In a triangle ABC, if overline{BC} = bar{i} - | vedclass

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(B) Given vectors are $\overline{BC} = \bar{i} - 2\bar{j} + 2\bar{k}$ and $\overline{CA} = 6\bar{i} + 3\bar{j} - 2\bar{k}$.In $	riangle ABC$, by the

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$18$

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(B) Given vectors are $\overline{BC} = \bar{i} - 2\bar{j} + 2\bar{k}$ and $\overline{CA} = 6\bar{i} + 3\bar{j} - 2\bar{k}$.In $	riangle ABC$, by the triangle law of vector addition, $\overline{AB} + \overline{BC} + \overline{CA} = \vec{0}$, so $\overline{AB} = -(\overline{BC} + \overline{CA})$.$\overline{AB} = -(\bar{i} - 2\bar{j} + 2\bar{k} + 6\bar{i} + 3\bar{j} - 2\bar{k}) = -(7\bar{i} + \bar{j} + 0\bar{k}) = -7\bar{i} - \bar{j}$.Now, calculate the magnitudes of the sides:$|\overline{BC}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.$|\overline{CA}| = \sqrt{6^2 + 3^2 + (-2)^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.$|\overline{AB}| = \sqrt{(-7)^2 + (-1)^2 + 0^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$.The perimeter is $|\overline{AB}| + |\overline{BC}| + |\overline{CA}| = 5\sqrt{2} + 3 + 7 = 10 + 5\sqrt{2} = 5(2 + \sqrt{2})$.Thus, the correct option is $B$.

In a triangle $ABC$, if $\overline{BC} = \bar{i} - 2\bar{j} + 2\bar{k}$ and $\overline{CA} = 6\bar{i} + 3\bar{j} - 2\bar{k}$, then the perimeter of the triangle is

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