If the position vectors of the vertices A, B | vedclass

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(B) Let the position vectors of vertices $A, B, C$ be $\vec{a} = 4\hat{i} + 7\hat{j} + 8\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$,and $\vec{

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$\frac{1}{3}(6\hat{i} + 13\hat{j} + 18\hat{k})$

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(B) Let the position vectors of vertices $A, B, C$ be $\vec{a} = 4\hat{i} + 7\hat{j} + 8\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$,and $\vec{c} = 2\hat{i} + 5\hat{j} + 7\hat{k}$.The angle bisector theorem states that the bisector of $\angle A$ divides the opposite side $BC$ in the ratio of the adjacent sides $AB:AC$.First,calculate the lengths of sides $AB$ and $AC$:$\vec{AB} = \vec{b} - \vec{a} = (2-4)\hat{i} + (3-7)\hat{j} + (4-8)\hat{k} = -2\hat{i} - 4\hat{j} - 4\hat{k}$.$|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.$\vec{AC} = \vec{c} - \vec{a} = (2-4)\hat{i} + (5-7)\hat{j} + (7-8)\hat{k} = -2\hat{i} - 2\hat{j} - 1\hat{k}$.$|\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.The ratio $AB:AC = 6:3 = 2:1$.The position vector of point $D$ dividing $BC$ in ratio $m:n$ is $\frac{m\vec{c} + n\vec{b}}{m+n}$.Here $m=2, n=1$:$\vec{D} = \frac{2\vec{c} + 1\vec{b}}{2+1} = \frac{2(2\hat{i} + 5\hat{j} + 7\hat{k}) + 1(2\hat{i} + 3\hat{j} + 4\hat{k})}{3}$.$\vec{D} = \frac{(4+2)\hat{i} + (10+3)\hat{j} + (14+4)\hat{k}}{3} = \frac{6\hat{i} + 13\hat{j} + 18\hat{k}}{3} = \frac{1}{3}(6\hat{i} + 13\hat{j} + 18\hat{k})$.

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

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