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 find the angle made by the vector \((\vec{A} - \vec{B})\) with \(\vec{A}\), we will follow these steps:

### 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:
\[ \vec{A} - \vec{B} = \hat{i} + \hat{j} + \hat{k} + \hat{i} + \hat{j} + \hat{k} = 2\hat{i} + 2\hat{j} + 2\hat{k} \]
Thus, we have:
\[ \vec{A} - \vec{B} = 2(\hat{i} + \hat{j} + \hat{k}) \]

### Step 3: Identify the vectors for angle calculation
Let:
\[ \vec{Y} = \vec{A} - \vec{B} = 2(\hat{i} + \hat{j} + \hat{k}) \quad \text{and} \quad \vec{A} = \hat{i} + \hat{j} + \hat{k} \]

### Step 4: Use the dot product to find the angle
The cosine of the angle \(\theta\) between two vectors can be found using the dot product formula:
\[ \vec{Y} \cdot \vec{A} = |\vec{Y}| |\vec{A}| \cos \theta \]

### Step 5: Calculate the dot product \(\vec{Y} \cdot \vec{A}\)
Calculating the dot product:
\[ \vec{Y} \cdot \vec{A} = (2\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 2(1) + 2(1) + 2(1) = 2 + 2 + 2 = 6 \]

### Step 6: Calculate the magnitudes of \(\vec{Y}\) and \(\vec{A}\)
Calculating the magnitudes:
\[ |\vec{Y}| = |2(\hat{i} + \hat{j} + \hat{k})| = 2|\hat{i} + \hat{j} + \hat{k}| = 2\sqrt{1^2 + 1^2 + 1^2} = 2\sqrt{3} \]
\[ |\vec{A}| = |\hat{i} + \hat{j} + \hat{k}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

### Step 7: Substitute into the dot product formula
Substituting into the formula:
\[ 6 = (2\sqrt{3})(\sqrt{3}) \cos \theta \]
\[ 6 = 6 \cos \theta \]
\[ \cos \theta = 1 \]

### Step 8: Solve for \(\theta\)
The angle \(\theta\) for which \(\cos \theta = 1\) is:
\[ \theta = 0^\circ \]

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