If the vectors overline{AB}=3 hat{i}+4 hat{k} | vedclass

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(D) Let $AD$ be the median of $	riangle ABC$ through vertex $A$. Since $D$ is the midpoint of $BC$,the vector $\overline{AD}$ is given by the formula

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$\sqrt{33}$ units.

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(D) Let $AD$ be the median of $	riangle ABC$ through vertex $A$. Since $D$ is the midpoint of $BC$,the vector $\overline{AD}$ is given by the formula $\overline{AD} = \frac{\overline{AB} + \overline{AC}}{2}$. Substituting the given vectors: $\overline{AD} = \frac{(3 \hat{i} + 4 \hat{k}) + (5 \hat{i} - 2 \hat{j} + 4 \hat{k})}{2}$ $\overline{AD} = \frac{(3+5) \hat{i} + (-2) \hat{j} + (4+4) \hat{k}}{2}$ $\overline{AD} = \frac{8 \hat{i} - 2 \hat{j} + 8 \hat{k}}{2}$ $\overline{AD} = 4 \hat{i} - \hat{j} + 4 \hat{k}$ Now,the length of the median is the magnitude of vector $\overline{AD}$: $|\overline{AD}| = \sqrt{4^2 + (-1)^2 + 4^2}$ $|\overline{AD}| = \sqrt{16 + 1 + 16}$ $|\overline{AD}| = \sqrt{33} 	ext{ units}$.

If the vectors $\overline{AB}=3 \hat{i}+4 \hat{k}$ and $\overline{AC}=5 \hat{i}-2 \hat{j}+4 \hat{k}$ are the sides of the triangle $ABC$,then the length of the median through $A$ is:

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