$\vec{c}$ is a vector along the bisector of the internal angle between the vectors $\vec{a}=4 \hat{i}+7…

$\vec{c}$ is a vector along the bisector of the internal angle between the vectors $\vec{a}=4 \hat{i}+7 \hat{j}-4 \hat{k}$ and $\vec{b}=12 \hat{i}-3 \hat{j}+4 \hat{k}$. If the magnitude of $\vec{c}$ is $3 \sqrt{13}$ then $\vec{c}=$
  1. $5 \hat{i}-8 \hat{j}+2 \sqrt{2} \hat{k}$
  2. $10 \hat{i}+4 \hat{j}-\hat{k}$
  3. $\hat{i}-10 \hat{j}+4 \hat{k}$
  4. $2 \sqrt{2} \hat{i}+5 \hat{j}-8 \hat{k}$

Solution

$\begin{aligned} & \text { (b) } \vec{a}=4 \hat{i}+7 \hat{j}-4 \hat{k}, \vec{b}=12 \hat{i}-3 \hat{j}+4 \hat{k} \\ & \mathrm{U}_{\vec{a}}=\frac{4 \hat{i}+7 \hat{j}-4 \hat{k}}{9}, \mathrm{U}_{\vec{b}}=\frac{12 \hat{i}-3 \hat{j}+4 \hat{k}}{13} \end{aligned}$ unit vector along bisector of vector $\vec{a}$ and $\vec{b}$ is $\begin{aligned} & \mathrm{U}_{\vec{c}}=k\left(\mathrm{U}_{\vec{a}}+\mathrm{U}_{\vec{b}}\right) \\ & \mathrm{U}_{\vec{c}}=k\left(\frac{160 \hat{i}+64 \hat{j}-16 \hat{k}}{117}\right) \text { or } k\left(\frac{-56 \hat{i}+64 \hat{j}-88 \hat{k}}{117}\right) \\ & \mathrm{U}_{\vec{c}}=\frac{16 k}{117}(10 \hat{i}+4 \hat{j}-\hat{k}) \text { or } \frac{8 k}{117}(-7 \hat{i}+8 \hat{j}-11 \hat{k}) \end{aligned}$
Since magnitude of $\vec{c}=3 \sqrt{13}$ $\therefore \vec{c}=10 \hat{i}+4 \hat{j}-\hat{k}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

Practice more Vectors questions on Aicharya