If $(\alpha, \beta, \gamma)$ are the Direction cosines of an angular bisector of two lines whose Direction…
If $(\alpha, \beta, \gamma)$ are the Direction cosines of an angular bisector of two lines whose Direction ratios are $(2,2,1)$ and $(2,-1,-2)$, then $(\alpha+\beta+\gamma)^2=$
$3$
$2$
$4$
$5$
Solution
Given, direction ratio are $(2,2,1)$ and $(2,-1,-2)$ Now, angle between the lines is given by $\cos \theta=\frac{4-2-2}{\sqrt{9} \cdot \sqrt{9}}=0 \Rightarrow \theta=\frac{\pi}{2}$
So, direction cosines of two lines are
$\left(\frac{2}{\sqrt{9}}, \frac{2}{\sqrt{9}}, \frac{1}{\sqrt{9}}\right)$ and $\left(\frac{2}{\sqrt{9}}, \frac{-1}{\sqrt{9}}, \frac{-2}{\sqrt{9}}\right)$
$=\left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)$ and $\left(\frac{2}{3}, \frac{-1}{3}, \frac{-2}{3}\right)$
So, required direction cosine is
$\left(\frac{l_1-l_2}{2 \sin \frac{\theta}{2}}, \frac{m_1-m_2}{2 \sin \frac{\theta}{2}}, \frac{n_1-n_2}{2 \sin \frac{\theta}{2}}\right)$
$\begin{aligned} & =\left(\frac{0}{2 \times \frac{1}{\sqrt{2}}}, \frac{1}{2 \times \frac{1}{\sqrt{2}}}, \frac{1}{2 \times \frac{1}{\sqrt{2}}}\right) \\ & =\left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)=(\alpha, \beta, \gamma)\end{aligned}$
Now, $(\alpha+\beta+\gamma) 2=\left(\frac{2}{\sqrt{2}}\right)^2=\frac{4}{2}=2$