If vec{AB} = 3 hat{i} + 5 hat{j} + 4 hat{k} a | vedclass

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

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$5$ units

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(B) Let $D$ be the midpoint of $BC$. 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 $\vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC})$.Given $\vec{AB} = 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$ and $\vec{AC} = 5 \hat{i} - 5 \hat{j} + 2 \hat{k}$.Substituting these values:$\vec{AD} = \frac{1}{2}((3 \hat{i} + 5 \hat{j} + 4 \hat{k}) + (5 \hat{i} - 5 \hat{j} + 2 \hat{k}))$$\vec{AD} = \frac{1}{2}(8 \hat{i} + 0 \hat{j} + 6 \hat{k})$$\vec{AD} = 4 \hat{i} + 3 \hat{k}$.The length of the median is the magnitude of vector $\vec{AD}$:$|\vec{AD}| = \sqrt{4^2 + 0^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ units.

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

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