If $3 \hat{\mathrm{j}}, 4 \hat{\mathrm{k}}$ and $3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ are the position…

If $3 \hat{\mathrm{j}}, 4 \hat{\mathrm{k}}$ and $3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ are the position vectors of the vertices $\mathrm{A}, \mathrm{B}$, $C$ respectively of $\triangle \mathrm{ABC}$, then the position vector of the point in which the bisector of $\angle \mathrm{A}$ meets $\mathrm{BC}$ is
  1. $\frac{5}{3} \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$
  2. $5 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$
  3. $5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
  4. $\frac{5}{3} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$

Solution

We have $\overline{\mathrm{a}}=3 \hat{\mathrm{j}}, \mathrm{b}=4 \hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ Angle bisector $\mathrm{AD}$ divides $\mathrm{BC}$ in the ratio $\mathrm{AB}: \mathrm{AC}$ $\mathrm{AB}=\sqrt{9+16}=5 \text { and } \mathrm{AC}=\sqrt{0+16}=4$ Thus $\mathrm{D}$ divides $\mathrm{BC}$ in the ratio $5: 4$ $\begin{aligned} & \therefore \mathrm{D}=\left[0, \frac{5(3)}{5+4}, \frac{5(4)+4(4)}{5+4}\right] \\ & =\left(0, \frac{15}{9}, \frac{20+16}{9}\right)=\left(0, \frac{5}{3}, 4\right)=\frac{5}{3} \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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