Question:

Step 1: Calculate the magnitudes of vectors \(\vec{A}\) and \(\vec{B}\). Given \(\vec{A} = 7\hat{i} - 4\hat{j} - 4\hat{k}\) and \(\vec{B} = -2\hat{i} - \hat{j} + 2\hat{k}\). $$|\vec{A}| = \sqrt{7^2 + (-4)^2 + (-4)^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$$ $$|\vec{B}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$ Step 2: Determine the unit vectors along \(\vec{A}\) and \(\vec{B}\). The unit vector along \(\vec{A}\) is \(\hat{A}\): $$\hat{A} = \frac{\vec{A}}{|\vec{A}|} = \frac{1}{9}(7\hat{i} - 4\hat{j} - 4\hat{k})$$ The unit vector along \(\vec{B}\) is \(\hat{B}\): $$\hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k})$$ Step 3: Find the direction vector for the internal angle bisector. A vector along the internal bisector of the angle between \(\vec{A}\) and \(\vec{B}\) is proportional to \(\hat{A} + \hat{B}\). $$\hat{A} + \hat{B} = \frac{1}{9}(7\hat{i} - 4\hat{j} - 4\hat{k}) + \frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k})$$ To combine these, find a common denominator: $$\hat{A} + \hat{B} = \frac{1}{9}(7\hat{i} - 4\hat{j} - 4\hat{k}) + \frac{3}{9}(-2\hat{i} - \hat{j} + 2\hat{k})$$ $$\hat{A} + \hat{B} = \frac{1}{9}(7\hat{i} - 4\hat{j} - 4\hat{k} - 6\hat{i} - 3\hat{j} + 6\hat{k})$$ $$\hat{A} + \hat{B} = \frac{1}{9}((7-6)\hat{i} + (-4-3)\hat{j} + (-4+6)\hat{k})$$ $$\hat{A} + \hat{B} = \frac{1}{9}(\hat{i} - 7\hat{j} + 2\hat{k})$$ Step 4: Calculate the magnitude of the direction vector \(\hat{A} + \hat{B}\). $$|\hat{A} + \hat{B}| = \left|\frac{1}{9}(\hat{i} - 7\hat{j} + 2\hat{k})\right| = \frac{1}{9}\sqrt{1^2 + (-7)^2 + 2^2}$$ $$|\hat{A} + \hat{B}| = \frac{1}{9}\sqrt{1 + 49 + 4} = \frac{1}{9}\sqrt{54}$$ Simplify \(\sqrt{54}\): \(\sqrt{54} = \sqrt{9 \cdot 6} = 3\sqrt{6}\). $$|\hat{A} + \hat{B}| = \frac{1}{9}(3\sqrt{6}) = \frac{\sqrt{6}}{3}$$ Step 5: Determine the unit vector along the bisector. The unit vector in the direction of the bisector is given by \(\frac{\hat{A} + \hat{B}}{|\hat{A} + \hat{B}|}\). $$\frac{\hat{A} + \hat{B}}{|\hat{A} + \hat{B}|} = \frac{\frac{1}{9}(\hat{i} - 7\hat{j} + 2\hat{k})}{\frac{\sqrt{6}}{3}}$$ $$ = \frac{1}{9} \cdot \frac{3}{\sqrt{6}}(\hat{i} - 7\hat{j} + 2\hat{k})$$ $$ = \frac{1}{3\sqrt{6}}(\hat{i} - 7\hat{j} + 2\hat{k})$$ Step 6: Construct vector \(\vec{C}\). Vector \(\vec{C}\) has a magnitude of \(|\vec{C}| = 5\sqrt{6}\) and is directed along the internal bisector. $$\vec{C} = |\vec{C}| \cdot \frac{\hat{A} + \hat{B}}{|\hat{A} + \hat{B}|}$$ $$\vec{C} = 5\sqrt{6} \cdot \frac{1}{3\sqrt{6}}(\hat{i} - 7\hat{j} + 2\hat{k})$$ $$\vec{C} = \frac{5}{3}(\hat{i} - 7\hat{j} + 2\hat{k})$$