Question

Find the angle between the pair of lines:
\( \vec{r} = 3\hat{i} + \hat{j} - 2\hat{k} + \lambda(\hat{i} - \hat{j} - 2\hat{k}) \) and
\( \vec{r} = 2\hat{i} - \hat{j} - 56\hat{k} + \mu(3\hat{i} - 5\hat{j} - 4\hat{k}) \).

Hint:

In the vector equation of a line \( \vec{r} = \vec{a} + \lambda\vec{b} \), \( \vec{a} \) is the position vector of a point on the line, and \( \vec{b} \) is the direction vector. The angle between lines depends only on their direction vectors.

Answer 1

Step 1: Understanding the Concept:
The angle between two lines in vector form is the angle between their direction vectors. The direction vectors are the vectors multiplied by the scalar parameters (\( \lambda \) and \( \mu \)). We use the dot product formula to find the angle.
Step 2: Key Formula or Approach:
If \( \vec{b_1} \) and \( \vec{b_2} \) are the direction vectors of the two lines, the angle \( \theta \) between them is given by:
\[ \cos\theta = \frac{\vec{b_1} \cdot \vec{b_2}}{|\vec{b_1}| |\vec{b_2}|} \] Step 3: Detailed Explanation or Calculation:
From the given line equations, we identify the direction vectors:
For the first line, the direction vector is \( \vec{b_1} = \hat{i} - \hat{j} - 2\hat{k} \).
For the second line, the direction vector is \( \vec{b_2} = 3\hat{i} - 5\hat{j} - 4\hat{k} \).
1. Calculate the dot product \( \vec{b_1} \cdot \vec{b_2} \):
\[ \vec{b_1} \cdot \vec{b_2} = (1)(3) + (-1)(-5) + (-2)(-4) = 3 + 5 + 8 = 16 \] 2. Calculate the magnitudes of the vectors:
\[ |\vec{b_1}| = \sqrt{1^2 + (-1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \] \[ |\vec{b_2}| = \sqrt{3^2 + (-5)^2 + (-4)^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2} \] 3. Calculate \( \cos\theta \):
\[ \cos\theta = \frac{16}{(\sqrt{6})(5\sqrt{2})} = \frac{16}{5\sqrt{12}} = \frac{16}{5(2\sqrt{3})} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}} \] Therefore, the angle is \( \theta = \cos^{-1}\left(\frac{8}{5\sqrt{3}}\right) \).
Step 4: Final Answer:
The angle between the pair of lines is \( \cos^{-1}\left(\frac{8}{5\sqrt{3}}\right) \).