Q75 / 125 · BITSAT 2012 · Mathematics

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
    $\sqrt{3}$
  • Correct answer
    B
    $\sqrt{6}$
  • C
    3
  • D
    None of these

Solution

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 $\theta=60^{\circ}$ $\therefore \cos 60^{\circ}=\frac{1 \times 1+1 \times(-1)+1 \times n}{\sqrt{h^{2}+1^{2}+1^{2} \times} \sqrt{1^{2}+1^{2}+y^{2}}}$ $\Rightarrow \frac{1}{2}=\frac{1}{\sqrt{3} \sqrt{2+n^{2}}} \Rightarrow 3\left(2+n^{2}\right)=4 n^{2}$ $\Rightarrow n^{2}=6 \Rightarrow n=\pm \sqrt{6}$

Hence, the correct answer is option: 'B'.

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