The angle between the vectors :
`vec(a)=hat(i)+2hat(j)-3hat(k)` and `3hat(i)-hat(j)+2hat(k)` is :

Options

A. `cos^(-1)((5)/(14))`
B. `cos^(-1)((9)/(14))`
C. `cos^(-1)(-(5)/(14))`
D. None of these

Answer

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

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

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

The dot product of two vectors \(\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\) and \(\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\) is given by:

\[ \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \]

For our vectors:

- \(a_1 = 1\), \(a_2 = 2\), \(a_3 = -3\)
- \(b_1 = 3\), \(b_2 = -1\), \(b_3 = 2\)

Calculating the dot product:

\[ \vec{a} \cdot \vec{b} = (1)(3) + (2)(-1) + (-3)(2) = 3 - 2 - 6 = -5 \]

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

The magnitude of a vector \(\vec{v} = x \hat{i} + y \hat{j} + z \hat{k}\) is given by:

\[ |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \]

Calculating the magnitude of \(\vec{a}\):

\[ |\vec{a}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

Calculating the magnitude of \(\vec{b}\):

\[ |\vec{b}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \]

### Step 3: Substitute into the formula to find \(\cos \theta\)

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

\[ -5 = |\vec{a}| |\vec{b}| \cos \theta \]

Substituting the magnitudes:

\[ -5 = \sqrt{14} \cdot \sqrt{14} \cdot \cos \theta \]

This simplifies to:

\[ -5 = 14 \cos \theta \]

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

Now, we can solve for \(\cos \theta\):

\[ \cos \theta = \frac{-5}{14} \]

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

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

\[ \theta = \cos^{-1}\left(\frac{-5}{14}\right) \]

### Final Answer

The angle \(\theta\) between the vectors \(\vec{a}\) and \(\vec{b}\) is:

\[ \theta = \cos^{-1}\left(\frac{-5}{14}\right) \]