Question

Find the angle between the following pair of lines : \( \frac{x-2}{3} = \frac{y+5}{2} = \frac{1-z}{-6} \) and \( \frac{x-7}{1} = \frac{y}{2} = \frac{6-z}{-2} \).

Hint:

Always ensure lines are in standard form (\( x, y, z \) coefficients are +1) before extracting direction ratios. A common trick is to hide a negative sign in the denominator like \( (1-z)/a = (z-1)/(-a) \).

Answer 1

Concept:
• The angle between two lines is the angle between their direction vectors.
• Standard form of a line: \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), where \( (a, b, c) \) are direction ratios.
• Formula for cosine of angle \( \theta \): \( \cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}||\vec{b_2}|} \).

Step 1: Write the lines in standard form and identify direction vectors
Line 1: \( \frac{x-2}{3} = \frac{y+5}{2} = \frac{z-1}{6} \) (Notice the sign change in the \( z \) term to keep \( z \) positive).
Direction vector \( \vec{b_1} = 3\hat{i} + 2\hat{j} + 6\hat{k} \).
Line 2: \( \frac{x-7}{1} = \frac{y}{2} = \frac{z-6}{2} \) (Notice the sign change in the \( z \) term).
Direction vector \( \vec{b_2} = 1\hat{i} + 2\hat{j} + 2\hat{k} \).

Step 2: Calculate dot product and magnitudes
Dot product \( \vec{b_1} \cdot \vec{b_2} \):
\[ (3)(1) + (2)(2) + (6)(2) = 3 + 4 + 12 = 19 \]
Magnitude \( |\vec{b_1}| \):
\[ \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \]
Magnitude \( |\vec{b_2}| \):
\[ \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

Step 3: Solve for the angle \( \theta \)
\[ \cos \theta = \frac{19}{7 \times 3} = \frac{19}{21} \]
\[ \theta = \cos^{-1} \left( \frac{19}{21} \right) \]