If bar{a}=bar{i}-2bar{j}+2bar{k}, bar{b}=6bar | vedclass

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(C) First, calculate the magnitudes of the individual vectors: $|\bar{a}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$. $|\bar{b}| =

Vedclass · Accepted Answer

$6$

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(C) First, calculate the magnitudes of the individual vectors: $|\bar{a}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$. $|\bar{b}| = \sqrt{6^2 + 3^2 + (-2)^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$. $|\bar{c}| = \sqrt{(-4)^2 + 3^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$. Sum of magnitudes: $|\bar{a}| + |\bar{b}| + |\bar{c}| = 3 + 7 + 13 = 23$. Next, calculate the sum of the vectors: $\bar{a} + \bar{b} + \bar{c} = (1+6-4)\bar{i} + (-2+3+3)\bar{j} + (2-2+12)\bar{k} = 3\bar{i} + 4\bar{j} + 12\bar{k}$. Calculate the magnitude of the resultant vector: $|\bar{a} + \bar{b} + \bar{c}| = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$. Finally, evaluate the expression: $\sqrt{(|\bar{a}| + |\bar{b}| + |\bar{c}|) + |\bar{a} + \bar{b} + \bar{c}|} = \sqrt{23 + 13} = \sqrt{36} = 6$.

If $\bar{a}=\bar{i}-2\bar{j}+2\bar{k}$, $\bar{b}=6\bar{i}+3\bar{j}-2\bar{k}$, and $\bar{c}=-4\bar{i}+3\bar{j}+12\bar{k}$ are three vectors, then $\sqrt{(|\bar{a}|+|\bar{b}|+|\bar{c}|)+|\bar{a}+\bar{b}+\bar{c}|}=$

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