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(B) Given vectors are $a = 3 \hat{i} - 4 \hat{k}$ and $b = 5 \hat{j} + 12 \hat{k}$.The magnitudes of the vectors are:$|a| = \sqrt{3^2 + (-4)^2} = \sqr

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$39 \hat{i} + 25 \hat{j} + 8 \hat{k}$

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(B) Given vectors are $a = 3 \hat{i} - 4 \hat{k}$ and $b = 5 \hat{j} + 12 \hat{k}$.The magnitudes of the vectors are:$|a| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$|b| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$The unit vectors along $a$ and $b$ are $\hat{a} = \frac{a}{|a|} = \frac{3 \hat{i} - 4 \hat{k}}{5}$ and $\hat{b} = \frac{b}{|b|} = \frac{5 \hat{j} + 12 \hat{k}}{13}$.$A$ vector that bisects the angle between $a$ and $b$ is given by the sum of the unit vectors: $\hat{a} + \hat{b}$.$\hat{a} + \hat{b} = \frac{3 \hat{i} - 4 \hat{k}}{5} + \frac{5 \hat{j} + 12 \hat{k}}{13}$Taking the common denominator $65$:$= \frac{13(3 \hat{i} - 4 \hat{k}) + 5(5 \hat{j} + 12 \hat{k})}{65}$$= \frac{39 \hat{i} - 52 \hat{k} + 25 \hat{j} + 60 \hat{k}}{65}$$= \frac{39 \hat{i} + 25 \hat{j} + 8 \hat{k}}{65}$Since any scalar multiple of this vector also bisects the angle,we can multiply by $65$ to get the vector $39 \hat{i} + 25 \hat{j} + 8 \hat{k}$.

$A$ vector which bisects the angle between $a = 3 \hat{i} - 4 \hat{k}$ and $b = 5 \hat{j} + 12 \hat{k}$ is

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