Question

The acute angle between the planes $2x - y + z = 6$ and $x + y + 2z = 3$ is:

• A. $\frac{\pi}{3}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{6}$
• D. $\frac{\pi}{2}$

Hint:

The angle between two planes is exactly the same as the angle between their normal vectors. Always extract the coefficients of $x, y, z$ as the normal components!

Answer 1

The Correct Option is A

Step 1: Concept
The angle $\theta$ between two planes $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$ is equal to the angle between their normal vectors $\vec{n_1} = (A_1, B_1, C_1)$ and $\vec{n_2} = (A_2, B_2, C_2)$, given by: \[ \cos\theta = \frac{|A_1A_2 + B_1B_2 + C_1C_2|}{\sqrt{A_1^2 + B_1^2 + C_1^2}\sqrt{A_2^2 + B_2^2 + C_2^2}} \]

Step 2: Meaning
For the given planes, the normal vectors are $\vec{n_1} = (2, -1, 1)$ and $\vec{n_2} = (1, 1, 2)$.

Step 3: Analysis
Calculate the terms: \[ A_1A_2 + B_1B_2 + C_1C_2 = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3 \] \[ \sqrt{A_1^2 + B_1^2 + C_1^2} = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6} \] \[ \sqrt{A_2^2 + B_2^2 + C_2^2} = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6} \] Now compute $\cos\theta$: \[ \cos\theta = \frac{3}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2} \] Since $\cos\theta = \frac{1}{2}$, the angle $\theta = \frac{\pi}{3}$ (or $60^\circ$).

Step 4: Conclusion
The acute angle between the two planes is $\frac{\pi}{3}$. Final Answer: (A)