The vectors vec{AB} = 3hat{i} + 4hat{k} and v | vedclass

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(C) Let $A$ be the origin $(0, 0, 0)$.Then $\vec{AB}$ and $\vec{AC}$ represent the position vectors of vertices $B$ and $C$ respectively.Let $M$ be th

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$\sqrt{33}$

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(C) Let $A$ be the origin $(0, 0, 0)$.Then $\vec{AB}$ and $\vec{AC}$ represent the position vectors of vertices $B$ and $C$ respectively.Let $M$ be the midpoint of side $BC$.The position vector of $M$ is given by $\vec{AM} = \frac{\vec{AB} + \vec{AC}}{2}$.Substituting the given vectors:$\vec{AM} = \frac{(3\hat{i} + 4\hat{k}) + (5\hat{i} - 2\hat{j} + 4\hat{k})}{2}$$\vec{AM} = \frac{(3+5)\hat{i} + (0-2)\hat{j} + (4+4)\hat{k}}{2}$$\vec{AM} = \frac{8\hat{i} - 2\hat{j} + 8\hat{k}}{2} = 4\hat{i} - \hat{j} + 4\hat{k}$.The length of the median $AM$ is the magnitude of vector $\vec{AM}$:$|\vec{AM}| = \sqrt{4^2 + (-1)^2 + 4^2}$$|\vec{AM}| = \sqrt{16 + 1 + 16} = \sqrt{33}$.

The vectors $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$ are the sides of a $\triangle ABC$. The length of the median through $A$ is:

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