If 4 hat{i}+7 hat{j}+8 hat{k},2 hat{i}+3 hat{ | vedclass

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(C) Let $\vec{a} = 4\hat{i}+7\hat{j}+8\hat{k}$,$\vec{b} = 2\hat{i}+3\hat{j}+4\hat{k}$,and $\vec{c} = 2\hat{i}+5\hat{j}+7\hat{k}$.By the Angle Bisector

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$\frac{1}{\sqrt{13}+6} (4\sqrt{13}+12)\hat{i} + (7\sqrt{13}+30)\hat{j} + (8\sqrt{13}+42)\hat{k}$

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(C) Let $\vec{a} = 4\hat{i}+7\hat{j}+8\hat{k}$,$\vec{b} = 2\hat{i}+3\hat{j}+4\hat{k}$,and $\vec{c} = 2\hat{i}+5\hat{j}+7\hat{k}$.By the Angle Bisector Theorem,the bisector of $\angle B$ divides the opposite side $AC$ in the ratio $BA : BC$.Calculate the lengths $BA$ and $BC$:$BA = |\vec{a} - \vec{b}| = |(4-2)\hat{i} + (7-3)\hat{j} + (8-4)\hat{k}| = |2\hat{i} + 4\hat{j} + 4\hat{k}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4+16+16} = \sqrt{36} = 6$.$BC = |\vec{c} - \vec{b}| = |(2-2)\hat{i} + (5-3)\hat{j} + (7-4)\hat{k}| = |0\hat{i} + 2\hat{j} + 3\hat{k}| = \sqrt{0^2 + 2^2 + 3^2} = \sqrt{4+9} = \sqrt{13}$.The point $D$ on $AC$ divides $AC$ in the ratio $BA : BC = 6 : \sqrt{13}$.Using the section formula,the position vector $\vec{d}$ is given by $\vec{d} = \frac{6\vec{c} + \sqrt{13}\vec{a}}{6+\sqrt{13}}$.$\vec{d} = \frac{6(2\hat{i}+5\hat{j}+7\hat{k}) + \sqrt{13}(4\hat{i}+7\hat{j}+8\hat{k})}{6+\sqrt{13}} = \frac{(12+4\sqrt{13})\hat{i} + (30+7\sqrt{13})\hat{j} + (42+8\sqrt{13})\hat{k}}{6+\sqrt{13}}$.

If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ are the position vectors of the vertices $A$,$B$ and $C$ respectively of triangle $ABC$,then the position vector of the point in which the bisector of $\angle B$ meets $CA$ is:

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