The vectors overrightarrow{AB} = 3hat{i} + 4h | vedclass

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(C) Let $D$ be the midpoint of $BC$. The median through $A$ is the vector $\overrightarrow{AD}$.Since $D$ is the midpoint of $BC$,we have $\overrighta

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$\sqrt{33}$

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(C) Let $D$ be the midpoint of $BC$. The median through $A$ is the vector $\overrightarrow{AD}$.Since $D$ is the midpoint of $BC$,we have $\overrightarrow{AD} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC})$.Substituting the given vectors:$\overrightarrow{AD} = \frac{1}{2}((3\hat{i} + 4\hat{k}) + (5\hat{i} - 2\hat{j} + 4\hat{k}))$$\overrightarrow{AD} = \frac{1}{2}(8\hat{i} - 2\hat{j} + 8\hat{k}) = 4\hat{i} - \hat{j} + 4\hat{k}$.The length of the median is the magnitude of $\overrightarrow{AD}$:$|\overrightarrow{AD}| = \sqrt{4^2 + (-1)^2 + 4^2} = \sqrt{16 + 1 + 16} = \sqrt{33}$.

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

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