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 $A$.The median vector $\overline{AD}$ is given by the formula $\overline{AD} = \frac{\ove

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$\sqrt{33}$ units

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(D) Let $AD$ be the median of $	riangle ABC$ through vertex $A$.The median 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}$$\overline{AD} = \frac{8 \hat{i} - 2 \hat{j} + 8 \hat{k}}{2}$$\overline{AD} = 4 \hat{i} - \hat{j} + 4 \hat{k}$Now,calculate the length (magnitude) of the median $\overline{AD}$:$|\overline{AD}| = \sqrt{4^2 + (-1)^2 + 4^2}$$|\overline{AD}| = \sqrt{16 + 1 + 16}$$|\overline{AD}| = \sqrt{33}$ units.

$(a)$ Clostridium butylicum	$(i)$ Cyclosporin-$A$
$(b)$ Trichoderma polysporum	$(ii)$ Butyric acid
$(c)$ Monascus purpureus	$(iii)$ Citric acid
$(d)$ Aspergillus niger	$(iv)$ Blood cholesterol lowering agent

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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