Find the angle between the vectors :
`vec(a)=2hat(i)-hat(j)+2hat(k) " and " vec(b)=6hat(i)+2hat(j)+3hat(k)`.

Answer

To find the angle between the vectors \(\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}\) and \(\vec{b} = 6\hat{i} + 2\hat{j} + 3\hat{k}\), we can use the formula involving the dot product of the vectors:

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

where \(\theta\) is the angle between the two vectors.

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

The dot product is calculated as follows:

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

Calculating each term:

- \(2 \cdot 6 = 12\)
- \(-1 \cdot 2 = -2\)
- \(2 \cdot 3 = 6\)

Now, summing these results:

\[ \vec{a} \cdot \vec{b} = 12 - 2 + 6 = 16 \]

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

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

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

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

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

### Step 3: Substitute the values into the dot product formula

Now we can substitute the values we found into the dot product formula:

\[ 16 = (3)(7) \cos \theta \]

This simplifies to:

\[ 16 = 21 \cos \theta \]

### Step 4: Solve for \(\cos \theta\)

Rearranging gives:

\[ \cos \theta = \frac{16}{21} \]

### Step 5: Find the angle \(\theta\)

To find \(\theta\), we take the inverse cosine:

\[ \theta = \cos^{-1}\left(\frac{16}{21}\right) \]

Thus, the angle between the vectors \(\vec{a}\) and \(\vec{b}\) is:

\[ \theta = \cos^{-1}\left(\frac{16}{21}\right) \]