Question:medium
The angle between the lines \( \frac{x-3}{1} = \frac{y+1}{-1} = \frac{z-2}{-1} \) and \( \frac{x+1}{2} = \frac{y-2}{2} = \frac{z+3}{-2} \) is
The angle between the lines \( \frac{x-3}{1} = \frac{y+1}{-1} = \frac{z-2}{-1} \) and \( \frac{x+1}{2} = \frac{y-2}{2} = \frac{z+3}{-2} \) is
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Extract direction ratios directly from symmetric form of line equations.
Updated On: May 10, 2026
- \( \cos^{-1}\left(\sqrt{\frac{2}{6}}\right) \)
- \( \cos^{-1}\left(\sqrt{\frac{6}{6}}\right) \)
- \( \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) \)
- \( \cos^{-1}\left(\frac{1}{3}\right) \)
- \( \cos^{-1}\left(\frac{\sqrt{2}}{3}\right) \)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
The angle between two lines in 3D space is the angle between their direction vectors. We can find this angle using the dot product formula for vectors.
Step 2: Key Formula or Approach:
If \(\vec{d_1}\) and \(\vec{d_2}\) are the direction vectors of two lines, the angle \(\theta\) between them is given by: \[ \cos\theta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{|\vec{d_1}| |\vec{d_2}|} \] The direction vector \(\langle a, b, c \rangle\) can be read directly from the denominators of the line equation \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\).
Step 3: Detailed Explanation:
1. Identify the direction vectors. For the first line, \(\frac{x-3}{1} = \frac{y+1}{-1} = \frac{z-2}{-1}\), the direction vector is: \[ \vec{d_1} = \langle 1, -1, -1 \rangle \] For the second line, \(\frac{x+1}{2} = \frac{y-2}{2} = \frac{z+3}{-2}\), the direction vector is: \[ \vec{d_2} = \langle 2, 2, -2 \rangle \] We can simplify \(\vec{d_2}\) by dividing by 2 to get \(\vec{d'_2} = \langle 1, 1, -1 \rangle\). This is a parallel vector and will give the same angle. Let's use \(\vec{d_2}\) as is for now. 2. Calculate the dot product. \[ \vec{d_1} \cdot \vec{d_2} = (1)(2) + (-1)(2) + (-1)(-2) = 2 - 2 + 2 = 2 \] 3. Calculate the magnitudes of the vectors. \[ |\vec{d_1}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \] \[ |\vec{d_2}| = \sqrt{2^2 + 2^2 + (-2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \] 4. Calculate \(\cos\theta\). \[ \cos\theta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{|\vec{d_1}| |\vec{d_2}|} = \frac{|2|}{\sqrt{3} \cdot 2\sqrt{3}} = \frac{2}{2 \cdot 3} = \frac{2}{6} = \frac{1}{3} \] Therefore, the angle \(\theta\) is \(\cos^{-1}\left(\frac{1}{3}\right)\).
Step 4: Final Answer:
The angle between the lines is \(\cos^{-1}(\frac{1}{3})\).
The angle between two lines in 3D space is the angle between their direction vectors. We can find this angle using the dot product formula for vectors.
Step 2: Key Formula or Approach:
If \(\vec{d_1}\) and \(\vec{d_2}\) are the direction vectors of two lines, the angle \(\theta\) between them is given by: \[ \cos\theta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{|\vec{d_1}| |\vec{d_2}|} \] The direction vector \(\langle a, b, c \rangle\) can be read directly from the denominators of the line equation \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\).
Step 3: Detailed Explanation:
1. Identify the direction vectors. For the first line, \(\frac{x-3}{1} = \frac{y+1}{-1} = \frac{z-2}{-1}\), the direction vector is: \[ \vec{d_1} = \langle 1, -1, -1 \rangle \] For the second line, \(\frac{x+1}{2} = \frac{y-2}{2} = \frac{z+3}{-2}\), the direction vector is: \[ \vec{d_2} = \langle 2, 2, -2 \rangle \] We can simplify \(\vec{d_2}\) by dividing by 2 to get \(\vec{d'_2} = \langle 1, 1, -1 \rangle\). This is a parallel vector and will give the same angle. Let's use \(\vec{d_2}\) as is for now. 2. Calculate the dot product. \[ \vec{d_1} \cdot \vec{d_2} = (1)(2) + (-1)(2) + (-1)(-2) = 2 - 2 + 2 = 2 \] 3. Calculate the magnitudes of the vectors. \[ |\vec{d_1}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \] \[ |\vec{d_2}| = \sqrt{2^2 + 2^2 + (-2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \] 4. Calculate \(\cos\theta\). \[ \cos\theta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{|\vec{d_1}| |\vec{d_2}|} = \frac{|2|}{\sqrt{3} \cdot 2\sqrt{3}} = \frac{2}{2 \cdot 3} = \frac{2}{6} = \frac{1}{3} \] Therefore, the angle \(\theta\) is \(\cos^{-1}\left(\frac{1}{3}\right)\).
Step 4: Final Answer:
The angle between the lines is \(\cos^{-1}(\frac{1}{3})\).
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