Question

If the direction cosines of two lines are given by $1+\mathrm{m}+\mathrm{n}=0$ and ${\mathrm{l}}^2-5{\mathrm{m}}^2+{\mathrm{n}}^2=0$, then the angle between them is:

• A. $\frac{\mathrm{\pi }}{2}$
• B. $\frac{\mathrm{\pi }}{6}$
• C. $\frac{\mathrm{\pi }}{4}$
• D. $\frac{\mathrm{\pi }}{3}$

Answer 1

The correct option is D

$\frac{\mathrm{\pi }}{3}$

Explanation for the correct option.

Step 1: Find the relation between direction cosines

Given the direction cosines of lines are, $1+\mathrm{m}+\mathrm{n}=0$ and ${\mathrm{l}}^2-5{\mathrm{m}}^2+{\mathrm{n}}^2=0$.

$$\begin{gathered} 1+\mathrm{m}+\mathrm{n}=0 \\ \Rightarrow 1+\mathrm{n}=-\mathrm{m}...\left(1\right) \end{gathered}$$

Substitute equation (1) in ${\mathrm{l}}^2-5{\mathrm{m}}^2+{\mathrm{n}}^2=0$ we get:

$\begin{array}{rcl}{l}^2+n^2-5(-l-n{)}^2& =& 0\\ l^2+n^2-5\left(l^2+n^2+2ln\right)& =& 0\\ -4l^2-4n^2-10ln& =& 0\\ 2l^2+2n^2+5ln& =& 0\\ 2l^2+5ln+2n^2& =& 0\\ l& =& \frac{-5n\pm n\sqrt{25-16}}{4}\end{array}$

Now solving we get:

$\begin{array}{rcl}l& =& \frac{-5n\pm 3n}{4}\\ l& =& -2\mathrm{n}\mathrm{and}\mathrm{l}=\frac{{\displaystyle -\mathrm{n}}}{{\displaystyle 2}}\end{array}$

Step 2: Find the angle

For $l=-2n,m=n$ and direction ratios are:

$\left(\frac{-\mathrm{n}}{2},\frac{-\mathrm{n}}{2},\mathrm{n}\right)\mathrm{and}(-2\mathrm{n},\mathrm{n},\mathrm{n})$

Now the angle between direction ratios is given by:

$\begin{array}{rcl}\cos \theta & =& \frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}\sqrt{{a_2}^2+{b_2}^2+{c_2}^2}}\\ \cos \theta & =& \frac{2-1+2}{\sqrt{6}\sqrt{6}}\\ & =& \frac{3}{6}\\ & =& \frac{1}{2}\end{array}$

$\Rightarrow \theta =\frac{\mathrm{\pi }}{3}$

Hence, option D is correct.