The vactors ` veca = 3hati -2hatj + 2hatk and vecb =- hati -2hatk` are the adjacent sides of a parallelogram. Then , the acute angle between ` veca and vecb` is

Options

A. `pi//4`
B. ` pi//3`
C. ` 3pi//4`
D. `2pi//3`

Explanation

To find the acute angle between the vectors \(\vec{a} = 3\hat{i} - 2\hat{j} + 2\hat{k}\) and \(\vec{b} = -\hat{i} - 2\hat{k}\), we can follow these steps:

### Step 1: Calculate the dot product of the vectors \(\vec{a}\) and \(\vec{b}\).

The dot product \(\vec{a} \cdot \vec{b}\) is given by:
\[ \vec{a} \cdot \vec{b} = (3)(-1) + (-2)(0) + (2)(-2) \]
Calculating this:
\[ \vec{a} \cdot \vec{b} = -3 + 0 - 4 = -7 \]

### Step 2: Calculate the magnitudes of the vectors \(\vec{a}\) and \(\vec{b}\).

The magnitude of \(\vec{a}\) is:
\[ |\vec{a}| = \sqrt{(3)^2 + (-2)^2 + (2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17} \]

The magnitude of \(\vec{b}\) is:
\[ |\vec{b}| = \sqrt{(-1)^2 + (0)^2 + (-2)^2} = \sqrt{1 + 0 + 4} = \sqrt{5} \]

### Step 3: Use the dot product and magnitudes to find the cosine of the angle \(\theta\).

The cosine of the angle \(\theta\) between the two vectors is given by:
\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \]
Substituting the values we found:
\[ \cos \theta = \frac{-7}{\sqrt{17} \cdot \sqrt{5}} = \frac{-7}{\sqrt{85}} \]

### Step 4: Find the angle \(\theta\).

Since we are interested in the acute angle, we take the absolute value of the cosine:
\[ \cos \theta = \left|\frac{-7}{\sqrt{85}}\right| = \frac{7}{\sqrt{85}} \]

Now, we can find \(\theta\) using the inverse cosine function:
\[ \theta = \cos^{-1}\left(\frac{7}{\sqrt{85}}\right) \]

### Step 5: Determine the acute angle.

Since \(\cos^{-1}\) will give us the angle in radians, we can calculate this value to find the acute angle.

### Final Answer:
The acute angle \(\theta\) between the vectors \(\vec{a}\) and \(\vec{b}\) is given by:
\[ \theta = \cos^{-1}\left(\frac{7}{\sqrt{85}}\right) \]