The angle between the lines whose direction cosines are given by the equations $l^2+m^2-n^2=0$ and $l+m+n=0$…
The angle between the lines whose direction cosines are given by the equations $l^2+m^2-n^2=0$ and $l+m+n=0$ is
- $\frac{\pi}{6}$
- $\frac{\pi}{4}$
- $\frac{\pi}{3}$
- $\frac{\pi}{2}$
Solution
Given,
$
\begin{aligned}
l^2+m^2-n^2 & =0 ...(i)\\
l+m+n & =0 ...(ii)
\end{aligned}
$
By Eq. (ii), $n=-(l+m)$
Put it Eq. (i), $l^2+m^2=n^2$
$
\begin{array}{lc}
\Rightarrow & l^2+m^2=[-(l+m)]^2 \\
\Rightarrow & l^2+m^2=l^2+m^2+2 l m \\
\Rightarrow & 2 l m=0 \\
\therefore & l=0 \text { or } m=0 \\
\Rightarrow & n=-m \text { or } n=-l \\
\because & l^2+m^2+n^2=l \text { and } l^2+m^2+n^2=1 \\
& 0+m^2+m^2=l \text { and } l^2+l^2=1
\end{array}
$
$
\begin{gathered}
2 m^2=1 \text { and } l^2=\frac{1}{2} \\
m= \pm \frac{1}{\sqrt{2}} \text { and } l= \pm \frac{1}{\sqrt{2}}
\end{gathered}
$
Direction cosine $\left(0, \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$ and $\left(0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
or $\left(\frac{1}{\sqrt{2}}, 0,-\frac{1}{\sqrt{2}}\right)$ and $\left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)$
$\Rightarrow$ Direction of two lines
$
\begin{array}{r}
\therefore \quad\left(l_1, m_1, n_1\right)=\left(0, \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right) \\
\left(l_2, m_2, n_2\right)=\left(\frac{1}{\sqrt{2}}, 0,-\frac{1}{\sqrt{2}}\right)
\end{array}
$
Let $\theta$ be the angle between the lines
$
\begin{aligned}
& \therefore \quad \cos \theta=\left|l_1 l_2+m_1 m_2+n_1 n_2\right|=\left|0+0+\frac{1}{2}\right| \\
& \cos \theta=1 / 2 \Rightarrow \theta=\pi / 3
\end{aligned}
$
Asked in: AP EAMCET 2021 (24 Aug Shift 1)
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