Find the angles between the planes `vecr(hati-2hatj-2hatk)=1` and `vecr(3hati-6hatj+2hatk)=0`

Answer

To find the angle between the two planes given by the equations \(\vec{r} \cdot (\hat{i} - 2\hat{j} - 2\hat{k}) = 1\) and \(\vec{r} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 0\), we first need to identify the normal vectors of these planes.

### Step 1: Identify the normal vectors
The normal vector of the first plane \(n_1\) can be directly read from the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) in the equation:
\[ n_1 = \hat{i} - 2\hat{j} - 2\hat{k} = (1, -2, -2) \]

The normal vector of the second plane \(n_2\) is:
\[ n_2 = 3\hat{i} - 6\hat{j} + 2\hat{k} = (3, -6, 2) \]

### Step 2: Calculate the dot product of the normal vectors
The dot product \(n_1 \cdot n_2\) is calculated as follows:
\[ n_1 \cdot n_2 = (1)(3) + (-2)(-6) + (-2)(2) = 3 + 12 - 4 = 11 \]

### Step 3: Calculate the magnitudes of the normal vectors
The magnitude of \(n_1\) is:
\[ |n_1| = \sqrt{1^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

The magnitude of \(n_2\) is:
\[ |n_2| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \]

### Step 4: Use the dot product to find the cosine of the angle
Using the formula for the cosine of the angle \(\theta\) between the two normal vectors:
\[ \cos \theta = \frac{n_1 \cdot n_2}{|n_1| |n_2|} = \frac{11}{3 \times 7} = \frac{11}{21} \]

### Step 5: Calculate the angle \(\theta\)
To find the angle \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{11}{21}\right) \]

Thus, the angle between the two planes is:
\[ \theta = \cos^{-1}\left(\frac{11}{21}\right) \]