Question

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

• A. cos^-1\((\frac {4}{9})\)
• B. cos^-1\((\frac {2}{9})\)
• C. cos^-1\((\frac {1}{9})\)
• D. cos^-1\((\frac {5}{9})\)

Answer 1

The Correct Option is D

To find the angle between two lines, we can use the direction vectors of the lines.For the first line, x + 1 = y + 3 = z - 4,the vector form is 
r₁ = (x, y, z) = (-1, -3, 4) + t(1, 1, 1)
The direction vector for this line is (1, 1, 1).
For the second line,  \(\frac {x-4}{1}\) = \(\frac {y+2}{2}\) = \(\frac {z+1}{2}\) 
the vector form is r₂ = (x, y, z) = (4, -4, -\(\frac {1}{2}\)) + s(1, 2, \(\frac {1}{2}\))
The direction vector for this line is (1, 2, \(\frac {1}{2}\)).To find the angle between two vectors, use the dot product formula:
cosθ =\( \frac {u · v}{|u| |v|}\)
calculate the dot product and magnitudes:
u · v = (1, 1, 1) · (1, 2, \(\frac {1}{2}\)) = 1 + 2 + \(\frac {1}{2}\) = \(\frac {9}{2}\)
|u| = \(\sqrt {1² + 1² + 1²}\) = \(\sqrt 3\)
|v| = \(\sqrt {1² + 1² + (\frac {1}{2})^2}\) = \(\sqrt {1 + 4 + \frac {1}{4}}\) = \(\sqrt {\frac {17}{4}}\) = \(\frac {\sqrt {17}}{2}\)
cosθ = \( \frac {u · v}{|u| |v|}\) =\(\frac {\frac {9}{2}}{\sqrt 3 \times \frac {\sqrt {17}}{2}}\) = \(\frac {9}{2}\) x \(\frac {2}{\sqrt 3 .\sqrt {17}}\)= \(\frac {9}{\sqrt 3 .\sqrt {17}}\) = \(\frac {9}{\sqrt {51}}\)
To determine the angle θ, we need to find the inverse cosine (arccos) of cosθ
θ = arccos\((\frac {9}{\sqrt {51}})\)
Therefore, the angle between the two lines is equal to cos^-1\((\frac {5}{9})\).