Q62 / 100 · BITSAT 2024 (Memory Based Paper 2) · Mathematics
The angle between the lines whose direction cosines are given by the equations $3 l+m+5 n=$ $0,6 \mathrm{~nm}-2 \mathrm{n} l+5 l \mathrm{m}=0$ is
- A$\cos ^{-1}\left(\frac{1}{6}\right)$
- Correct answerB$\cos ^{-1}\left(-\frac{1}{6}\right)$
- C$\cos ^{-1}\left(\frac{2}{3}\right)$
- D$\cos ^{-1}\left(-\frac{5}{6}\right)$
Solution
The given equations are
$3 l+m+5 n=0$ ...(i)
and $6 \mathrm{mn}-2 \mathrm{n} l+5 l \mathrm{m}=0$ ...(ii)
From (i), we have $m=-3 l-5 n$.
Putting $\mathrm{m}=-3 l-5 \mathrm{n}$ in (ii),
we get $6(-3 l-5 n) n-2 n l+5 l(-3 l-5 n)=0$
$\Rightarrow(n+l)(2 n+l)=0$
$\Rightarrow$ either $l=-\mathrm{n}$ or $l=-2 \mathrm{n}$.
If $l=-n$, then putting $l=-n$ in (i), we obtain $m=-2 n$.
If $l=-2 \mathrm{n}$, then putting $l=-2 \mathrm{n}$ in (i), we obtain $\mathrm{m}=\mathrm{n}$.
Thus, the direction ratios of two lines are $-\mathrm{n},-$ $2 \mathrm{n}, \mathrm{n}$ and $-2 \mathrm{n}, \mathrm{n}, \mathrm{n}$ i.e., $1,2,-1$ and $-2,1,1$.
Hence, the direction cosines are
$\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}$ or $\frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}$. The angle $\theta$
between the lines is given by
$\begin{aligned}
& \cos \theta=\frac{1}{\sqrt{6}} \times \frac{-2}{\sqrt{6}}+\frac{2}{\sqrt{6}} \times \frac{1}{\sqrt{6}}+\frac{-1}{\sqrt{6}} \times \frac{1}{\sqrt{6}}=\frac{-1}{6} \\
& \Rightarrow \theta=\cos ^{-1}\left(\frac{-1}{6}\right)
\end{aligned}$
Hence, the correct answer is option: 'B'.