Question:
Find the angle between the following pair of lines :
\(\frac{x - 2}{3} = \frac{y + 5}{2} = \frac{1 - z}{-6}\) and \(\frac{x - 7}{1} = \frac{y}{2} = \frac{6 - z}{-2}\).
Find the angle between the following pair of lines :
\(\frac{x - 2}{3} = \frac{y + 5}{2} = \frac{1 - z}{-6}\) and \(\frac{x - 7}{1} = \frac{y}{2} = \frac{6 - z}{-2}\).
\(\frac{x - 2}{3} = \frac{y + 5}{2} = \frac{1 - z}{-6}\) and \(\frac{x - 7}{1} = \frac{y}{2} = \frac{6 - z}{-2}\).
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Be extremely careful with terms like \(1 - z\). You must rewrite them as \(z - 1\) and adjust the denominator's sign to extract correct DRs.
If the dot product is zero, the lines are perpendicular (angle is \(90^\circ\)).
If the dot product is zero, the lines are perpendicular (angle is \(90^\circ\)).
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• Direction ratios (DRs) of a line are the denominators when the equation is in standard form \(\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}\).
• Angle \(\theta\) between two lines with DRs \((a_1, b_1, c_1)\) and \((a_2, b_2, c_2)\) is \(\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\).
Step 1: Convert equations to standard form and find DRs
Line 1: \(\frac{x - 2}{3} = \frac{y + 5}{2} = \frac{z - 1}{6}\). Direction Ratios \(\vec{b_1} = (3, 2, 6)\).
Line 2: \(\frac{x - 7}{1} = \frac{y}{2} = \frac{z - 6}{2}\). Direction Ratios \(\vec{b_2} = (1, 2, 2)\).
Step 2: Evaluate the numerator for the cosine formula
The dot product of direction vectors: \[ \vec{b_1} \cdot \vec{b_2} = (3)(1) + (2)(2) + (6)(2) \] \[ = 3 + 4 + 12 = 19 \]
Step 3: Evaluate the magnitudes of the direction vectors
\[ |\vec{b_1}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \] \[ |\vec{b_2}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
Step 4: Calculate the angle \(\theta\)
\[ \cos \theta = \frac{19}{7 \times 3} = \frac{19}{21} \] \[ \theta = \cos^{-1}\left( \frac{19}{21} \right) \]
• Direction ratios (DRs) of a line are the denominators when the equation is in standard form \(\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}\).
• Angle \(\theta\) between two lines with DRs \((a_1, b_1, c_1)\) and \((a_2, b_2, c_2)\) is \(\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\).
Step 1: Convert equations to standard form and find DRs
Line 1: \(\frac{x - 2}{3} = \frac{y + 5}{2} = \frac{z - 1}{6}\). Direction Ratios \(\vec{b_1} = (3, 2, 6)\).
Line 2: \(\frac{x - 7}{1} = \frac{y}{2} = \frac{z - 6}{2}\). Direction Ratios \(\vec{b_2} = (1, 2, 2)\).
Step 2: Evaluate the numerator for the cosine formula
The dot product of direction vectors: \[ \vec{b_1} \cdot \vec{b_2} = (3)(1) + (2)(2) + (6)(2) \] \[ = 3 + 4 + 12 = 19 \]
Step 3: Evaluate the magnitudes of the direction vectors
\[ |\vec{b_1}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \] \[ |\vec{b_2}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
Step 4: Calculate the angle \(\theta\)
\[ \cos \theta = \frac{19}{7 \times 3} = \frac{19}{21} \] \[ \theta = \cos^{-1}\left( \frac{19}{21} \right) \]
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