If ` vec(a) = hat(i) - 2 hat(j) + 3 hat(k) and vec(b) = 2 hat(i) - 3 hat(j) + 5 hat(k)` , then angle between `vec(a) and vec(b)` is

Options

A. `(pi)/(2)`
B. `(pi)/(2)`
C. `(pi)/(6)`
D. `pi`

Answer

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

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

### Step 1: Identify the vectors
Given:
\[ \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \]
\[ \vec{b} = 2\hat{i} - 3\hat{j} + 5\hat{k} \]

### Step 2: Calculate the dot product \( \vec{a} \cdot \vec{b} \)
The dot product is calculated as follows:
\[ \vec{a} \cdot \vec{b} = (1)(2) + (-2)(-3) + (3)(5) \]
Calculating each term:
- \( 1 \cdot 2 = 2 \)
- \( -2 \cdot -3 = 6 \)
- \( 3 \cdot 5 = 15 \)

Adding these results:
\[ \vec{a} \cdot \vec{b} = 2 + 6 + 15 = 23 \]

### Step 3: Calculate the magnitudes \( |\vec{a}| \) and \( |\vec{b}| \)
The magnitude of \( \vec{a} \) is calculated as follows:
\[ |\vec{a}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

The magnitude of \( \vec{b} \) is calculated as follows:
\[ |\vec{b}| = \sqrt{2^2 + (-3)^2 + 5^2} = \sqrt{4 + 9 + 25} = \sqrt{38} \]

### Step 4: Substitute into the cosine formula
Now we can substitute the dot product and magnitudes into the cosine formula:
\[ \cos \theta = \frac{23}{|\vec{a}| |\vec{b}|} = \frac{23}{\sqrt{14} \cdot \sqrt{38}} \]

### Step 5: Calculate \( \sqrt{14} \cdot \sqrt{38} \)
Calculating the product:
\[ \sqrt{14} \cdot \sqrt{38} = \sqrt{14 \cdot 38} = \sqrt{532} \]

### Step 6: Calculate \( \cos \theta \)
Thus, we have:
\[ \cos \theta = \frac{23}{\sqrt{532}} \]

### Step 7: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{23}{\sqrt{532}}\right) \]

### Final Result
Using a calculator to find the angle:
\[ \theta \approx \cos^{-1}(0.99) \approx 8.1^\circ \]