Question

The angle between the lines \[ \mathbf{r_1} = (i + 2j + 3k) + \lambda (i + j + 2k) \quad \text{and} \quad \mathbf{r_2} = (3i + k) + \lambda' (2i + j - k), \quad \lambda, \lambda' \in \mathbb{R} \] is

• A. \( \cos^{-1} \left( \frac{1}{6} \right) \)
• B. \( \cos^{-1} \left( \frac{1}{5} \right) \)
• C. \( \cos^{-1} \left( \frac{1}{3} \right) \)
• D. \( \cos^{-1} \left( \frac{2}{3} \right) \)

Hint:

To find the angle between two lines, calculate the dot product of their direction ratios and divide by the product of their magnitudes.

Answer 1

The Correct Option is A

Step 1: Understanding the formula for the angle between two lines.
The angle \( \theta \) between two lines can be found using the formula: \[ \cos \theta = \frac{\mathbf{a_1} \cdot \mathbf{a_2}}{|\mathbf{a_1}| |\mathbf{a_2}|} \] where \( \mathbf{a_1} \) and \( \mathbf{a_2} \) are the direction ratios of the two lines.

Step 2: Finding the direction ratios of the lines.
From the given equations of the lines, the direction ratios of \( \mathbf{r_1} \) and \( \mathbf{r_2} \) are: \[ \mathbf{a_1} = (1, 1, 2) \quad \text{and} \quad \mathbf{a_2} = (2, 1, -1) \]
Step 3: Calculating the angle.
Now, calculate the dot product \( \mathbf{a_1} \cdot \mathbf{a_2} \) and the magnitudes of the vectors: \[ \mathbf{a_1} \cdot \mathbf{a_2} = 1 \times 2 + 1 \times 1 + 2 \times (-1) = 1 \] \[ |\mathbf{a_1}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}, \quad |\mathbf{a_2}| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{6} \] Thus: \[ \cos \theta = \frac{1}{6} \] The angle between the lines is \( \cos^{-1} \left( \frac{1}{6} \right) \), which makes option (A) the correct answer.