Find the acute angle between the plane :
`vec(r). (hati - 2hatj - 2 hatk) = 1 and vec(r). (3 hati - 6 hatj + 2 hatk)` = 0 `

Answer

To find the acute angle between the two given planes, we need to follow these steps:

### Step 1: Identify the normal vectors of the planes
The equations of the planes are given in vector form:

1. Plane \( P_1: \vec{r} \cdot (\hat{i} - 2\hat{j} - 2\hat{k}) = 1 \)
- The normal vector \( \vec{n_1} = \hat{i} - 2\hat{j} - 2\hat{k} \)
- Therefore, \( \vec{n_1} = (1, -2, -2) \)

2. Plane \( P_2: \vec{r} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 0 \)
- The normal vector \( \vec{n_2} = 3\hat{i} - 6\hat{j} + 2\hat{k} \)
- Therefore, \( \vec{n_2} = (3, -6, 2) \)

### Step 2: Calculate the dot product of the normal vectors
The dot product \( \vec{n_1} \cdot \vec{n_2} \) is calculated as follows:

\[ \vec{n_1} \cdot \vec{n_2} = (1)(3) + (-2)(-6) + (-2)(2) \]
\[ = 3 + 12 - 4 = 11 \]

### Step 3: Calculate the magnitudes of the normal vectors
Next, we calculate the magnitudes of \( \vec{n_1} \) and \( \vec{n_2} \):

\[ |\vec{n_1}| = \sqrt{1^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

\[ |\vec{n_2}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \]

### Step 4: Use the dot product to find the cosine of the angle
The cosine of the angle \( \theta \) between the two planes can be found using the formula:

\[ \cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \]

Substituting the values we calculated:

\[ \cos \theta = \frac{11}{3 \cdot 7} = \frac{11}{21} \]

### Step 5: Calculate the angle \( \theta \)
Now, we find the angle \( \theta \):

\[ \theta = \cos^{-1}\left(\frac{11}{21}\right) \]

This gives us the acute angle between the two planes.

### Summary of Steps:
1. Identify the normal vectors of the planes.
2. Calculate the dot product of the normal vectors.
3. Calculate the magnitudes of the normal vectors.
4. Use the dot product to find the cosine of the angle.
5. Calculate the angle using the inverse cosine function.