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(A) The unit vectors along $a$ and $b$ are given by:$\hat{a} = \frac{a}{|a|} = \frac{7i - 4j - 4k}{\sqrt{7^2 + (-4)^2 + (-4)^2}} = \frac{7i - 4j - 4k}

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$\frac{5}{3}(i - 7j + 2k)$

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(A) The unit vectors along $a$ and $b$ are given by:$\hat{a} = \frac{a}{|a|} = \frac{7i - 4j - 4k}{\sqrt{7^2 + (-4)^2 + (-4)^2}} = \frac{7i - 4j - 4k}{\sqrt{49 + 16 + 16}} = \frac{7i - 4j - 4k}{9}$$\hat{b} = \frac{b}{|b|} = \frac{-2i - j + 2k}{\sqrt{(-2)^2 + (-1)^2 + 2^2}} = \frac{-2i - j + 2k}{\sqrt{4 + 1 + 4}} = \frac{-2i - j + 2k}{3} = \frac{-6i - 3j + 6k}{9}$The internal bisector vector is proportional to $\hat{a} + \hat{b} = \frac{1}{9}(7i - 4j - 4k - 6i - 3j + 6k) = \frac{1}{9}(i - 7j + 2k)$.Let $c = \lambda(\hat{a} + \hat{b}) = \frac{\lambda}{9}(i - 7j + 2k)$.Given $|c| = 5\sqrt{6}$,we have $|c|^2 = 25 	imes 6 = 150$.$|c|^2 = \frac{\lambda^2}{81}(1^2 + (-7)^2 + 2^2) = \frac{\lambda^2}{81}(1 + 49 + 4) = \frac{54\lambda^2}{81} = \frac{2\lambda^2}{3}$.Equating the two: $\frac{2\lambda^2}{3} = 150 \Rightarrow \lambda^2 = 225 \Rightarrow \lambda = 15$ (for internal bisector).Thus,$c = \frac{15}{9}(i - 7j + 2k) = \frac{5}{3}(i - 7j + 2k)$.

The vector $c$ directed along the internal bisector of the angle between the vectors $a = 7i - 4j - 4k$ and $b = -2i - j + 2k$ with $|c| = 5\sqrt{6}$ is

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