The length of the projection of the line segm | vedclass

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(A) Let the points be $A(3, 4, 5)$, $B(4, 6, 3)$, $C(-1, 2, 4)$, and $D(1, 0, 5)$.First, we find the vector $\overrightarrow{AB}$ representing the lin

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$\frac{4}{3}$

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(A) Let the points be $A(3, 4, 5)$, $B(4, 6, 3)$, $C(-1, 2, 4)$, and $D(1, 0, 5)$.First, we find the vector $\overrightarrow{AB}$ representing the line segment joining $A$ and $B$:$\overrightarrow{AB} = (4-3)\hat{i} + (6-4)\hat{j} + (3-5)\hat{k} = \hat{i} + 2\hat{j} - 2\hat{k}$.Next, we find the vector $\overrightarrow{CD}$ representing the line joining $C$ and $D$:$\overrightarrow{CD} = (1 - (-1))\hat{i} + (0-2)\hat{j} + (5-4)\hat{k} = 2\hat{i} - 2\hat{j} + \hat{k}$.The length of the projection of vector $\overrightarrow{AB}$ on vector $\overrightarrow{CD}$ is given by the formula $\left| \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{CD}|} \right|$.Calculate the dot product $\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(2) + (2)(-2) + (-2)(1) = 2 - 4 - 2 = -4$.Calculate the magnitude of $\overrightarrow{CD} = |\overrightarrow{CD}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.Therefore, the length of the projection is $\left| \frac{-4}{3} \right| = \frac{4}{3}$.

The length of the projection of the line segment joining the points $(3, 4, 5)$ and $(4, 6, 3)$ on the line joining the points $(-1, 2, 4)$ and $(1, 0, 5)$ is

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