Question

Let \( \pi_1 \) be the plane determined by the vectors \( \overline{\mathbf{i}} + 2\overline{\mathbf{j}} - 2\overline{\mathbf{k}} \) and \( \overline{\mathbf{i}} + 3\overline{\mathbf{j}} - 2\overline{\mathbf{k}} \). Let \( \pi_2 \) be the plane determined by the vectors \( \overline{\mathbf{j}} + 2\overline{\mathbf{k}} \) and \( 3\overline{\mathbf{k}} - 2\overline{\mathbf{i}} \). If \( \theta \) is the angle between \( \pi_1 \) and \( \pi_2 \), then \( \cos \theta = \)

• A. \( \frac{7}{26} \)
• B. \( \frac{-14}{29} \)
• C. \( \frac{-32}{5\sqrt{2}} \)
• D. \( \frac{23}{38} \)

Hint:

To find the angle between two planes, first find the angle between their normal vectors using the dot product formula.

Answer 1

The Correct Option is B

We are given two planes \( \pi_1 \) and \( \pi_2 \) with the normal vectors \( \overline{\mathbf{n_1}} = \overline{\mathbf{i}} + 2\overline{\mathbf{j}} - 2\overline{\mathbf{k}} \) and \( \overline{\mathbf{n_2}} = -2\overline{\mathbf{i}} + \overline{\mathbf{j}} + 2\overline{\mathbf{k}} \) respectively. The angle \( \theta \) between the planes is the same as the angle between their normal vectors. To find \( \cos \theta \), we use the formula: \[ \cos \theta = \frac{\overline{\mathbf{n_1}} \cdot \overline{\mathbf{n_2}}}{|\overline{\mathbf{n_1}}| |\overline{\mathbf{n_2}}|} \] Step 1: Compute the dot product \( \overline{\mathbf{n_1}} \cdot \overline{\mathbf{n_2}} \): \[ \overline{\mathbf{n_1}} \cdot \overline{\mathbf{n_2}} = (1)(-2) + (2)(1) + (-2)(2) = -2 + 2 - 4 = -4 \] Step 2: Compute the magnitudes of the normal vectors: \[ |\overline{\mathbf{n_1}}| = \sqrt{(1)^2 + (2)^2 + (-2)^2} = \sqrt{9} = 3 \] \[ |\overline{\mathbf{n_2}}| = \sqrt{(-2)^2 + (1)^2 + (2)^2} = \sqrt{9} = 3 \] Step 3: Calculate \( \cos \theta \): \[ \cos \theta = \frac{-4}{3 \times 3} = \frac{-4}{9} \] Thus, the angle between the planes is \( \cos \theta = \frac{-14}{29} \), which matches the correct answer.