Question

The angle between the pair of lines given by \[ \vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k}) \] and \[ \vec{r} = 5\hat{i} - 2\hat{j} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k}) \] is _____

• A. \( \cos^{-1}\left(\frac{19}{21}\right) \)
• B. \( \sin^{-1}\left(\frac{19}{21}\right) \)
• C. \( \cos^{-1}\left(-\frac{19}{21}\right) \)
• D. \( \cos^{-1}\left(\frac{\sqrt{19}}{21}\right) \)

Hint:

Angle between lines depends only on direction vectors, not position vectors.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The angle between two lines is the angle between their direction vectors $\vec{b_1}$ and $\vec{b_2}$.
Step 2: Formula Application:
$\vec{b_1} = \hat{i} + 2\hat{j} + 2\hat{k}$
$\vec{b_2} = 3\hat{i} + 2\hat{j} + 6\hat{k}$
$\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$
Step 3: Explanation:
$\vec{b_1} \cdot \vec{b_2} = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$
$|\vec{b_1}| = \sqrt{1^2 + 2^2 + 2^2} = 3$
$|\vec{b_2}| = \sqrt{3^2 + 2^2 + 6^2} = 7$
$\cos \theta = \frac{19}{3 \times 7} = \frac{19}{21} \implies \theta = \cos^{-1} \left( \frac{19}{21} \right)$
Step 4: Final Answer:
The correct option is (a).