Q209 / 270 · NDA 2017 (Phase 1) · Mathematics

What is the acute angle between the pair of straight lines $\sqrt{2} x+\sqrt{3} y=1$ and $\sqrt{3} x+\sqrt{2} y=2 ?$

  • Correct answer
    A
    $\tan ^{-1}\left(\frac{1}{2 \sqrt{6}}\right)$
  • B
    $\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
  • C
    $\tan ^{-1}(3)$
  • D
    $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Solution

Given lines, $\sqrt{2} x+\sqrt{3} y=1$ and $\sqrt{3} x+\sqrt{2} y=2$ We know, Slope $(\mathrm{m})=\frac{-\mathrm{a}}{\mathrm{b}}$ $\therefore$ Slope of $\sqrt{2} x+\sqrt{3} y=1$ is $m_1=\frac{-\sqrt{2}}{\sqrt{3}}$ Slope of $\sqrt{3} x+\sqrt{2} y=2$ is $m_2=\frac{-\sqrt{3}}{\sqrt{2}}$ $\begin{aligned} & \tan \theta=\left|\frac{\mathrm{m}_1-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|=\left|\frac{\frac{-\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{2}}{\sqrt{3}}}{1+\left(\frac{-\sqrt{2}}{\sqrt{3}}\right)\left(\frac{-\sqrt{3}}{\sqrt{2}}\right)}\right| \\ & =\left|\frac{\frac{-3+2}{\sqrt{6}}}{2}\right|=\left|\frac{-1}{2 \sqrt{6}}\right|=\frac{1}{2 \sqrt{6}} \\ & \therefore \theta=\tan ^{-1}\left(\frac{1}{2 \sqrt{6}}\right) \end{aligned}$

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

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