Given :
`vec A = hati + hatj +hatk and vec B =-hati-hatj-hatk` What is the angle between `(vec A - vec B) and vec A` ?

Options

A. `0^(@)`
B. `180^(@)`
C. `90^(@)`
D. `60^(@)`

Explanation

To find the angle between the vectors \( \vec{A} - \vec{B} \) and \( \vec{A} \), we will follow these steps:

### Step 1: Calculate \( \vec{A} - \vec{B} \)
Given:
\[ \vec{A} = \hat{i} + \hat{j} + \hat{k} \]
\[ \vec{B} = -\hat{i} - \hat{j} - \hat{k} \]
We can calculate \( \vec{A} - \vec{B} \):
\[ \vec{A} - \vec{B} = (\hat{i} + \hat{j} + \hat{k}) - (-\hat{i} - \hat{j} - \hat{k}) \]
\[ = \hat{i} + \hat{j} + \hat{k} + \hat{i} + \hat{j} + \hat{k} \]
\[ = 2\hat{i} + 2\hat{j} + 2\hat{k} \]

### Step 2: Find the dot product \( (\vec{A} - \vec{B}) \cdot \vec{A} \)
Now we need to find the dot product of \( \vec{A} - \vec{B} \) with \( \vec{A} \):
\[ (\vec{A} - \vec{B}) \cdot \vec{A} = (2\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) \]
Calculating the dot product:
\[ = 2(\hat{i} \cdot \hat{i}) + 2(\hat{j} \cdot \hat{j}) + 2(\hat{k} \cdot \hat{k}) \]
\[ = 2(1) + 2(1) + 2(1) = 2 + 2 + 2 = 6 \]

### Step 3: Calculate the magnitudes of \( \vec{A} - \vec{B} \) and \( \vec{A} \)
1. **Magnitude of \( \vec{A} - \vec{B} \)**:
\[ |\vec{A} - \vec{B}| = |2\hat{i} + 2\hat{j} + 2\hat{k}| = \sqrt{(2)^2 + (2)^2 + (2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]

2. **Magnitude of \( \vec{A} \)**:
\[ |\vec{A}| = |\hat{i} + \hat{j} + \hat{k}| = \sqrt{(1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 4: Use the dot product to find the angle \( \theta \)
Using the formula:
\[ (\vec{A} - \vec{B}) \cdot \vec{A} = |\vec{A} - \vec{B}| |\vec{A}| \cos \theta \]
Substituting the values we found:
\[ 6 = (2\sqrt{3})(\sqrt{3}) \cos \theta \]
\[ 6 = 2 \cdot 3 \cos \theta \]
\[ 6 = 6 \cos \theta \]
\[ \cos \theta = 1 \]

### Step 5: Find \( \theta \)
Since \( \cos \theta = 1 \):
\[ \theta = 0^\circ \]

### Final Answer:
The angle between \( \vec{A} - \vec{B} \) and \( \vec{A} \) is \( 0^\circ \).

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