Question

If the vector $19\hat{i}+22\hat{j}+5\hat{k}$ bisects an angle between the vectors $a$ and $6\hat{i}+8\hat{j}$, then the unit vector in the direction of $a$ is

• A. $\frac{1}{5}(4\hat{i}+3\hat{k})$
• B. $\frac{1}{3}(2\hat{i}+2\hat{j}+\hat{k})$
• C. $\frac{1}{3}(\hat{i}+2\hat{j}+2\hat{k})$
• D. $\frac{1}{3}(2\hat{i}+2\hat{j}-\hat{k})$

Answer 1

Answer: (B)

Let vector $a=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$
Now, vector bisector of angle between $a$ and $6\hat{i}+8\hat{j}$ is
$\lambda \left(\frac{a}{|a|}+\frac{6\hat{i}+8\hat{j}}{\sqrt{6^2+8^2}}\right)=\lambda \left(\frac{a}{|a|}+\frac{6\hat{i}+8\hat{j}}{10}\right)$
$=19\hat{i}+22\hat{j}+5\hat{k}($ given $)$
$\Rightarrow \frac{a}{|a|}=\frac{19\hat{i}+22\hat{j}+5\hat{k}}{\lambda }-\left(\frac{3}{5}\hat{i}+\frac{4}{5}\hat{j}\right)$
$=\left(\frac{19}{\lambda }-\frac{3}{5}\right)\hat{i}+\left(\frac{22}{\lambda }-\frac{4}{5}\right)\hat{j}+\frac{5}{\lambda }\hat{k}$
$\because \frac{a}{|a|}$ is unit vector along $a$, so
${\left(\frac{19}{\lambda }-\frac{3}{5}\right)}^2+{\left(\frac{22}{\lambda }-\frac{4}{5}\right)}^2+{\left(\frac{5}{\lambda }\right)}^2=1$
$\frac{361}{{\lambda }^2}+\frac{484}{{\lambda }^2}+\frac{25}{{\lambda }^2}+\frac{9}{25}+\frac{16}{25}-\frac{114}{5\lambda }-\frac{176}{5\lambda }=1$
$\Rightarrow \frac{870}{{\lambda }^2}-\frac{290}{5\lambda }=0$
$\Rightarrow \lambda =15$
So, $\frac{a}{|a|}=\left(\frac{19}{15}-\frac{3}{5}\right)\hat{i}+\left(\frac{22}{15}-\frac{4}{5}\right)\hat{j}+\frac{\hat{k}}{3}$
$=\frac{10}{15}\hat{i}+\frac{10}{15}\hat{j}+\frac{\hat{k}}{3}=\frac{1}{3}(2\hat{i}+2\hat{j}+\hat{k})$