Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

The angle between lines whose direction cosines are given by $\ell+m+n=0$, $\ell^2+m^2-n^2=0$, is :

Options:

$\frac{\pi}{2}$

$\frac{\pi}{3}$

$\frac{\pi}{6}$

None of these

Correct Answer:

None of these

Explanation:

$\ell+m+n=0, \ell^2+m^2-n^2=0$

We also have

$\ell^2+m^2+n^2=1 $

$\Rightarrow 2 n^2=1 $

$\Rightarrow n=\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$

Also, $\ell^2+m^2=n^2=(-(\ell+m))^2$

$\Rightarrow \ell m=0$ and $\ell+m= \pm \frac{1}{\sqrt{2}}$

Hence, direction cosines are lines are

$\left(\frac{1}{\sqrt{2}}, 0,-\frac{1}{\sqrt{2}}\right),\left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right), $

$\left(0, \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\left(0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$

Angle between these lines in both cases is zero.