Question

The angle between the pair of lines \(\frac{x+3}{3} = \frac{y-1}{5} = \frac{z+3}{4}\) and \(\frac{x+1}{1} = \frac{y-4}{4} = \frac{z-5}{2}\) is

• A. \(\theta = \cos^{-1}\left(\frac{27}{5}\right)\)
• B. \(\theta = \cos^{-1}\left(\frac{19}{21}\right)\)
• C. \(\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)\)
• D. \(\theta = \cos^{-1}\left(\frac{5\sqrt{3}}{16}\right)\)

Answer 1

The Correct Option is C

For the first line, the direction vector is given by the coefficients of x, y, and z, which is \(\left( \frac{1}{3}, \frac{1}{5}, \frac{1}{4} \right)\) 
For the second line, the direction vector is also given by the coefficients of x, y, and z, which is \(\left( \frac{1}{1}, \frac{1}{4}, \frac{1}{2} \right)\), or simply \((1, \frac{1}{4}, \frac{1}{2})\)
Now, we can calculate the dot product of the two direction vectors: 
\(\left(\frac{1}{3}\right)(1) + \left(\frac{1}{5}\right)\left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)\left(\frac{1}{2}\right)\)

\(=\frac{1}{3} + \frac{1}{20} + \frac{1}{8}\)
\(=\frac{27}{40}\)

The magnitude of the first direction vector is 
\(\sqrt{(\frac{1}{3})^2 + (\frac{1}{5})^2 + (\frac{1}{4})^2}\)

= \(\sqrt{\frac{1}{9} + \frac{1}{25} + \frac{1}{16}} = \sqrt{\frac{25}{225} + \frac{9}{225} + \frac{14}{225}}\)

= \(\sqrt{\frac{48}{225}}\)

= \(\sqrt{\frac{16}{75}}\)
= \(\frac{4}{\sqrt{75}}\)

The magnitude of the second direction vector is
\(\sqrt{1^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{2}\right)^2}\)

= \(\sqrt{1 + \frac{1}{16} + \frac{1}{4}}\)

= \(\sqrt{\frac{16}{16} + \frac{1}{16} + \frac{4}{16}}\)

= \(\sqrt{\frac{21}{16}}\)

= \(\frac{\sqrt{21}}{4}\)

Using the dot product formula, we have: 
\(\cos(\theta) = \frac{\frac{27}{40}}{\left(\frac{4}{\sqrt{75}}\right) \cdot \left(\frac{\sqrt{21}}{4}\right)}\)

= \(\frac{27}{40} \times \frac{\sqrt{75}}{4\sqrt{21}}\)

= \(\frac{27\sqrt{75}}{160\sqrt{21}}\)

= \(\frac{27}{160} \times \frac{\sqrt{75}}{\sqrt{21}}\)

= \(\frac{27}{160} \times \frac{\sqrt{25 \times 3}}{\sqrt{3 \times 7}}\)

= \(\frac{27}{160} \times \frac{\sqrt{3}}{\sqrt{7}}\)

= \(\frac{27\sqrt{3}}{160\sqrt{7}}\)
Therefore, the correct option is (C) \(\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)\)