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

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(A) First,we find the vectors $\overline{PQ}$ and $\overline{AB}$:$\overline{PQ} = (3 - (-2)) \hat{i} + (2 - 1) \hat{j} + (5 - 3) \hat{k} = 5 \hat{i}

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

Vedclass · Answer

(A) First,we find the vectors $\overline{PQ}$ and $\overline{AB}$:$\overline{PQ} = (3 - (-2)) \hat{i} + (2 - 1) \hat{j} + (5 - 3) \hat{k} = 5 \hat{i} + \hat{j} + 2 \hat{k}$$\overline{AB} = (7 - 4) \hat{i} + (-5 - (-3)) \hat{j} + (-1 - 5) \hat{k} = 3 \hat{i} - 2 \hat{j} - 6 \hat{k}$The vector projection of $\overline{PQ}$ on $\overline{AB}$ is given by the formula:$	ext{Vector Projection} = \frac{(\overline{PQ} \cdot \overline{AB}) \overline{AB}}{|\overline{AB}|^2}$Calculate the dot product $\overline{PQ} \cdot \overline{AB}$:$\overline{PQ} \cdot \overline{AB} = (5)(3) + (1)(-2) + (2)(-6) = 15 - 2 - 12 = 1$Calculate the magnitude squared $|\overline{AB}|^2$:$|\overline{AB}|^2 = 3^2 + (-2)^2 + (-6)^2 = 9 + 4 + 36 = 49$Thus,the vector projection is:$\frac{1}{49}(3 \hat{i} - 2 \hat{j} - 6 \hat{k})$

The vector projection of $\overline{PQ}$ on $\overline{AB}$,where $P \equiv (-2, 1, 3)$,$Q \equiv (3, 2, 5)$,$A \equiv (4, -3, 5)$ and $B \equiv (7, -5, -1)$ is

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