The vectors overrightarrow{AB} = 3hat{i} + 5h | vedclass

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(C) Let $D$ be the midpoint of $BC$. The median through $A$ is the vector $\overrightarrow{AD}$.By the triangle law of vector addition,$\overrightarro

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$5$

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(C) Let $D$ be the midpoint of $BC$. The median through $A$ is the vector $\overrightarrow{AD}$.By the triangle law of vector addition,$\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD}$.Since $D$ is the midpoint of $BC$,$\overrightarrow{BD} = \frac{1}{2}\overrightarrow{BC}$.We know that $\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB} = (5\hat{i} - 5\hat{j} + 2\hat{k}) - (3\hat{i} + 5\hat{j} + 4\hat{k}) = 2\hat{i} - 10\hat{j} - 2\hat{k}$.Therefore,$\overrightarrow{BD} = \frac{1}{2}(2\hat{i} - 10\hat{j} - 2\hat{k}) = \hat{i} - 5\hat{j} - \hat{k}$.Now,$\overrightarrow{AD} = (3\hat{i} + 5\hat{j} + 4\hat{k}) + (\hat{i} - 5\hat{j} - \hat{k}) = 4\hat{i} + 0\hat{j} + 3\hat{k} = 4\hat{i} + 3\hat{k}$.The length of the median is the magnitude of $\overrightarrow{AD}$:$|\overrightarrow{AD}| = \sqrt{4^2 + 0^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 	ext{ units}$.

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

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