ABCD is a tetrahedron. bar{i}-2bar{j}+3bar{k} | vedclass

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(C) Let the position vectors of $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Given $\vec{a} = \bar{i}-2\bar{j}+3\bar{k}$,$\vec{b}

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$\frac{\sqrt{213}}{\sqrt{2}}$

Vedclass · Answer

(C) Let the position vectors of $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Given $\vec{a} = \bar{i}-2\bar{j}+3\bar{k}$,$\vec{b} = -2\bar{i}+\bar{j}+3\bar{k}$,and $\vec{c} = 3\bar{i}+2\bar{j}-\bar{k}$.The centroid of the face $BCD$ is $\vec{g}_{BCD} = \frac{\vec{b}+\vec{c}+\vec{d}}{3} = -\bar{i}+2\bar{j}-3\bar{k}$.Thus,$\vec{b}+\vec{c}+\vec{d} = 3(-\bar{i}+2\bar{j}-3\bar{k}) = -3\bar{i}+6\bar{j}-9\bar{k}$.Substituting $\vec{b}$ and $\vec{c}$:$(-2\bar{i}+\bar{j}+3\bar{k}) + (3\bar{i}+2\bar{j}-\bar{k}) + \vec{d} = -3\bar{i}+6\bar{j}-9\bar{k}$.$(\bar{i}+3\bar{j}+2\bar{k}) + \vec{d} = -3\bar{i}+6\bar{j}-9\bar{k}$.$\vec{d} = -4\bar{i}+3\bar{j}-11\bar{k}$.The centroid $G$ of the tetrahedron is $\vec{g} = \frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}$.$\vec{g} = \frac{(\bar{i}-2\bar{j}+3\bar{k}) + (-2\bar{i}+\bar{j}+3\bar{k}) + (3\bar{i}+2\bar{j}-\bar{k}) + (-4\bar{i}+3\bar{j}-11\bar{k})}{4} = \frac{-2\bar{i}+4\bar{j}-6\bar{k}}{4} = -0.5\bar{i}+\bar{j}-1.5\bar{k}$.$GD = |\vec{d} - \vec{g}| = |(-4\bar{i}+3\bar{j}-11\bar{k}) - (-0.5\bar{i}+\bar{j}-1.5\bar{k})| = |-3.5\bar{i}+2\bar{j}-9.5\bar{k}|$.$GD = \sqrt{(-3.5)^2 + 2^2 + (-9.5)^2} = \sqrt{12.25 + 4 + 90.25} = \sqrt{106.5} = \sqrt{\frac{213}{2}} = \frac{\sqrt{213}}{\sqrt{2}}$.

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