Question

What is the value of n so that the angle between the lines having direction ratios $(1, 1, 1)$ and $(1, -1, n)$ is $60^{\circ}$ ?

• A. $\sqrt{3}$
• B. $\sqrt{6}$
• C. 3
• D. None of these

Answer 1

Answer: (B)

$\sqrt{6}$

Answer 2

If $\left(l_{1}, m_{1}, n_{1}\right)$ and $\left(l_{2}, m_{2}, n_{2}\right)$ are the direction ratios then
angle between the lines is $\cos \theta=\frac{l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}}{\sqrt{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}} \sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}}$
Here $l_{1}=1, m_{1}=1, n_{1}=1$
and $l_{2}=1, m_{2}=-1, n_{2}=n$ and $q=60^{\circ}$
$\therefore \cos 60^{\circ}=\frac{1 \times 1+1 \times(-1)+1 \times n}{\sqrt{1^{2}+1^{2}+1^{2}} \times \sqrt{1^{2}+1^{2}+n^{2}}}$
$\Rightarrow \frac{1}{2}=\frac{n}{\sqrt{3} \sqrt{2+n^{2}}}$
$\Rightarrow 3\left(2+n^{2}\right)=4 n^{2}$
$\Rightarrow n^{2}=6$
$\Rightarrow n=\pm \sqrt{6}$