Question

The angle between the straight lines \( x - 1 = \frac{2y + 3}{3} = \frac{z + 5}{2} \) and \( x = 3r + 2; y = -2r - 1; z = 2 \), where \( r \) is a parameter, is:

• A. \( \frac{\pi}{4} \)
• B. \( \cos^{-1}\left(\frac{-3}{\sqrt{182}}\right) \)
• C. \( \sin^{-1}\left(\frac{-3}{\sqrt{182}}\right) \)
• D. \( \frac{\pi}{2} \)
• E. \( 0 \)

Hint:

Always normalize the coefficients of $x, y,$ and $z$ to 1 in symmetric equations before identifying direction ratios. A common trap is having $2y$ instead of $y$.

Answer 1

The Correct Option is D

Concept: The angle \( \theta \) between two lines is the angle between their direction vectors, \( \vec{b_1} \) and \( \vec{b_2} \). Two lines are perpendicular (\( \theta = \pi/2 \)) if their dot product is zero: \( \vec{b_1} \cdot \vec{b_2} = 0 \).

Step 1: Find the direction vector \( \vec{b_1} \) of the first line.
The first line is \( x - 1 = \frac{2y + 3}{3} = \frac{z + 5}{2} \). To find the standard form \( \frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n} \), we must ensure the coefficients of \( x, y, z \) are 1: \[ \frac{x - 1}{1} = \frac{y + 3/2}{3/2} = \frac{z + 5}{2} \] The direction ratios are \( (1, 3/2, 2) \). Multiplying by 2 to clear fractions, we get \( \vec{b_1} = (2, 3, 4) \).

Step 2: Find the direction vector \( \vec{b_2} \) of the second line.
The second line is given parametrically: \( x = 3r + 2, y = -2r - 1, z = 2 \). This can be written as \( \frac{x - 2}{3} = \frac{y + 1}{-2} \) and \( z = 2 \). The direction vector is formed by the coefficients of the parameter \( r \): \( \vec{b_2} = (3, -2, 0) \).

Step 3: Calculate the dot product.
\[ \vec{b_1} \cdot \vec{b_2} = (2)(3) + (3)(-2) + (4)(0) \] \[ \vec{b_1} \cdot \vec{b_2} = 6 - 6 + 0 = 0 \] Since the dot product is zero, the lines are perpendicular.