Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

The direction cosines of two mutually perpendicular lines are $\ell_1, m_1, n_1$ and $\ell_2, m_2, n_2$. The direction cosines of the line perpendicular to both the given lines will be :

Options:

$\ell_1+\ell_2, m_1+m_2, n_1+n_2$

$\ell_1-\ell_2, m_1-m_2, n_1-n_2$

$\ell_1 \ell_2, m_1 m_2, n_1 n_2$

$m_1 n_2-m_2 n_1, n_1 \ell_2-n_2 \ell_1, \ell_1 m_2-\ell_2 m_1$

Correct Answer:

$m_1 n_2-m_2 n_1, n_1 \ell_2-n_2 \ell_1, \ell_1 m_2-\ell_2 m_1$

Explanation:

Let the direction cosine of the required line be $\ell$, m and n.

We must have,

$\ell \ell_1+mm_1+nn_1=0, \ell \ell_2+mm_2+nn_2=0$

$\Rightarrow \frac{\ell}{m_1 n_2-n_1 m_2}=\frac{m}{n_1 \ell_2-n_2 \ell_1}=\frac{n}{\ell_1 m_2-\ell_2 m_1}$

$=\frac{\sqrt{\ell^2+m^2+n^2}}{\sqrt{\sum\left(m_1 n_2-n_2 m_2\right)^2}}$

We have $\ell_1 \ell_2+m_1 m_2+n_1 n_2=0$

Thus

$\sum\left(m_1 n_2-n_1 m_2\right)^2=\left(\sum \ell_1\right)^2\left(\sum \ell_2\right)^2-\sum \ell_1 \ell_2=1$

$\Rightarrow \ell=m_1 n_2-n_1 m_2, m=n_1 \ell_2-n_2 \ell_1, n=\ell_1 m_2-\ell_2 m_1$