Question

The angle between two planes \( 2x + y - 2z = 5 \) and \( 3x - 6y - 2z = 7 \) is:

• A. \( \frac{2}{\pi} \)
• B. \( \frac{4}{\pi} \)
• C. \( \cos^{-1} \left( \frac{4}{21} \right) \)
• D. \( \cos^{-1} \left( \frac{16}{61} \right) \)

Hint:

The angle between two planes can be found using the dot product of their normal vectors. Remember, the formula for the cosine of the angle is: \[ \cos \theta = \frac{| \vec{N_1} \cdot \vec{N_2} |}{|\vec{N_1}| |\vec{N_2}|} \]

Answer 1

The Correct Option is C

The angle \( \theta \) between two planes is given by the formula: \[ \cos \theta = \frac{| \vec{N_1} \cdot \vec{N_2} |}{|\vec{N_1}| |\vec{N_2}|} \] where \( \vec{N_1} \) and \( \vec{N_2} \) are the normal vectors of the two planes. For the first plane \( 2x + y - 2z = 5 \), the normal vector \( \vec{N_1} = (2, 1, -2) \). For the second plane \( 3x - 6y - 2z = 7 \), the normal vector \( \vec{N_2} = (3, -6, -2) \). Now, compute the dot product of the two normal vectors: \[ \vec{N_1} \cdot \vec{N_2} = 2 \times 3 + 1 \times (-6) + (-2) \times (-2) = 6 - 6 + 4 = 4 \] Next, compute the magnitudes of the normal vectors: \[ |\vec{N_1}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \] \[ |\vec{N_2}| = \sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \] Now, substitute these values into the formula for \( \cos \theta \): \[ \cos \theta = \frac{|4|}{3 \times 7} = \frac{4}{21} \] Thus, the angle between the planes is: \[ \theta = \cos^{-1} \left( \frac{4}{21} \right) \] Hence, the correct answer is: \[ \boxed{\cos^{-1} \left( \frac{4}{21} \right)} \]