The angle between the planes
`vecr.(2hati-hatj+hatk)=6` and `vecr.(hati+hatj+2hatk)=5` is

Options

A. `(pi)/3`
B. `(2pi)/3`
C. `(pi)/6`
D. `(5pi)/6`

Explanation

To find the angle between the two planes given by the equations:

1. \(\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 6\)
2. \(\vec{r} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 5\)

we will follow these steps:

### Step 1: Identify the normal vectors of the planes

The normal vector of the first plane can be extracted from the equation:

- For Plane 1: \(\vec{n_1} = 2\hat{i} - \hat{j} + \hat{k}\)

The normal vector of the second plane is:

- For Plane 2: \(\vec{n_2} = \hat{i} + \hat{j} + 2\hat{k}\)

### Step 2: Use the dot product to find the cosine of the angle between the normals

The angle \(\theta\) between the two planes is the same as the angle between their normal vectors. We can use the dot product formula:

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

### Step 3: Calculate the dot product \(\vec{n_1} \cdot \vec{n_2}\)

Calculating the dot product:

\[ \vec{n_1} \cdot \vec{n_2} = (2\hat{i} - \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j} + 2\hat{k}) \]

Expanding this:

\[ = 2(1) + (-1)(1) + 1(2) = 2 - 1 + 2 = 3 \]

### Step 4: Calculate the magnitudes of \(\vec{n_1}\) and \(\vec{n_2}\)

Calculating the magnitude of \(\vec{n_1}\):

\[ |\vec{n_1}| = \sqrt{(2^2) + (-1^2) + (1^2)} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

Calculating the magnitude of \(\vec{n_2}\):

\[ |\vec{n_2}| = \sqrt{(1^2) + (1^2) + (2^2)} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

### Step 5: Substitute into the cosine formula

Now substituting back into the cosine formula:

\[ \cos \theta = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2} \]

### Step 6: Find the angle \(\theta\)

To find \(\theta\), we take the inverse cosine:

\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \]

This gives us:

\[ \theta = 60^\circ \]

### Conclusion

The angle between the two planes is \(60^\circ\).

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