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: Use normals to find the angle.
The angle between two flat planes is the same as the angle between the arrows that stick straight out of them, called normal vectors. The numbers in front of x, y, z in a plane equation give that normal arrow.

Step 2: Read the normal vectors.
For $2x - y + z = 6$ the normal is $\vec{n_1} = (2, -1, 1)$. For $x + y + 2z = 3$ the normal is $\vec{n_2} = (1, 1, 2)$.

Step 3: Write the angle formula.
The cosine of the angle uses the dot product on top and the lengths on the bottom.
\[ \cos\theta = \frac{|\vec{n_1}\cdot\vec{n_2}|}{|\vec{n_1}||\vec{n_2}|} \]

Step 4: Find the dot product.
Multiply matching parts and add.
\[ \vec{n_1}\cdot\vec{n_2} = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3 \]

Step 5: Find the lengths.
Each length is the square root of the sum of squares.
\[ |\vec{n_1}| = \sqrt{4 + 1 + 1} = \sqrt{6}, \quad |\vec{n_2}| = \sqrt{1 + 1 + 4} = \sqrt{6} \]

Step 6: Put it all together.
Now divide.
\[ \cos\theta = \frac{3}{\sqrt{6}\cdot\sqrt{6}} = \frac{3}{6} = \frac{1}{2} \] The angle whose cosine is $\frac{1}{2}$ is $\frac{\pi}{3}$.
\[ \boxed{\dfrac{\pi}{3}} \]