hat{i}-2 hat{j}+hat{k},2 hat{i}+hat{j}-hat{k} | vedclass

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(B) Given position vectors: $\vec{A} = \hat{i}-2 \hat{j}+\hat{k}$,$\vec{B} = 2 \hat{i}+\hat{j}-\hat{k}$,$\vec{C} = \hat{i}-\hat{j}-2 \hat{k}$.$D$ is t

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$\frac{1}{\sqrt{14}}(-\hat{i}-3 \hat{j}+2 \hat{k})$

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(B) Given position vectors: $\vec{A} = \hat{i}-2 \hat{j}+\hat{k}$,$\vec{B} = 2 \hat{i}+\hat{j}-\hat{k}$,$\vec{C} = \hat{i}-\hat{j}-2 \hat{k}$.$D$ is the midpoint of $BC$,so $\vec{D} = \frac{\vec{B}+\vec{C}}{2} = \frac{(2+1)\hat{i} + (1-1)\hat{j} + (-1-2)\hat{k}}{2} = \frac{3}{2}\hat{i} - \frac{3}{2}\hat{k}$.$E$ is the midpoint of $CA$,so $\vec{E} = \frac{\vec{C}+\vec{A}}{2} = \frac{(1+1)\hat{i} + (-1-2)\hat{j} + (-2+1)\hat{k}}{2} = \hat{i} - \frac{3}{2}\hat{j} - \frac{1}{2}\hat{k}$.Now,$\overrightarrow{DE} = \vec{E} - \vec{D} = (1 - \frac{3}{2})\hat{i} + (-\frac{3}{2} - 0)\hat{j} + (-\frac{1}{2} - (-\frac{3}{2}))\hat{k} = -\frac{1}{2}\hat{i} - \frac{3}{2}\hat{j} + \hat{k}$.To find the unit vector along $\overrightarrow{DE}$,we calculate its magnitude: $|\overrightarrow{DE}| = \sqrt{(-\frac{1}{2})^2 + (-\frac{3}{2})^2 + 1^2} = \sqrt{\frac{1}{4} + \frac{9}{4} + 1} = \sqrt{\frac{10}{4} + 1} = \sqrt{\frac{14}{4}} = \frac{\sqrt{14}}{2}$.The unit vector is $\frac{\overrightarrow{DE}}{|\overrightarrow{DE}|} = \frac{-\frac{1}{2}\hat{i} - \frac{3}{2}\hat{j} + \hat{k}}{\frac{\sqrt{14}}{2}} = \frac{-\hat{i} - 3\hat{j} + 2\hat{k}}{\sqrt{14}}$.

$\hat{i}-2 \hat{j}+\hat{k}$,$2 \hat{i}+\hat{j}-\hat{k}$,and $\hat{i}-\hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$,and $C$ of a triangle $ABC$ respectively. If $D$ and $E$ are the midpoints of $BC$ and $CA$ respectively,then the unit vector along $\overrightarrow{DE}$ is

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