatrix}3 \\\\ 4 \\end{bmatrix}$, construct a new vector, $\\vec{n}$ , that is pointing along the same direction but is of unit length;\n \n ## Response Area: 1\n \n (Secret) Expected Answer: 0.6,0.8;\n My Latest Submission: none made;\n No direct answer\n \n ## Worked Solution 1: \n To construct a vector pointing in the same direction as $\\vec{N}$ but with unit length, we need to normalize it. Normalising a vector involves scaling the vector with the reciprocal of its length.\n; ;\n***\n\nFirst, let's calculate the length of $\\vec{N}$\n; ;\n$$\n\\begin{array}{rl}|\\vec{N}|&=\\sqrt{\\vec{N} \\cdot \\vec{N}} \\\\\\\\&= \\sqrt{\\begin{bmatrix}3 \\\\ 4\\end{bmatrix} \\cdot \\begin{bmatrix} 3 \\\\ 4 \\end{bmatrix}} \\\\\\\\&= \\sqrt{3^2+4^2}\\\\\\\\&= \\sqrt{25} \\\\\\\\ &= 5\\end{array}\n$$\n; ;\n***\n; ;\nNow, we can normalise $\\vec{N}$ by multiplying it by the reciprocal of its length\n; ;\n$$\n\\begin{array}{rl}\\vec{n}&=\\frac{1}{|\\vec{N}|} \\ \\vec{N}\\\\\\\\ &=\\frac{1}{5}\\begin{bmatrix}3 \\\\ 4\\end{bmatrix} \\\\\\\\ &= \\begin{bmatrix}0.6 \\\\ 0.8\\end{bmatrix}\\end{array}\n$$\n \n \n # Part (d):\n Part Content: Between the two vectors $\\vec{U}$ and $\\vec{W}$, which one of them has an orientation that is closer to $\\vec{V}$?\n; ;\n$$\n\\vec{U}=\\begin{bmatrix}9 \\\\ -4 \\\\ -5 \\\\ 3\\end{bmatrix} \\quad \\vec{W}=\\begin{bmatrix}-4 \\\\ 7 \\\\ -2 \\\\ -5\\end{bmatrix} \\quad \\vec{V}=\\begin{bmatrix}3 \\\\ 9 \\\\ -1 \\\\ -2\\end{bmatrix}\n$$;\n \n ## Response Area: 1\n \n (Secret) Expected Answer: [false] U,[true] W;\n My Latest Submission: none made;\n No direct answer\n \n ## Worked Solution 1: \n No content\n \n ## Worked Solution 1: Stupid but more conceptually involved method\n The dot product provides information about the alignment of two vectors. Intuitively, a larger dot product indicates a closer alignment between the vectors. However, the dot product is also influenced by the magnitudes of the vectors. To ensure a fair comparison of orientations, it is necessary to normalize the vectors involved ( we don't have to normalise $\\vec{V}$ because its magnitude will influence both dot products to the same extent ). By normalising the vectors and then computing the dot product, we can accurately assess their relative orientations independent of their magnitudes.