The acute angle between two lines such that the direction cosines $l, m, n$, of each of them satisfy the…
The acute angle between two lines such that the direction cosines $l, m, n$, of each of them satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is :
$15^{\circ}$
$30^{\circ}$
$60^{\circ}$
$45^{\circ}$
Solution
Let $l_1, m_1, n_1$ and $l_2, m_2, n_2$ be the d.c of line 1 and 2 respectively, then as given $l_1+m_1+n_1=0$
and $l_2+m_2+n_2=0$
and $l_1{ }^2+m_1{ }^2-n_1^2=0$ and $l_2^2+m_2^2-n_2^2=0$
$\left(\because l+m+n=0\right.$ and $\left.l^2+m^2-n^2=0\right)$
Angle between lines, $\theta$ is
$
\cos \theta=l_1 l_2+m_1 m_2+n_1 n_2
$
As given $l^2+m^2=n^2$ and $l+m=-n$ $\Rightarrow(-n)^2-2 l m=n^2 \Rightarrow 2 l m=0$ or $l m=0$
So $l_1 m_1=0, l_2 m_2=0$
If $l_1=0, m_1 \neq 0$ then $l_1 m_2=0$
If $m_1=0, l_1 \neq 0$ then $l_2 m_1=0$
If $l_2=0, m_2 \neq 0$ then $l_2 m_1=0$
If $m_2=0, l_2 \neq 0$ then $l_1 m_2=0$
Also $l_1 l_2=0$ and $m_1 m_2=0$
$
\begin{aligned}
& l^2+m^2-n^2=l^2+m^2+n^2-2 n^2=0 \\
& \Rightarrow 1-2 n^2=0 \Rightarrow n=\pm \frac{1}{\sqrt{2}} \\
& \therefore n_1=\pm \frac{1}{\sqrt{2}}, n_2=\pm \frac{1}{\sqrt{2}} \\
& \therefore \cos \theta=\frac{1}{2} \theta=60^{\circ} \text { (acute angle) }
\end{aligned}
$