Question

Two lines \(2x=3y=-z\) and \(6z=-y=-4z\) intersect at a point, then find the angle between them.

Answer 1

The direction vectors of the two lines can be found by looking at the coefficients of \( x \), \( y \), and \( z \) in their respective equations. For the first line, the direction vector is \( \langle 2, 3, -1 \rangle \), and for the second line, it is \( \langle 6, -1, -4 \rangle \).

The angle between two lines with direction vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be found using the dot product formula:

                                                             \( \cos \theta = \frac{(\mathbf{a} \cdot \mathbf{b})}{(\left | \mathbf{a} \right | \left | \mathbf{b} \right |)} \)

Where \( \theta \) is the angle between the two lines.

Substituting the direction vectors for the two lines, we get:

\[ \cos \theta = \frac{(\langle 2, 3, -1 \rangle \cdot \langle 6, -1, -4 \rangle)}{(\left | \langle 2, 3, -1 \rangle \right | \left | \langle 6, -1, -4 \rangle \right |)} \]

Evaluating the dot product and magnitudes, we get:

\[ \cos \theta = \frac{(12 - 3 + 4)}{\sqrt{14} \times \sqrt{53}} \]

Simplifying, we get:

\[ \cos \theta = \frac{13}{\sqrt{14} \times \sqrt{53}} \]

Using a calculator, we can find the angle \( \theta \) to be approximately 54.5 degrees. Therefore, the angle between the two lines is approximately 54.5 degrees.