If the vector bar{i}-7 bar{j}+2 bar{k} is alo | vedclass

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(B) Let $\bar{u} = \bar{i} - 7\bar{j} + 2\bar{k}$ be the vector along the internal bisector. The magnitude $|\bar{u}| = \sqrt{1^2 + (-7)^2 + 2^2} = \s

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$\frac{7}{9}$

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(B) Let $\bar{u} = \bar{i} - 7\bar{j} + 2\bar{k}$ be the vector along the internal bisector. The magnitude $|\bar{u}| = \sqrt{1^2 + (-7)^2 + 2^2} = \sqrt{1 + 49 + 4} = \sqrt{54} = 3\sqrt{6}$.Let $\bar{b} = -2\bar{i} - \bar{j} + 2\bar{k}$. The magnitude $|\bar{b}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = 3$.The unit vector along $\bar{b}$ is $\hat{b} = \frac{-2\bar{i} - \bar{j} + 2\bar{k}}{3} = -\frac{2}{3}\bar{i} - \frac{1}{3}\bar{j} + \frac{2}{3}\bar{k}$.Let $\hat{a} = x\bar{i} + y\bar{j} + z\bar{k}$ be the unit vector along $\bar{a}$.The internal bisector is along $\hat{a} + \hat{b}$. Thus,$\hat{a} + \hat{b} = k\bar{u}$ for some scalar $k > 0$.$\hat{a} = k(\bar{i} - 7\bar{j} + 2\bar{k}) - (-\frac{2}{3}\bar{i} - \frac{1}{3}\bar{j} + \frac{2}{3}\bar{k}) = (k + \frac{2}{3})\bar{i} + (-7k + \frac{1}{3})\bar{j} + (2k - \frac{2}{3})\bar{k}$.Since $|\hat{a}| = 1$,we have $(k + \frac{2}{3})^2 + (-7k + \frac{1}{3})^2 + (2k - \frac{2}{3})^2 = 1$.$(k^2 + \frac{4}{3}k + \frac{4}{9}) + (49k^2 - \frac{14}{3}k + \frac{1}{9}) + (4k^2 - \frac{8}{3}k + \frac{4}{9}) = 1$.$54k^2 - 6k + 1 = 1 \implies 54k^2 - 6k = 0 \implies 6k(9k - 1) = 0$.Since $k > 0$,$k = \frac{1}{9}$.Then $x = k + \frac{2}{3} = \frac{1}{9} + \frac{6}{9} = \frac{7}{9}$.

If the vector $\bar{i}-7 \bar{j}+2 \bar{k}$ is along the internal bisector of the angle between the vectors $\bar{a}$ and $-2 \bar{i}-\bar{j}+2 \bar{k}$ and the unit vector along $\bar{a}$ is $x \bar{i}+y \bar{j}+z \bar{k}$ then $x=$

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