If `vec(A) = hat(i) + hat(j) + hat(k)` and `B = -hat(i) - hat(j) - hat(k)`. Then angle made by `(vec(A) - vec(B))` with `vec(A)` is :

Options

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

Answer

To solve the problem, we need to find the angle made by the vector \( \vec{A} - \vec{B} \) with the vector \( \vec{A} \).

### Step 1: Define the vectors
Given:
\[ \vec{A} = \hat{i} + \hat{j} + \hat{k} \]
\[ \vec{B} = -\hat{i} - \hat{j} - \hat{k} \]

### Step 2: Calculate \( \vec{A} - \vec{B} \)
We can calculate \( \vec{A} - \vec{B} \) as follows:
\[ \vec{A} - \vec{B} = (\hat{i} + \hat{j} + \hat{k}) - (-\hat{i} - \hat{j} - \hat{k}) \]
This simplifies to:
\[ \vec{A} - \vec{B} = \hat{i} + \hat{j} + \hat{k} + \hat{i} + \hat{j} + \hat{k} \]
\[ = 2\hat{i} + 2\hat{j} + 2\hat{k} \]

### Step 3: Express \( \vec{A} - \vec{B} \) in terms of \( \vec{A} \)
Notice that:
\[ \vec{A} - \vec{B} = 2(\hat{i} + \hat{j} + \hat{k}) = 2\vec{A} \]

### Step 4: Find the angle between \( \vec{A} - \vec{B} \) and \( \vec{A} \)
The angle \( \theta \) between two vectors \( \vec{X} \) and \( \vec{Y} \) can be found using the formula:
\[ \cos \theta = \frac{\vec{X} \cdot \vec{Y}}{|\vec{X}| |\vec{Y}|} \]

In our case, let \( \vec{X} = \vec{A} - \vec{B} = 2\vec{A} \) and \( \vec{Y} = \vec{A} \).

### Step 5: Calculate the dot product and magnitudes
1. **Dot Product**:
\[ \vec{X} \cdot \vec{Y} = (2\vec{A}) \cdot \vec{A} = 2(\vec{A} \cdot \vec{A}) = 2|\vec{A}|^2 \]

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

3. **Magnitude of \( \vec{X} \)**:
\[ |\vec{X}| = |2\vec{A}| = 2|\vec{A}| = 2\sqrt{3} \]

### Step 6: Substitute into the cosine formula
Now substituting into the cosine formula:
\[ \cos \theta = \frac{2|\vec{A}|^2}{|\vec{X}| |\vec{Y}|} = \frac{2 \cdot 3}{(2\sqrt{3})(\sqrt{3})} \]
\[ = \frac{6}{2 \cdot 3} = 1 \]

### Step 7: Determine the angle
Since \( \cos \theta = 1 \), it implies:
\[ \theta = 0^\circ \]

### Final Answer
The angle made by \( \vec{A} - \vec{B} \) with \( \vec{A} \) is \( 0^\circ \).

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