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

Answer 2

Given:
Line 1: \( \frac{x+3}{3} = \frac{y-1}{5} = \frac{z+3}{4} \)
Line 2: \( \frac{x+1}{1} = \frac{y-4}{4} = \frac{z-5}{2} \)

Step 1: Direction ratios
From Line 1: direction vector \( \vec{a}_1 = \langle 3, 5, 4 \rangle \)
From Line 2: direction vector \( \vec{a}_2 = \langle 1, 4, 2 \rangle \)

Step 2: Dot product
\( \vec{a}_1 \cdot \vec{a}_2 = (3)(1) + (5)(4) + (4)(2) = 3 + 20 + 8 = 31 \)

Step 3: Magnitudes
\( |\vec{a}_1| = \sqrt{3^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} \)
\( |\vec{a}_2| = \sqrt{1^2 + 4^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21} \) 

Step 4: Cosine of angle
 \(\cos \theta = \frac{\vec{a}_1 \cdot \vec{a}_2}{|\vec{a}_1||\vec{a}_2|}\) = \(\frac{27}{160} \times \frac{\sqrt{75}}{\sqrt{21}}\)

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

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

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

Answer: \(\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)\)

Answer 3

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 \(\left( 1, \frac{1}{4}, \frac{1}{2} \right)\), 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:

\[ \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{5}\right)^2 + \left(\frac{1}{4}\right)^2} \]

\[ = \sqrt{\frac{1}{9} + \frac{1}{25} + \frac{1}{16}} = \sqrt{\frac{25}{225} + \frac{9}{225} + \frac{14}{225}} \]

\[ = \sqrt{\frac{48}{225}} = \sqrt{\frac{16}{75}} = \frac{4}{\sqrt{75}} \]

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) \(\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)\).