What is the interior acute angle of the parallelogram whose sides are represented by the vectors `(1)/(sqrt2) hat(i) +(2)/(sqrt2) hat(j) + hat(k) and (1)/(sqrt2) hat(i) - (1)/(sqrt2) hat(j) + hat(k)`?

Options

A. `60^(@)`
B. `45^(@)`
C. `30^(@)`
D. `15^(@)`

Answer

To find the interior acute angle of the parallelogram whose sides are represented by the vectors \( \mathbf{A} = \frac{1}{\sqrt{2}} \hat{i} + \frac{2}{\sqrt{2}} \hat{j} + \hat{k} \) and \( \mathbf{B} = \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} + \hat{k} \), we can follow these steps:

### Step 1: Identify the vectors
Let:
\[ \mathbf{A} = \frac{1}{\sqrt{2}} \hat{i} + \frac{2}{\sqrt{2}} \hat{j} + \hat{k} \]
\[ \mathbf{B} = \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} + \hat{k} \]

### Step 2: Calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \)
The dot product is calculated as follows:
\[ \mathbf{A} \cdot \mathbf{B} = \left( \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \right) + \left( \frac{2}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} \right) + \left( 1 \cdot 1 \right) \]
Calculating each term:
\[ \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} \]
\[ \frac{2}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} = -\frac{2}{2} = -1 \]
\[ 1 \cdot 1 = 1 \]
Thus,
\[ \mathbf{A} \cdot \mathbf{B} = \frac{1}{2} - 1 + 1 = \frac{1}{2} \]

### Step 3: Calculate the magnitudes \( |\mathbf{A}| \) and \( |\mathbf{B}| \)
The magnitude of \( \mathbf{A} \) is:
\[ |\mathbf{A}| = \sqrt{\left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{2}{\sqrt{2}} \right)^2 + 1^2} \]
Calculating each term:
\[ \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}, \quad \left( \frac{2}{\sqrt{2}} \right)^2 = \frac{4}{2} = 2, \quad 1^2 = 1 \]
Thus,
\[ |\mathbf{A}| = \sqrt{\frac{1}{2} + 2 + 1} = \sqrt{\frac{1}{2} + \frac{4}{2} + \frac{2}{2}} = \sqrt{\frac{7}{2}} = \frac{\sqrt{7}}{\sqrt{2}} \]

The magnitude of \( \mathbf{B} \) is:
\[ |\mathbf{B}| = \sqrt{\left( \frac{1}{\sqrt{2}} \right)^2 + \left( -\frac{1}{\sqrt{2}} \right)^2 + 1^2} \]
Calculating each term:
\[ |\mathbf{B}| = \sqrt{\frac{1}{2} + \frac{1}{2} + 1} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 4: Use the cosine formula to find the angle \( \theta \)
The cosine of the angle \( \theta \) between the vectors is given by:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]
Substituting the values:
\[ \cos \theta = \frac{\frac{1}{2}}{\left(\frac{\sqrt{7}}{\sqrt{2}}\right)(\sqrt{2})} = \frac{\frac{1}{2}}{\sqrt{7}} = \frac{1}{2\sqrt{7}} \]

### Step 5: Find \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{2\sqrt{7}}\right) \]
However, we need to find the acute angle, and we know that:
\[ \cos(60^\circ) = \frac{1}{2} \]
Thus, we can conclude that the interior acute angle of the parallelogram is:
\[ \theta = 60^\circ \]

### Final Answer
The interior acute angle of the parallelogram is \( 60^\circ \).