The vector projection of overline{AB} on over | vedclass

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(C) First,we find the vectors $\overline{AB}$ and $\overline{CD}$:$\overline{AB} = (1-2)\hat{i} + (-4-(-3))\hat{j} + (-2-0)\hat{k} = -\hat{i} - \hat{j

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$-\frac{1}{49}(3 \hat{i}-6 \hat{j}+2 \hat{k})$

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(C) First,we find the vectors $\overline{AB}$ and $\overline{CD}$:$\overline{AB} = (1-2)\hat{i} + (-4-(-3))\hat{j} + (-2-0)\hat{k} = -\hat{i} - \hat{j} - 2\hat{k}$$\overline{CD} = (7-4)\hat{i} + (0-6)\hat{j} + (10-8)\hat{k} = 3\hat{i} - 6\hat{j} + 2\hat{k}$The vector projection of $\overline{AB}$ on $\overline{CD}$ is given by the formula:$	ext{Vector Projection} = (\overline{AB} \cdot \hat{CD}) \hat{CD} = (\overline{AB} \cdot \overline{CD}) \frac{\overline{CD}}{|\overline{CD}|^2}$Calculate the dot product $\overline{AB} \cdot \overline{CD}$:$\overline{AB} \cdot \overline{CD} = (-1)(3) + (-1)(-6) + (-2)(2) = -3 + 6 - 4 = -1$Calculate the magnitude squared $|\overline{CD}|^2$:$|\overline{CD}|^2 = 3^2 + (-6)^2 + 2^2 = 9 + 36 + 4 = 49$Substitute these values into the formula:$	ext{Vector Projection} = (-1) \frac{3\hat{i} - 6\hat{j} + 2\hat{k}}{49} = -\frac{1}{49}(3\hat{i} - 6\hat{j} + 2\hat{k})$

The vector projection of $\overline{AB}$ on $\overline{CD}$,where $A \equiv(2,-3,0), B \equiv(1,-4,-2), C \equiv(4,6,8)$ and $D \equiv(7,0,10)$,is

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