In a parallelogram ABCD,the position vectors | vedclass

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(C) In a parallelogram $ABCD$,the diagonals $AC$ and $DB$ bisect each other at the same point $M$.Therefore,$M$ is the midpoint of $AC$ as well.The po

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$\frac{7}{\sqrt{51}}$

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(C) In a parallelogram $ABCD$,the diagonals $AC$ and $DB$ bisect each other at the same point $M$.Therefore,$M$ is the midpoint of $AC$ as well.The position vector of $M$ is given by $\vec{OM} = \frac{\vec{OA} + \vec{OC}}{2}$.Given $\vec{OA} = 3\hat{i} + 3\hat{j} + 5\hat{k}$ and $\vec{OC} = \hat{i} - 5\hat{j} - 5\hat{k}$.$\vec{OM} = \frac{(3+1)\hat{i} + (3-5)\hat{j} + (5-5)\hat{k}}{2} = \frac{4\hat{i} - 2\hat{j} + 0\hat{k}}{2} = 2\hat{i} - \hat{j}$.The projection of $\vec{OM}$ on $\vec{OC}$ is given by $\frac{\vec{OM} \cdot \vec{OC}}{|\vec{OC}|}$.$\vec{OM} \cdot \vec{OC} = (2)(1) + (-1)(-5) + (0)(-5) = 2 + 5 + 0 = 7$.$|\vec{OC}| = \sqrt{1^2 + (-5)^2 + (-5)^2} = \sqrt{1 + 25 + 25} = \sqrt{51}$.Thus,the projection is $\frac{7}{\sqrt{51}}$.

In a parallelogram $ABCD$,the position vectors of vertices $A$ and $C$ are $3\hat{i} + 3\hat{j} + 5\hat{k}$ and $\hat{i} - 5\hat{j} - 5\hat{k}$ respectively. If $M$ is the midpoint of the diagonal $DB$,find the projection of $\overline{OM}$ on $\overline{OC}$,where $O$ is the origin.

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