ABCD is a parallelogram. The position vectors | vedclass

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(D) In a parallelogram,the diagonals bisect each other. Therefore,the midpoint $M$ of the diagonal $DB$ is the same as the midpoint of the diagonal $A

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) In a parallelogram,the diagonals bisect each other. Therefore,the midpoint $M$ of the diagonal $DB$ is the same as the midpoint of the diagonal $AC$.The position vector of $M$ is given by $\vec{OM} = \frac{\vec{OA} + \vec{OC}}{2} = \frac{(3\hat{i} + 3\hat{j} + 5\hat{k}) + (\hat{i} - 5\hat{j} - 5\hat{k})}{2} = \frac{4\hat{i} - 2\hat{j} + 0\hat{k}}{2} = 2\hat{i} - \hat{j}$.The projection of vector $\vec{OM}$ on $\vec{OC}$ is given by the formula $\frac{\vec{OM} \cdot \vec{OC}}{|\vec{OC}|}$.We have $\vec{OM} = 2\hat{i} - \hat{j}$ and $\vec{OC} = \hat{i} - 5\hat{j} - 5\hat{k}$.The dot product $\vec{OM} \cdot \vec{OC} = (2)(1) + (-1)(-5) + (0)(-5) = 2 + 5 + 0 = 7$.The magnitude of $\vec{OC}$ is $|\vec{OC}| = \sqrt{1^2 + (-5)^2 + (-5)^2} = \sqrt{1 + 25 + 25} = \sqrt{51}$.Therefore,the magnitude of the projection is $\frac{\vec{OM} \cdot \vec{OC}}{|\vec{OC}|} = \frac{7}{\sqrt{51}}$.

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

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