If 3 hat{j},4 hat{k} and 3 hat{j}+4 hat{k} ar | vedclass

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

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$\frac{5}{3} \hat{j}+4 \hat{k}$

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(D) Let the position vectors of vertices $A$,$B$,and $C$ be $\vec{a} = 3 \hat{j}$,$\vec{b} = 4 \hat{k}$,and $\vec{c} = 3 \hat{j} + 4 \hat{k}$ respectively.According to the Angle Bisector Theorem,the bisector of $\angle A$ divides the opposite side $BC$ in the ratio of the lengths of the sides adjacent to the angle,i.e.,$AB : AC$.First,calculate the lengths of sides $AB$ and $AC$:$AB = |\vec{b} - \vec{a}| = |4 \hat{k} - 3 \hat{j}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = 5$.$AC = |\vec{c} - \vec{a}| = |(3 \hat{j} + 4 \hat{k}) - 3 \hat{j}| = |4 \hat{k}| = 4$.Thus,the bisector of $\angle A$ divides $BC$ in the ratio $m : n = 5 : 4$.The position vector $\vec{d}$ of the point $D$ on $BC$ is given by the section formula:$\vec{d} = \frac{m \vec{c} + n \vec{b}}{m + n} = \frac{5(3 \hat{j} + 4 \hat{k}) + 4(4 \hat{k})}{5 + 4}$.$\vec{d} = \frac{15 \hat{j} + 20 \hat{k} + 16 \hat{k}}{9} = \frac{15 \hat{j} + 36 \hat{k}}{9}$.$\vec{d} = \frac{15}{9} \hat{j} + \frac{36}{9} \hat{k} = \frac{5}{3} \hat{j} + 4 \hat{k}$.

If $3 \hat{j}$,$4 \hat{k}$ and $3 \hat{j}+4 \hat{k}$ 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 $\angle A$ meets $BC$ is

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