If `theta` is the angle between the vectors `a=2hati+2hatj-hatk and b=6hati-3hatj+2hatk`, then

Options

A. `costheta=(4)/(21)`
B. `costheta=(3)/(19)`
C. `costheta=(2)/(19)`
D. `costheta=(5)/(21)`

Answer

To find the angle \( \theta \) between the vectors \( \mathbf{a} \) and \( \mathbf{b} \), we will use the formula for the cosine of the angle between two vectors:

\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \]

### Step 1: Calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \)

Given:
\[ \mathbf{a} = 2\hat{i} + 2\hat{j} - \hat{k} \]
\[ \mathbf{b} = 6\hat{i} - 3\hat{j} + 2\hat{k} \]

The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as follows:

\[ \mathbf{a} \cdot \mathbf{b} = (2)(6) + (2)(-3) + (-1)(2) \]
\[ = 12 - 6 - 2 \]
\[ = 4 \]

### Step 2: Calculate the magnitudes \( |\mathbf{a}| \) and \( |\mathbf{b}| \)

The magnitude of \( \mathbf{a} \) is given by:

\[ |\mathbf{a}| = \sqrt{(2)^2 + (2)^2 + (-1)^2} \]
\[ = \sqrt{4 + 4 + 1} \]
\[ = \sqrt{9} = 3 \]

The magnitude of \( \mathbf{b} \) is given by:

\[ |\mathbf{b}| = \sqrt{(6)^2 + (-3)^2 + (2)^2} \]
\[ = \sqrt{36 + 9 + 4} \]
\[ = \sqrt{49} = 7 \]

### Step 3: Substitute into the cosine formula

Now substituting the values into the cosine formula:

\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{4}{3 \cdot 7} \]
\[ = \frac{4}{21} \]

### Final Result

Thus, the value of \( \cos \theta \) is:

\[ \cos \theta = \frac{4}{21} \]