Find the angle between the planes `2x-3y+6z=9` and xy plane

Answer

To find the angle between the plane given by the equation \(2x - 3y + 6z = 9\) and the xy-plane, we can follow these steps:

### Step 1: Identify the normal vector of the given plane
The equation of the plane can be written in the form \(Ax + By + Cz = D\), where \(A = 2\), \(B = -3\), and \(C = 6\). The normal vector \(\mathbf{n_1}\) of the given plane is given by the coefficients of \(x\), \(y\), and \(z\):
\[ \mathbf{n_1} = (2, -3, 6) \]

### Step 2: Identify the normal vector of the xy-plane
The equation of the xy-plane is \(z = 0\), which can also be expressed as \(0x + 0y + 1z = 0\). The normal vector \(\mathbf{n_2}\) of the xy-plane is:
\[ \mathbf{n_2} = (0, 0, 1) \]

### Step 3: Use the formula for the angle between two planes
The angle \(\theta\) between two planes can be found using the formula:
\[ \cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{|\mathbf{n_1}| |\mathbf{n_2}|} \]
where \(\mathbf{n_1} \cdot \mathbf{n_2}\) is the dot product of the normal vectors and \(|\mathbf{n_1}|\) and \(|\mathbf{n_2}|\) are the magnitudes of the normal vectors.

### Step 4: Calculate the dot product \(\mathbf{n_1} \cdot \mathbf{n_2}\)
The dot product is calculated as follows:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = (2)(0) + (-3)(0) + (6)(1) = 0 + 0 + 6 = 6 \]

### Step 5: Calculate the magnitudes of the normal vectors
The magnitude of \(\mathbf{n_1}\) is:
\[ |\mathbf{n_1}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]
The magnitude of \(\mathbf{n_2}\) is:
\[ |\mathbf{n_2}| = \sqrt{0^2 + 0^2 + 1^2} = \sqrt{1} = 1 \]

### Step 6: Substitute into the cosine formula
Now substituting the values into the cosine formula:
\[ \cos \theta = \frac{6}{7 \cdot 1} = \frac{6}{7} \]

### Step 7: Find the angle \(\theta\)
To find the angle \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{6}{7}\right) \]

### Final Answer
Thus, the angle between the plane \(2x - 3y + 6z = 9\) and the xy-plane is:
\[ \theta = \cos^{-1}\left(\frac{6}{7}\right) \]