\n; ;\n***\n; ;\n### Normalisation\n; ;\nWe first have to find the length of each vector\n; ;\n$$\n\\begin{matrix}\n\\small\\boxed{\\begin{array}{rl}|\\vec{U}|&=\\sqrt{\\vec{U} \\cdot \\vec{U}} \\\\\\\\&= \\sqrt{\\begin{bmatrix}9 \\\\ -4 \\\\ -5 \\\\ 3\\end{bmatrix} \\cdot \\begin{bmatrix}9 \\\\ -4 \\\\ -5 \\\\ 3\\end{bmatrix}} \\\\\\\\&= \\sqrt{9^2+(-4)^2+(-5)^2+3^2}\\\\\\\\&= \\sqrt{131}\\end{array}}\n&&\n\\small\\boxed{\\begin{array}{rl}|\\vec{W}|&=\\sqrt{\\vec{W} \\cdot \\vec{W}} \\\\\\\\&= \\sqrt{\\begin{bmatrix}-4 \\\\ 7 \\\\ -2 \\\\ -5\\end{bmatrix} \\cdot \\begin{bmatrix}-4 \\\\ 7 \\\\ -2 \\\\ -5\\end{bmatrix}} \\\\\\\\&= \\sqrt{(-4)^2+7^2+(-2)^2+(-5)^2}\\\\\\\\&= \\sqrt{94}\\end{array}}\n\\end{matrix}\n$$\n; ;\n***\n; ;\nWe can then normalise them\n; ;\n$$\n\\begin{matrix}\\boxed{\\begin{array}{rl}\\vec{u}&=\\frac{1}{|\\vec{U}|} \\ \\vec{U}\\\\\\\\ &=\\frac{1}{\\sqrt{131}}\\scriptsize\\begin{bmatrix}9 \\\\ -4 \\\\ -5 \\\\ 3\\end{bmatrix} \\end{array}} \\quad \\boxed{\\begin{array}{rl}\\vec{w}&=\\frac{1}{|\\vec{W}|} \\ \\vec{W}\\\\\\\\ &=\\frac{1}{\\sqrt{94}}\\scriptsize\\begin{bmatrix}-4 \\\\ 7 \\\\ -2 \\\\ -5\\end{bmatrix} \\end{array}}\\end{matrix}\n$$\n; ;\n***\n; ;\n### Dot Product\n; ;\nWe can now compute the dot product between the normalised vectors and $\\vec{V}$\n\n$$\n\\begin{matrix}\n\\boxed{\\begin{array}{rl}\n\\vec{u}\\cdot\\vec{V}&=\n\\left(\\frac{1}{\\sqrt{131}}\\scriptsize\\begin{bmatrix}9 \\\\ -4 \\\\ -5 \\\\ 3\\end{bmatrix}\\right)•\\scriptsize \\begin{bmatrix}3 \\\\ 9 \\\\ -1 \\\\ -2\\end{bmatrix}\n\\\\\\\\\n&=\\frac{1}{\\sqrt{131}}\\left(\\scriptsize\\begin{bmatrix}9 \\\\ -4 \\\\ -5 \\\\ 3\\end{bmatrix}•\\begin{bmatrix}3 \\\\ 9 \\\\ -1 \\\\ -2\\end{bmatrix}\\right)\n\\\\\\\\\n&=\\frac{1}{\\sqrt{131}}\\scriptsize[(9\\cdot3)+(-4\\cdot9)+(-5\\cdot-1)+(3\\cdot-2)]\n\\\\\\\\\n&≈-0.8737\n \\end{array}}\n\\\\\\\\\n\\boxed{\\begin{array}{rl}\n\\vec{w}\\cdot\\vec{V}&=\n\\left(\\frac{1}{\\sqrt{94}}\\scriptsize\\begin{bmatrix}-4 \\\\ 7 \\\\ -2 \\\\ -5\\end{bmatrix}\\right)•\\scriptsize \\begin{bmatrix}3 \\\\ 9 \\\\ -1 \\\\ -2\\end{bmatrix}\n\\\\\\\\\n&=\\frac{1}{\\sqrt{94}}\\left(\\scriptsize\\begin{bmatrix}-4 \\\\ 7 \\\\ -2 \\\\ -5\\end{bmatrix}•\\begin{bmatrix}3 \\\\ 9 \\\\ -1 \\\\ -2\\end{bmatrix}\\right)\n\\\\\\\\\n&=\\frac{1}{\\sqrt{94}}\\scriptsize[(-4\\cdot3)+(7\\cdot9)+(-2\\cdot-1)+(-5\\cdot-2)]\n\\\\\\\\\n&≈6.4980\n \\end{array}}\n\\end{matrix}\n$$\n; ;\nComparing the magnitudes of $\\vec{w}\\cdot\\vec{V}$ and $\\vec{u}\\cdot\\vec{V}$, we find that $\\vec{w}\\cdot\\vec{V}$ is larger than $\\vec{w}\\cdot\\vec{V}$.