If the vectors vec{AB} = -3hat{i} + 4hat{k} a | vedclass

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(B) Let $D$ be the midpoint of side $BC$. The median through $A$ is the vector $\vec{AD}$.Since $D$ is the midpoint of $BC$, the position vector of $D

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$\sqrt{18}$

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(B) Let $D$ be the midpoint of side $BC$. The median through $A$ is the vector $\vec{AD}$.Since $D$ is the midpoint of $BC$, the position vector of $D$ relative to $A$ is given by the average of the vectors $\vec{AB}$ and $\vec{AC}$:$\vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC})$$\vec{AD} = \frac{1}{2}((-3\hat{i} + 4\hat{k}) + (5\hat{i} - 2\hat{j} + 4\hat{k}))$$\vec{AD} = \frac{1}{2}(2\hat{i} - 2\hat{j} + 8\hat{k})$$\vec{AD} = \hat{i} - \hat{j} + 4\hat{k}$The length of the median $\vec{AD}$ is the magnitude of the vector $\vec{AD}$:$|\vec{AD}| = \sqrt{(1)^2 + (-1)^2 + (4)^2}$$|\vec{AD}| = \sqrt{1 + 1 + 16}$$|\vec{AD}| = \sqrt{18}$

If 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 $\triangle ABC$, then the length of the median through $A$ is

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