If hat{i}+4 hat{j}+3 hat{k},hat{i}+2 hat{j}+3 | vedclass

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(C) Given position vectors are $\vec{A} = \hat{i}+4 \hat{j}+3 \hat{k}$,$\vec{B} = \hat{i}+2 \hat{j}+3 \hat{k}$,and $\vec{C} = 3 \hat{i}+2 \hat{j}+\hat

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$\hat{j}$

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(C) Given position vectors are $\vec{A} = \hat{i}+4 \hat{j}+3 \hat{k}$,$\vec{B} = \hat{i}+2 \hat{j}+3 \hat{k}$,and $\vec{C} = 3 \hat{i}+2 \hat{j}+\hat{k}$.Since $D$ is the midpoint of $BC$,its position vector $\vec{D}$ is given by $\vec{D} = \frac{\vec{B} + \vec{C}}{2} = \frac{(\hat{i}+2 \hat{j}+3 \hat{k}) + (3 \hat{i}+2 \hat{j}+\hat{k})}{2} = \frac{4 \hat{i}+4 \hat{j}+4 \hat{k}}{2} = 2 \hat{i}+2 \hat{j}+2 \hat{k}$.Since $E$ is the midpoint of $AC$,its position vector $\vec{E}$ is given by $\vec{E} = \frac{\vec{A} + \vec{C}}{2} = \frac{(\hat{i}+4 \hat{j}+3 \hat{k}) + (3 \hat{i}+2 \hat{j}+\hat{k})}{2} = \frac{4 \hat{i}+6 \hat{j}+4 \hat{k}}{2} = 2 \hat{i}+3 \hat{j}+2 \hat{k}$.Now,$\overrightarrow{DE} = \vec{E} - \vec{D} = (2 \hat{i}+3 \hat{j}+2 \hat{k}) - (2 \hat{i}+2 \hat{j}+2 \hat{k}) = \hat{j}$.

If $\hat{i}+4 \hat{j}+3 \hat{k}$,$\hat{i}+2 \hat{j}+3 \hat{k}$,and $3 \hat{i}+2 \hat{j}+\hat{k}$ are position vectors of $A$,$B$,and $C$ respectively,and if $D$ and $E$ are midpoints of sides $BC$ and $AC$,then $\overrightarrow{DE}$ is equal to:

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