\n; ;\nTherefore, the vector $\\vec{W}$ has an orientation that is closer to $\\vec{V}$.\n \n ## Worked Solution 2: Plugging in numbers\n To determine which vector, $\\vec{U}$ or $\\vec{W}$, has an orientation closer to $\\vec{V}$ using the dot product-cosine formula\n; ;\n***\n1. **Calculate the dot products:**\n ; ;\n $$\n \\vec{U}\\cdot\\vec{V}=(9 \\cdot 3) + (-4 \\cdot 9) + (-5 \\cdot -1) + (3 \\cdot -2) = -10 \\\\ \\ \\\\ \\vec{W} \\cdot \\vec{V} = (-4 \\cdot 3) + (7 \\cdot 9) + (-2 \\cdot -1) + (-5 \\cdot -2) = 63\n $$\n ; ;\n ***\n2. **Calculate the magnitudes:**\n ; ;\n $$\n \\begin{matrix}\n \\small\\boxed{\\begin{array}{rl}|\\vec{U}|&=\\sqrt{\\vec{U} \\cdot \\vec{U}} \\\\\\\\&= \\sqrt{\\begin{bmatrix}9 \\\\ -4 \\\\ -5 \\\\ 3\\end{bmatrix} \\cdot \\begin{bmatrix}9 \\\\ -4 \\\\ -5 \\\\ 3\\end{bmatrix}} \\\\\\\\&= \\sqrt{9^2+(-4)^2+(-5)^2+3^2}\\\\\\\\&= \\sqrt{131}\\end{array}}\n &&\n \\small\\boxed{\\begin{array}{rl}|\\vec{W}|&=\\sqrt{\\vec{W} \\cdot \\vec{W}} \\\\\\\\&= \\sqrt{\\begin{bmatrix}-4 \\\\ 7 \\\\ -2 \\\\ -5\\end{bmatrix} \\cdot \\begin{bmatrix}-4 \\\\ 7 \\\\ -2 \\\\ -5\\end{bmatrix}} \\\\\\\\&= \\sqrt{(-4)^2+7^2+(-2)^2+(-5)^2}\\\\\\\\&= \\sqrt{94}\\end{array}}\n \\end{matrix}\n \\\\ \\ \\\\\n \\small\\boxed{\\begin{array}{rl}|\\vec{V}|&=\\sqrt{\\vec{V} \\cdot \\vec{V}} \\\\\\\\&= \\sqrt{\\begin{bmatrix}3 \\\\ 9 \\\\ -1 \\\\ -2\\end{bmatrix} \\cdot \\begin{bmatrix}3 \\\\ 9 \\\\ -1 \\\\ -2\\end{bmatrix}} \\\\\\\\&= \\sqrt{(3)^2+9^2+(-1)^2+(-2)^2}\\\\\\\\&= \\sqrt{95}\\end{array}}\n\n $$\n ; ;\n ; ;\n ***\n3. **Calculate the cosines of the angles:**\n \n $$\n \\cos(\\theta_{\\vec{U}\\vec{V}}) = \\frac{\\vec{U}\\cdot\\vec{V}}{|\\vec{U}| |\\vec{V}|} = \\frac{-10}{\\sqrt{131} \\cdot \\sqrt{95}}\n \\\\ \\ \\\\ \\ \\\\\n \\cos(\\theta_{\\vec{W}\\vec{V}}) = \\frac{\\vec{W}\\cdot\\vec{V}}{|\\vec{W}| |\\vec{V}|} = \\frac{62}{\\sqrt{94} \\cdot \\sqrt{95}}\n $$\n ; ;\n ***\n4. **Calculate and compare the angles**\n \n $$\n \\theta_{\\vec{U}\\vec{V}}=\\cos^{-1}\\left(\\frac{-10}{\\sqrt{131} \\cdot \\sqrt{95}}\\right)≈95.1^\\circ\n \\\\ \\ \\\\ \\ \\\\\n \\theta_{\\vec{W}\\vec{V}}=\\cos^{-1}\\left(\\frac{62}{\\sqrt{94} \\cdot \\sqrt{95}}\\right)≈49.0^\\circ\n $$\n \n Since $\\theta_{\\vec{W}\\vec{V}}<\\theta_{\\vec{U}\\vec{V}}$ , the orientation of $\\vec{W}$ is closer to $\\vec{V}$ then that of \\$\\vec{U}\\$\n \n ",
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