Question

The angle between the two lines: \[ \frac{x + 1}{2} = \frac{y + 3}{2} = \frac{z - 4}{-1} \] \[ \frac{x - 4}{1} = \frac{y + 4}{2} = \frac{z + 1}{2} \] is:

• A. \( \cos^{-1} \frac{1}{9} \)
• B. \( \cos^{-1} \frac{4}{9} \)
• C. \( \cos^{-1} \frac{2}{9} \)
• D. \( \cos^{-1} \frac{3}{9} \)

Hint:

The angle between two lines with direction vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \]

Answer 1

The Correct Option is B

Step 1: Extracting direction vectors.
For a line in symmetric form: \[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \] the direction vector is \( (a, b, c) \). For the first line: \[ \frac{x + 1}{2} = \frac{y + 3}{2} = \frac{z - 4}{-1} \] The direction vector is: \[ \mathbf{d_1} = (2, 2, -1) \] For the second line: \[ \frac{x - 4}{1} = \frac{y + 4}{2} = \frac{z + 1}{2} \] The direction vector is: \[ \mathbf{d_2} = (1, 2, 2) \] Step 2: Using the angle formula.
The angle \( \theta \) between two lines with direction vectors \( \mathbf{d_1} = (a_1, b_1, c_1) \) and \( \mathbf{d_2} = (a_2, b_2, c_2) \) is given by: \[ \cos \theta = \frac{\mathbf{d_1} \cdot \mathbf{d_2}}{|\mathbf{d_1}| |\mathbf{d_2}|} \] Step 3: Computing the dot product.
\[ \mathbf{d_1} \cdot \mathbf{d_2} = (2 \times 1) + (2 \times 2) + (-1 \times 2) \] \[ = 2 + 4 - 2 = 4 \] Step 4: Computing magnitudes.
\[ |\mathbf{d_1}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \] \[ |\mathbf{d_2}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \] Step 5: Computing \( \cos \theta \).
\[ \cos \theta = \frac{4}{3 \times 3} = \frac{4}{9} \] Step 6: Computing the angle.
\[ \theta = \cos^{-1} \frac{4}{9} \] Thus, the correct answer is (B) \( \cos^{-1} \frac{4}{9} \).