If 2 hat{i}+4 hat{j}-5 hat{k}, hat{i}+hat{j}+ | vedclass

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(B) Let the position vectors of vertices $A$, $B$, and $C$ be $\vec{a} = 2 \hat{i}+4 \hat{j}-5 \hat{k}$, $\vec{b} = \hat{i}+\hat{j}+\hat{k}$, and $\ve

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$\frac{1}{\sqrt{214}}(3 \hat{i}+6 \hat{j}-13 \hat{k})$

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(B) Let the position vectors of vertices $A$, $B$, and $C$ be $\vec{a} = 2 \hat{i}+4 \hat{j}-5 \hat{k}$, $\vec{b} = \hat{i}+\hat{j}+\hat{k}$, and $\vec{c} = \hat{j}+2 \hat{k}$.Let $D$ be the midpoint of side $BC$. The position vector of $D$ is $\vec{d} = \frac{\vec{b}+\vec{c}}{2} = \frac{(\hat{i}+\hat{j}+\hat{k}) + (0\hat{i}+\hat{j}+2\hat{k})}{2} = \frac{\hat{i}+2\hat{j}+3\hat{k}}{2}$.The vector along the median $AD$ is $\vec{AD} = \vec{d} - \vec{a} = \left(\frac{1}{2}\hat{i}+\hat{j}+\frac{3}{2}\hat{k}\right) - (2\hat{i}+4\hat{j}-5\hat{k}) = (\frac{1}{2}-2)\hat{i} + (1-4)\hat{j} + (\frac{3}{2}+5)\hat{k} = -\frac{3}{2}\hat{i} - 3\hat{j} + \frac{13}{2}\hat{k}$.To find the unit vector, we calculate the magnitude $|\vec{AD}| = \sqrt{(-\frac{3}{2})^2 + (-3)^2 + (\frac{13}{2})^2} = \sqrt{\frac{9}{4} + 9 + \frac{169}{4}} = \sqrt{\frac{9+36+169}{4}} = \sqrt{\frac{214}{4}} = \frac{\sqrt{214}}{2}$.The unit vector is $\frac{\vec{AD}}{|\vec{AD}|} = \frac{-\frac{3}{2}\hat{i} - 3\hat{j} + \frac{13}{2}\hat{k}}{\frac{\sqrt{214}}{2}} = \frac{-3\hat{i} - 6\hat{j} + 13\hat{k}}{\sqrt{214}}$.Note: The direction of the median vector can be taken as $\vec{DA}$ or $\vec{AD}$. Since the options involve $3\hat{i}+6\hat{j}-13\hat{k}$, we take the vector $\vec{DA} = \frac{3}{2}\hat{i} + 3\hat{j} - \frac{13}{2}\hat{k}$. The unit vector is $\frac{3\hat{i}+6\hat{j}-13\hat{k}}{\sqrt{214}}$.

If $2 \hat{i}+4 \hat{j}-5 \hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, and $\hat{j}+2 \hat{k}$ are the position vectors of the vertices $A$, $B$, and $C$ of a triangle respectively, then a unit vector along the median drawn through the vertex $A$ is

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