Question

Find the angle between the two lines having direction ratio $(1,1,2)$ and $((\sqrt{3}-1),(-\sqrt{3}-1),4)$.

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{6}$
• D. None of these

Answer 1

Answer: (A)

$\frac{\pi }{3}$

Let $\overrightarrow{m_1}$ and $\overrightarrow{m_2}$ be vectors parallel to the two given lines.

Then, angle between the two given lines is same as the angle between $\overrightarrow{m_1}$ and $\overrightarrow{m_2}$.

$\overrightarrow{m_1}=$ Vector parallel to the line with direction ratios $(1,1,2)=i+j+2k$ and

$\overrightarrow{m_2}=$ Vector parallel to the line with direction ratio $((\sqrt{3}-1),(-\sqrt{3}-1),4)=(\sqrt{3}-1)i+(-\sqrt{3}-1)j+4k$

Let $\theta$ be the angle between the given lines.

Then $\cos \theta =\frac{\overrightarrow{m_1}.\overrightarrow{m_2}}{\left|\overrightarrow{m_1}\right|\left|\overrightarrow{m_2}\right|}=\frac{(\sqrt{3}-1)-(\sqrt{3}+1)+8}{\sqrt{1+1+4}\sqrt{{(\sqrt{3}-1)}^2+{(\sqrt{3}+1)}^2+16}}$

$\Rightarrow \cos \theta =\frac{6}{\sqrt{6}\sqrt{24}}=\frac{1}{2}$

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