If 3 hat{i}-5 hat{j}+2 hat{k}, 7 hat{i}+2 hat | vedclass

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(D) Given position vectors are $\vec{a} = 3\hat{i}-5\hat{j}+2\hat{k}$, $\vec{b} = 7\hat{i}+2\hat{j}-4\hat{k}$, $\vec{c} = \hat{i}-3\hat{j}+4\hat{k}$,

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$\pi$

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(D) Given position vectors are $\vec{a} = 3\hat{i}-5\hat{j}+2\hat{k}$, $\vec{b} = 7\hat{i}+2\hat{j}-4\hat{k}$, $\vec{c} = \hat{i}-3\hat{j}+4\hat{k}$, and $\vec{d} = -7\hat{i}-17\hat{j}+16\hat{k}$.First, we find the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$:$\overrightarrow{AB} = \vec{b} - \vec{a} = (7-3)\hat{i} + (2-(-5))\hat{j} + (-4-2)\hat{k} = 4\hat{i} + 7\hat{j} - 6\hat{k}$.$\overrightarrow{CD} = \vec{d} - \vec{c} = (-7-1)\hat{i} + (-17-(-3))\hat{j} + (16-4)\hat{k} = -8\hat{i} - 14\hat{j} + 12\hat{k}$.Notice that $\overrightarrow{CD} = -2(4\hat{i} + 7\hat{j} - 6\hat{k}) = -2\overrightarrow{AB}$.Since $\overrightarrow{CD}$ is a negative scalar multiple of $\overrightarrow{AB}$, the vectors are anti-parallel.Therefore, the angle $	heta$ between $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is $\pi$ radians.

If $3 \hat{i}-5 \hat{j}+2 \hat{k}, 7 \hat{i}+2 \hat{j}-4 \hat{k}, \hat{i}-3 \hat{j}+4 \hat{k}$ and $-7 \hat{i}-17 \hat{j}+16 \hat{k}$ are position vectors of the points $A, B, C$ and $D$ respectively, then the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is

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