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Mathematics Question on Differential equations

The angle between the pair of lines x+33=y−15=z+34\frac{x+3}{3} = \frac{y-1}{5} = \frac{z+3}{4} and x+11=y−44=z−52\frac{x+1}{1} = \frac{y-4}{4} = \frac{z-5}{2} is

A

θ=cos⁡−1(275)\theta = \cos^{-1}\left(\frac{27}{5}\right)

B

θ=cos⁡−1(1921)\theta = \cos^{-1}\left(\frac{19}{21}\right)

C

θ=cos⁡−1(8315)\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)

D

θ=cos⁡−1(5316)\theta = \cos^{-1}\left(\frac{5\sqrt{3}}{16}\right)

Answer

θ=cos⁡−1(8315)\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)

Explanation

Solution

For the first line, the direction vector is given by the coefficients of x, y, and z, which is (13,15,14)\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 (11,14,12)\left( \frac{1}{1}, \frac{1}{4}, \frac{1}{2} \right), or simply (1,14,12)(1, \frac{1}{4}, \frac{1}{2})
Now, we can calculate the dot product of the two direction vectors:
(13)(1)+(15)(14)+(14)(12)\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)

=13+120+18=\frac{1}{3} + \frac{1}{20} + \frac{1}{8}
=2740=\frac{27}{40}

The magnitude of the first direction vector is
(13)2+(15)2+(14)2\sqrt{(\frac{1}{3})^2 + (\frac{1}{5})^2 + (\frac{1}{4})^2}

= 19+125+116=25225+9225+14225\sqrt{\frac{1}{9} + \frac{1}{25} + \frac{1}{16}} = \sqrt{\frac{25}{225} + \frac{9}{225} + \frac{14}{225}}

= 48225\sqrt{\frac{48}{225}}

= 1675\sqrt{\frac{16}{75}}
= 475\frac{4}{\sqrt{75}}

The magnitude of the second direction vector is
12+(14)2+(12)2\sqrt{1^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{2}\right)^2}

= 1+116+14\sqrt{1 + \frac{1}{16} + \frac{1}{4}}

= 1616+116+416\sqrt{\frac{16}{16} + \frac{1}{16} + \frac{4}{16}}

= 2116\sqrt{\frac{21}{16}}

= 214\frac{\sqrt{21}}{4}

Using the dot product formula, we have:
cos⁡(θ)=2740(475)⋅(214)\cos(\theta) = \frac{\frac{27}{40}}{\left(\frac{4}{\sqrt{75}}\right) \cdot \left(\frac{\sqrt{21}}{4}\right)}

= 2740×75421\frac{27}{40} \times \frac{\sqrt{75}}{4\sqrt{21}}

= 277516021\frac{27\sqrt{75}}{160\sqrt{21}}

= 27160×7521\frac{27}{160} \times \frac{\sqrt{75}}{\sqrt{21}}

= 27160×25×33×7\frac{27}{160} \times \frac{\sqrt{25 \times 3}}{\sqrt{3 \times 7}}

= 27160×37\frac{27}{160} \times \frac{\sqrt{3}}{\sqrt{7}}

= 2731607\frac{27\sqrt{3}}{160\sqrt{7}}
Therefore, the correct option is (C) θ=cos⁡−1(8315)\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)