Find the vector vec{c} which is in the direct | vedclass

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(A) The unit vector along the internal angle bisector of $\vec{a}$ and $\vec{b}$ is given by $\hat{u} = \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\v

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The unit vector along the internal angle bisector of $\vec{a}$ and $\vec{b}$ is given by $\hat{u} = \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|}$.First,calculate the magnitudes: $|\vec{a}| = \sqrt{7^2 + (-4)^2 + (-4)^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$.$|\vec{b}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.Now,$\frac{\vec{a}}{|\vec{a}|} = \frac{1}{9}(7\hat{i} - 4\hat{j} - 4\hat{k})$ and $\frac{\vec{b}}{|\vec{b}|} = \frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k}) = \frac{1}{9}(-6\hat{i} - 3\hat{j} + 6\hat{k})$.Adding these: $\hat{u} = \frac{1}{9}(7-6)\hat{i} + \frac{1}{9}(-4-3)\hat{j} + \frac{1}{9}(-4+6)\hat{k} = \frac{1}{9}(\hat{i} - 7\hat{j} + 2\hat{k})$.The magnitude of this vector is $\sqrt{(\frac{1}{9})^2 + (-\frac{7}{9})^2 + (\frac{2}{9})^2} = \sqrt{\frac{1+49+4}{81}} = \sqrt{\frac{54}{81}} = \frac{3\sqrt{6}}{9} = \frac{\sqrt{6}}{3}$.Since $\vec{c} = k\hat{u}$ and $|\vec{c}| = 5\sqrt{6}$,we have $|k| \cdot \frac{\sqrt{6}}{3} = 5\sqrt{6}$,which gives $|k| = 15$.Thus,$\vec{c} = 15 \cdot \frac{1}{9}(\hat{i} - 7\hat{j} + 2\hat{k}) = \frac{5}{3}(\hat{i} - 7\hat{j} + 2\hat{k})$.

Find the vector $\vec{c}$ which is in the direction of the internal angle bisector of the vectors $\vec{a} = 7\hat{i} - 4\hat{j} - 4\hat{k}$ and $\vec{b} = -2\hat{i} - \hat{j} + 2\hat{k}$ with $|\vec{c}| = 5\sqrt{6}$.

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