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, $\frac{ a }{| a |}=\left(\frac{19}{15}-\frac{3}{5}\right) \hat{ i }+\left(\frac{22}{15}-\frac{4}{5}\right) \hat{ j }+\frac{\hat{ k }}{3}$
$=\frac{10}{15} \hat{ i }+\frac{10}{15} \hat{ j }+\frac{\hat{ k }}{3}=\frac{1}{3}(2 \hat{ i }+2 \hat{ j }+\hat{ k })$