Let $\overline{O A}=-4 \bar{i}+3 \bar{k}, \overline{O B}=14 \bar{i}+2 \bar{j}-5 \bar{k} \cdot \overline{O…

Let $\overline{O A}=-4 \bar{i}+3 \bar{k}, \overline{O B}=14 \bar{i}+2 \bar{j}-5 \bar{k} \cdot \overline{O D}$ bisects $\angle A O B$ and $|\overline{O D}|=\sqrt{6}$, then $\overline{O D}=$
  1. $\pm(\bar{i}+\bar{j}+2 \bar{k})$
  2. $\pm(\bar{i}+2 \bar{j}+\bar{k})$
  3. $\pm(2 \bar{i}+\bar{j}+\bar{k})$
  4. $\pm \frac{1}{\sqrt{2}}(2 \bar{i}+\bar{j}+\sqrt{7} \bar{k})$

Solution

Considering $\overrightarrow{O A}$ and $\overrightarrow{O B}$ as adjacent sides of a parallogram. $\overrightarrow{O D}$ will be along diagonal. Equation of diagonal $\begin{aligned} & = \pm \lambda\left(\frac{\vec{a}}{|\vec{a}|}+\frac{\vec{b}}{|\vec{b}|}\right) \\ & = \pm \lambda\left(\frac{-4 \hat{i}+3 \hat{k}}{\sqrt{4^2+3^2}}+\frac{14 \hat{i}+2 \hat{j}-5 \hat{k}}{\sqrt{14^2+2^2+5^2}}\right) \\ & = \pm \lambda\left(\frac{-4 \hat{i}+3 \hat{k}}{5}+\frac{14 \hat{i}+2 \hat{j}-5 \hat{k}}{15}\right)\end{aligned}$ $\begin{aligned} & = \pm \frac{\lambda}{15}(2 \hat{i}+2 \hat{j}+4 \hat{k}) \\ & = \pm \frac{2 \lambda}{15}(\hat{i}+\hat{j}+2 \hat{k})\end{aligned}$ $\because|\overrightarrow{O D}|=\sqrt{6}$ $\frac{2 \lambda}{15} \sqrt{1^2+1^2+2^2}=\sqrt{6}$ $\therefore \frac{2 \lambda}{15}=1$ $\therefore$ Equation of $\overrightarrow{O D}$ is $\pm(\hat{i}+\hat{j}+2 \hat{k})$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

Practice more Vectors questions on Aicharya