vec{c} is a vector along the bisector of the | vedclass

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(B) First,find the unit vectors along $\vec{a}$ and $\vec{b}$.$|\vec{a}| = \sqrt{4^2 + 7^2 + (-4)^2} = \sqrt{16 + 49 + 16} = \sqrt{81} = 9$.$|\vec{b}|

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$10 \hat{i}+4 \hat{j}-\hat{k}$

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(B) First,find the unit vectors along $\vec{a}$ and $\vec{b}$.$|\vec{a}| = \sqrt{4^2 + 7^2 + (-4)^2} = \sqrt{16 + 49 + 16} = \sqrt{81} = 9$.$|\vec{b}| = \sqrt{12^2 + (-3)^2 + 4^2} = \sqrt{144 + 9 + 16} = \sqrt{169} = 13$.Unit vectors are $\hat{a} = \frac{4 \hat{i} + 7 \hat{j} - 4 \hat{k}}{9}$ and $\hat{b} = \frac{12 \hat{i} - 3 \hat{j} + 4 \hat{k}}{13}$.The vector along the internal angle bisector is given by $\vec{v} = \lambda(\hat{a} + \hat{b})$.$\hat{a} + \hat{b} = \frac{13(4 \hat{i} + 7 \hat{j} - 4 \hat{k}) + 9(12 \hat{i} - 3 \hat{j} + 4 \hat{k})}{117} = \frac{(52 + 108) \hat{i} + (91 - 27) \hat{j} + (-52 + 36) \hat{k}}{117} = \frac{160 \hat{i} + 64 \hat{j} - 16 \hat{k}}{117} = \frac{16}{117}(10 \hat{i} + 4 \hat{j} - \hat{k})$.Let $\vec{c} = k(10 \hat{i} + 4 \hat{j} - \hat{k})$. The magnitude is $|\vec{c}| = |k| \sqrt{10^2 + 4^2 + (-1)^2} = |k| \sqrt{100 + 16 + 1} = |k| \sqrt{117} = |k| \sqrt{9 	imes 13} = 3|k| \sqrt{13}$.Given $|\vec{c}| = 3 \sqrt{13}$,we have $3|k| \sqrt{13} = 3 \sqrt{13}$,so $k = 1$.Thus,$\vec{c} = 10 \hat{i} + 4 \hat{j} - \hat{k}$.

$\vec{c}$ is a vector along the bisector of the internal angle between the vectors $\vec{a}=4 \hat{i}+7 \hat{j}-4 \hat{k}$ and $\vec{b}=12 \hat{i}-3 \hat{j}+4 \hat{k}$. If the magnitude of $\vec{c}$ is $3 \sqrt{13}$,then $\vec{c}=$

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