The acute angle that the vector `2hati-2hatj+hatk` makes with the plane determined by the vectors `2hati+3hatj-hatk` and `hati-hatj+2hatk` is

Options

A. `cos^(-1)(1/(sqrt(3)))`
B. `sin^(-1)(1/(sqrt(3)))`
C. `tan^(-1)(sqrt(2))`
D. `cot^(-1)(sqrt(3))`

Explanation

To find the acute angle that the vector \( \mathbf{A} = 2\hat{i} - 2\hat{j} + \hat{k} \) makes with the plane determined by the vectors \( \mathbf{B} = 2\hat{i} + 3\hat{j} - \hat{k} \) and \( \mathbf{C} = \hat{i} - \hat{j} + 2\hat{k} \), we can follow these steps:

### Step 1: Find the normal vector to the plane formed by vectors \( \mathbf{B} \) and \( \mathbf{C} \)
The normal vector \( \mathbf{n} \) to the plane formed by two vectors \( \mathbf{B} \) and \( \mathbf{C} \) can be found using the cross product:
\[ \mathbf{n} = \mathbf{B} \times \mathbf{C} \]

### Step 2: Calculate the cross product \( \mathbf{B} \times \mathbf{C} \)
Using the determinant method:
\[ \mathbf{B} \times \mathbf{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & -1 & 2 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{B} \times \mathbf{C} = \hat{i}(3 \cdot 2 - (-1) \cdot (-1)) - \hat{j}(2 \cdot 2 - (-1) \cdot 1) + \hat{k}(2 \cdot (-1) - 3 \cdot 1) \]
\[ = \hat{i}(6 - 1) - \hat{j}(4 + 1) + \hat{k}(-2 - 3) \]
\[ = 5\hat{i} - 5\hat{j} - 5\hat{k} \]
Thus, \( \mathbf{n} = 5\hat{i} - 5\hat{j} - 5\hat{k} \).

### Step 3: Find the magnitude of the normal vector \( \mathbf{n} \)
\[ |\mathbf{n}| = \sqrt{(5)^2 + (-5)^2 + (-5)^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} \]

### Step 4: Calculate the dot product \( \mathbf{A} \cdot \mathbf{n} \)
\[ \mathbf{A} \cdot \mathbf{n} = (2\hat{i} - 2\hat{j} + \hat{k}) \cdot (5\hat{i} - 5\hat{j} - 5\hat{k}) \]
Calculating the dot product:
\[ = 2 \cdot 5 + (-2) \cdot (-5) + 1 \cdot (-5) = 10 + 10 - 5 = 15 \]

### Step 5: Find the magnitude of vector \( \mathbf{A} \)
\[ |\mathbf{A}| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

### Step 6: Use the sine formula to find \( \sin \theta \)
The sine of the angle \( \theta \) between the vector \( \mathbf{A} \) and the normal vector \( \mathbf{n} \) is given by:
\[ \sin \theta = \frac{|\mathbf{A} \cdot \mathbf{n}|}{|\mathbf{A}| \cdot |\mathbf{n}|} \]
Substituting the values:
\[ \sin \theta = \frac{15}{3 \cdot 5\sqrt{3}} = \frac{15}{15\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Step 7: Find the angle \( \theta \)
To find \( \theta \):
\[ \theta = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

### Conclusion
Thus, the acute angle that the vector \( 2\hat{i} - 2\hat{j} + \hat{k} \) makes with the plane determined by the vectors \( 2\hat{i} + 3\hat{j} - \hat{k} \) and \( \hat{i} - \hat{j} + 2\hat{k} \) is \( \theta = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) \).

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