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(B) Let the bisector of angle $A$ meet $BC$ at point $D$. According to the Angle Bisector Theorem,$D$ divides $BC$ in the ratio $AB : AC$.Given positi

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

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(B) Let the bisector of angle $A$ meet $BC$ at point $D$. According to the Angle Bisector Theorem,$D$ divides $BC$ in the ratio $AB : AC$.Given position vectors: $\vec{A} = 4\hat{i} + 7\hat{j} + 8\hat{k}$,$\vec{B} = 2\hat{i} + 3\hat{j} + 4\hat{k}$,$\vec{C} = 2\hat{i} + 5\hat{j} + 7\hat{k}$.Calculate vectors $\vec{AB}$ and $\vec{AC}$:$\vec{AB} = \vec{B} - \vec{A} = (2-4)\hat{i} + (3-7)\hat{j} + (4-8)\hat{k} = -2\hat{i} - 4\hat{j} - 4\hat{k}$.$\vec{AC} = \vec{C} - \vec{A} = (2-4)\hat{i} + (5-7)\hat{j} + (7-8)\hat{k} = -2\hat{i} - 2\hat{j} - 1\hat{k}$.Calculate magnitudes:$|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.$|\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.The ratio $AB : AC = 6 : 3 = 2 : 1$.Using the section formula,the position vector of $D$ is:$\vec{D} = \frac{1(\vec{B}) + 2(\vec{C})}{1+2} = \frac{(2\hat{i} + 3\hat{j} + 4\hat{k}) + 2(2\hat{i} + 5\hat{j} + 7\hat{k})}{3}$.$\vec{D} = \frac{(2+4)\hat{i} + (3+10)\hat{j} + (4+14)\hat{k}}{3} = \frac{6\hat{i} + 13\hat{j} + 18\hat{k}}{3} = \frac{1}{3}(6\hat{i} + 13\hat{j} + 18\hat{k})$.

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

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