Find the angle between the planes `vecr.(hati+hatj-2hatk)=3 and vecr.(2hati-2hatj+hatk)=2`2

Answer

To find the angle between the two given planes, we can follow these steps:

### Step 1: Identify the normal vectors of the planes
The equations of the planes are given as:
1. \( \vec{r} \cdot (\hat{i} + \hat{j} - 2\hat{k}) = 3 \)
2. \( \vec{r} \cdot (2\hat{i} - 2\hat{j} + \hat{k}) = 2 \)

From these equations, we can identify the normal vectors:
- For Plane 1, the normal vector \( \vec{n_1} = \hat{i} + \hat{j} - 2\hat{k} \)
- For Plane 2, the normal vector \( \vec{n_2} = 2\hat{i} - 2\hat{j} + \hat{k} \)

### Step 2: Calculate the dot product of the normal vectors
Next, we calculate the dot product \( \vec{n_1} \cdot \vec{n_2} \):
\[ \vec{n_1} \cdot \vec{n_2} = (1)(2) + (1)(-2) + (-2)(1) = 2 - 2 - 2 = -2 \]

### Step 3: Calculate the magnitudes of the normal vectors
Now, we find the magnitudes of \( \vec{n_1} \) and \( \vec{n_2} \):
\[ |\vec{n_1}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]
\[ |\vec{n_2}| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

### Step 4: Use the formula for the cosine of the angle between the planes
The cosine of the angle \( \theta \) between the two planes is given by:
\[ \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} \]
Substituting the values we calculated:
\[ \cos \theta = \frac{|-2|}{\sqrt{6} \cdot 3} = \frac{2}{3\sqrt{6}} \]

### Step 5: Find the angle \( \theta \)
To find the angle \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{2}{3\sqrt{6}}\right) \]

### Final Answer
Thus, the angle between the two planes is:
\[ \theta = \cos^{-1}\left(\frac{2}{3\sqrt{6}}\right) \]