If vec{a} = 7hat{j} + 10hat{k}, vec{b} = -hat | vedclass

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Question

(A) Let $\vec{AB} = \vec{b} - \vec{a} = -\hat{i} - \hat{j} - 4\hat{k}$, so $c = |\vec{AB}| = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{18} = 3\sqrt{2}$.

Vedclass · Accepted Answer

$(2 - 3\sqrt{2})\hat{i} + (3 + 3\sqrt{2})\hat{j} + 6\hat{k}$

Vedclass · Answer

(A) Let $\vec{AB} = \vec{b} - \vec{a} = -\hat{i} - \hat{j} - 4\hat{k}$, so $c = |\vec{AB}| = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{18} = 3\sqrt{2}$.Let $\vec{AC} = \vec{c} - \vec{a} = -4\hat{i} + 2\hat{j} - 4\hat{k}$, so $b = |\vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-4)^2} = \sqrt{36} = 6$.The angle bisector of $\angle A$ divides the opposite side $BC$ in the ratio $c:b = 3\sqrt{2}:6 = 1:\sqrt{2}$.Using the section formula, the position vector $\vec{p}$ of the point dividing $BC$ in ratio $m:n$ is $\frac{m\vec{c} + n\vec{b}}{m+n}$.Here $m=1, n=\sqrt{2}$, so $\vec{p} = \frac{1\vec{c} + \sqrt{2}\vec{b}}{1+\sqrt{2}} = \frac{(-4\hat{i} + 9\hat{j} + 6\hat{k}) + \sqrt{2}(-\hat{i} + 6\hat{j} + 6\hat{k})}{1+\sqrt{2}} = \frac{(-4-\sqrt{2})\hat{i} + (9+6\sqrt{2})\hat{j} + (6+6\sqrt{2})\hat{k}}{1+\sqrt{2}}$.Rationalizing the coefficients: $\frac{-4-\sqrt{2}}{1+\sqrt{2}} = \frac{(-4-\sqrt{2})(\sqrt{2}-1)}{1} = -4\sqrt{2} + 4 - 2 + \sqrt{2} = 2 - 3\sqrt{2}$.$\frac{9+6\sqrt{2}}{1+\sqrt{2}} = \frac{(9+6\sqrt{2})(\sqrt{2}-1)}{1} = 9\sqrt{2} - 9 + 12 - 6\sqrt{2} = 3 + 3\sqrt{2}$.$\frac{6+6\sqrt{2}}{1+\sqrt{2}} = 6$. Thus, $\vec{p} = (2 - 3\sqrt{2})\hat{i} + (3 + 3\sqrt{2})\hat{j} + 6\hat{k}$.

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