Let $\overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and…

Let $\overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{OB}}=-2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$ be the position vectors of two points $A$ and $B$. If $C$ is a point on the bisector $\angle \mathrm{AOB}$ and $\mathrm{OC}=\sqrt{42}$, then $\overrightarrow{\mathrm{OC}}=$
  1. $4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}}$
  2. $\hat{i}+5 \hat{j}+4 \hat{k}$
  3. $5 \hat{i}+4 \hat{j}+\hat{k}$
  4. $\hat{i}-4 \hat{j}+5 \hat{k}$

Solution

Given $\overrightarrow{O A}=\vec{i}+2 \vec{j}-2 \vec{k}$ and $\overrightarrow{O B}=-2 \vec{i}-3 \vec{j}+6 \vec{k}$ Now unit vector bisector $\angle A O B$ $\begin{aligned} & =\frac{\vec{i}+2 \vec{j}-2 \vec{k}}{3} \pm \frac{(-2 i-3 j+6 k)}{7} \\ & =\frac{\vec{i}+5 \vec{j}+4 \vec{k}}{21} \text { or } \frac{13 \vec{i}+23 \vec{j}-32 \vec{k}}{21} \end{aligned}$ Since $\frac{(1+25+16)}{441} x^2=42 \Rightarrow x=21$ So required vector $=\vec{i}+5 \vec{j}+4 \vec{k}$

Asked in: AP EAMCET 2023 (15 May Shift 2)

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