The acute angle that the vector `2hati-2hatj+2hatk` 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 \( \mathbf{A} = 2\hat{i} - 2\hat{j} + 2\hat{k} \) 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
The normal vector \( \mathbf{N} \) to the plane formed by vectors \( \mathbf{B} \) and \( \mathbf{C} \) can be found using the cross product \( \mathbf{B} \times \mathbf{C} \).

\[ \mathbf{B} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}, \quad \mathbf{C} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \]

Calculating the cross product:

\[ \mathbf{N} = \mathbf{B} \times \mathbf{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & -1 & 2 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{N} = \hat{i} \begin{vmatrix} 3 & -1 \\ -1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -1 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

\[ \begin{vmatrix} 3 & -1 \\ -1 & 2 \end{vmatrix} = (3)(2) - (-1)(-1) = 6 - 1 = 5 \]

\[ \begin{vmatrix} 2 & -1 \\ 1 & 2 \end{vmatrix} = (2)(2) - (-1)(1) = 4 + 1 = 5 \]

\[ \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} = (2)(-1) - (3)(1) = -2 - 3 = -5 \]

Putting it all together:

\[ \mathbf{N} = 5\hat{i} - 5\hat{j} - 5\hat{k} = 5(\hat{i} - \hat{j} - \hat{k}) \]

### Step 2: Calculate the dot product of \( \mathbf{A} \) and \( \mathbf{N} \)

Now, we calculate the dot product \( \mathbf{A} \cdot \mathbf{N} \):

\[ \mathbf{A} = 2\hat{i} - 2\hat{j} + 2\hat{k} \]

\[ \mathbf{A} \cdot \mathbf{N} = (2)(5) + (-2)(-5) + (2)(-5) = 10 + 10 - 10 = 10 \]

### Step 3: Find the magnitudes of \( \mathbf{A} \) and \( \mathbf{N} \)

Now we find the magnitudes:

\[ |\mathbf{A}| = \sqrt{2^2 + (-2)^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]

\[ |\mathbf{N}| = |5(\hat{i} - \hat{j} - \hat{k})| = 5\sqrt{1^2 + (-1)^2 + (-1)^2} = 5\sqrt{3} \]

### Step 4: Calculate the cosine of the angle

Using the formula for the angle \( \theta \) between the vector \( \mathbf{A} \) and the normal vector \( \mathbf{N} \):

\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{N}}{|\mathbf{A}| |\mathbf{N}|} \]

Substituting the values:

\[ \cos \theta = \frac{10}{(2\sqrt{3})(5\sqrt{3})} = \frac{10}{30} = \frac{1}{3} \]

### Step 5: Find the acute angle with the plane

The acute angle \( \phi \) that \( \mathbf{A} \) makes with the plane is given by:

\[ \phi = 90^\circ - \theta \]

Using \( \theta = \cos^{-1}\left(\frac{1}{3}\right) \):

Thus, the acute angle that the vector \( \mathbf{A} \) makes with the plane is:

\[ \phi = 90^\circ - \cos^{-1}\left(\frac{1}{3}\right) \]