The projection of overline{AB} on overline{CD | vedclass

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(B) 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

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$\frac{1}{7}$

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(B) 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 scalar projection of vector $\overline{AB}$ on vector $\overline{CD}$ is given by the formula:$	ext{Projection} = \frac{\overline{AB} \cdot \overline{CD}}{|\overline{CD}|}$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 of $\overline{CD}$:$|\overline{CD}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$Thus,the scalar projection is:$\frac{-1}{7}$Note: The magnitude of the projection is $|\frac{-1}{7}| = \frac{1}{7}$.

The 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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