Question

The angle between the pair of lines \( \frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3} \) and \( \frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4} \) is:

• A. \( \cos^{-1} \left( \frac{21}{9\sqrt{38}} \right) \)
• B. \( \cos^{-1} \left( \frac{23}{9\sqrt{38}} \right) \)
• C. \( \cos^{-1} \left( \frac{24}{9\sqrt{38}} \right) \)
• D. \( \cos^{-1} \left( \frac{25}{9\sqrt{38}} \right) \)
• E. \( \cos^{-1} \left( \frac{26}{9\sqrt{38}} \right) \)

Hint:

Direction ratios are always found in the denominators of the symmetric form of line equations, provided the coefficients of \( x, y, \) and \( z \) are positive unity.

Answer 1

Answer: (E)

Concept: The angle \( \theta \) between two lines with direction ratios \( (a_1, b_1, c_1) \) and \( (a_2, b_2, c_2) \) is given by the formula: \[ \cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \]

Step 1: Identify the direction ratios (D.R.s) of the lines.
From Line 1: \( (a_1, b_1, c_1) = (2, 5, -3) \) From Line 2: \( (a_2, b_2, c_2) = (-1, 8, 4) \)

Step 2: Calculate the numerator and denominator components.
Numerator \( (a_1a_2 + b_1b_2 + c_1c_2) \): \[ 2(-1) + 5(8) + (-3)(4) = -2 + 40 - 12 = 26 \] Denominator Part 1 \( (\sqrt{a_1^2 + b_1^2 + c_1^2}) \): \[ \sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38} \] Denominator Part 2 \( (\sqrt{a_2^2 + b_2^2 + c_2^2}) \): \[ \sqrt{(-1)^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9 \]

Step 3: Substitute into the cosine formula.
\[ \cos \theta = \frac{26}{9\sqrt{38}} \] \[ \theta = \cos^{-1} \left( \frac{26}{9\sqrt{38}} \right) \]