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

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

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$\frac{1}{3}(6 \hat{\imath} + 13 \hat{\jmath} + 18 \hat{k})$

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(D) Let the position vectors of vertices $A, B, C$ be $\vec{a} = 4 \hat{\imath} + 7 \hat{\jmath} + 8 \hat{k}$,$\vec{b} = 2 \hat{\imath} + 3 \hat{\jmath} + 4 \hat{k}$,and $\vec{c} = 2 \hat{\imath} + 5 \hat{\jmath} + 7 \hat{k}$.According to the Angle Bisector Theorem,the bisector of $\angle A$ divides the opposite side $BC$ in the ratio of the sides $AB$ and $AC$.Let $D$ be the point where the bisector of $\angle A$ meets $BC$. Then $D$ divides $BC$ in the ratio $AB : AC$.First,calculate the lengths of sides $AB$ and $AC$:$\vec{AB} = \vec{b} - \vec{a} = (2-4)\hat{\imath} + (3-7)\hat{\jmath} + (4-8)\hat{k} = -2\hat{\imath} - 4\hat{\jmath} - 4\hat{k}$.$AB = |\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.$\vec{AC} = \vec{c} - \vec{a} = (2-4)\hat{\imath} + (5-7)\hat{\jmath} + (7-8)\hat{k} = -2\hat{\imath} - 2\hat{\jmath} - 1\hat{k}$.$AC = |\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.The ratio $AB : AC = 6 : 3 = 2 : 1$.Using the section formula,the position vector of $D$ is given by:$\vec{d} = \frac{AB \cdot \vec{c} + AC \cdot \vec{b}}{AB + AC} = \frac{6(2\hat{\imath} + 5\hat{\jmath} + 7\hat{k}) + 3(2\hat{\imath} + 3\hat{\jmath} + 4\hat{k})}{6 + 3}$$\vec{d} = \frac{(12\hat{\imath} + 30\hat{\jmath} + 42\hat{k}) + (6\hat{\imath} + 9\hat{\jmath} + 12\hat{k})}{9} = \frac{18\hat{\imath} + 39\hat{\jmath} + 54\hat{k}}{9} = 2\hat{\imath} + \frac{13}{3}\hat{\jmath} + 6\hat{k}$.Wait,re-evaluating the ratio division: $D$ divides $BC$ in ratio $c:b$ (where $c=AB, b=AC$).$\vec{d} = \frac{AC \cdot \vec{b} + AB \cdot \vec{c}}{AC + AB} = \frac{3(2\hat{\imath} + 3\hat{\jmath} + 4\hat{k}) + 6(2\hat{\imath} + 5\hat{\jmath} + 7\hat{k})}{3 + 6} = \frac{6\hat{\imath} + 9\hat{\jmath} + 12\hat{k} + 12\hat{\imath} + 30\hat{\jmath} + 42\hat{k}}{9} = \frac{18\hat{\imath} + 39\hat{\jmath} + 54\hat{k}}{9} = 2\hat{\imath} + \frac{13}{3}\hat{\jmath} + 6\hat{k}$.Checking options: $\frac{1}{3}(6\hat{\imath} + 13\hat{\jmath} + 18\hat{k}) = 2\hat{\imath} + \frac{13}{3}\hat{\jmath} + 6\hat{k}$. This matches option $D$.

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