If there are two vectors `vec(A)` and `vec(B)` such that `vec(A)+vec(B)=hat(i)+2hat(j)+hat(k)` and `vec(A)-vec(B)=(hat(i)-vec(k))`, then choose the correct options.

Options

A. the angle between `vec(A)` and `vec(B)` is `60^(@)`
B. `vec(A) = hat(i) + hat(j)`
C. `vec(B) = hat(j) + hat(k)`
D. the angle between `vec(A)` and `vec(B)` is `120^(@)`

Explanation

To solve the problem, we need to find the vectors \(\vec{A}\) and \(\vec{B}\) based on the given equations:

1. \(\vec{A} + \vec{B} = \hat{i} + 2\hat{j} + \hat{k}\) (Equation 1)
2. \(\vec{A} - \vec{B} = \hat{i} - \hat{k}\) (Equation 2)

### Step 1: Add the two equations

We can add Equation 1 and Equation 2 to eliminate \(\vec{B}\):

\[ (\vec{A} + \vec{B}) + (\vec{A} - \vec{B}) = (\hat{i} + 2\hat{j} + \hat{k}) + (\hat{i} - \hat{k}) \]

This simplifies to:

\[ 2\vec{A} = 2\hat{i} + 2\hat{j} \]

### Step 2: Solve for \(\vec{A}\)

Now, divide both sides by 2:

\[ \vec{A} = \hat{i} + \hat{j} \]

### Step 3: Substitute \(\vec{A}\) back into one of the original equations

We can substitute \(\vec{A}\) into Equation 1 to find \(\vec{B}\):

\[ (\hat{i} + \hat{j}) + \vec{B} = \hat{i} + 2\hat{j} + \hat{k} \]

### Step 4: Solve for \(\vec{B}\)

Rearranging gives us:

\[ \vec{B} = (\hat{i} + 2\hat{j} + \hat{k}) - (\hat{i} + \hat{j}) \]

This simplifies to:

\[ \vec{B} = \hat{j} + \hat{k} \]

### Step 5: Find the angle between \(\vec{A}\) and \(\vec{B}\)

We need to find the angle \(\theta\) between the vectors \(\vec{A}\) and \(\vec{B}\). The formula for the cosine of the angle between two vectors is:

\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]

### Step 6: Calculate \(\vec{A} \cdot \vec{B}\)

First, calculate the dot product:

\[ \vec{A} \cdot \vec{B} = (\hat{i} + \hat{j}) \cdot (\hat{j} + \hat{k}) = (1)(0) + (1)(1) + (0)(1) = 1 \]

### Step 7: Calculate the magnitudes of \(\vec{A}\) and \(\vec{B}\)

Now, calculate the magnitudes:

\[ |\vec{A}| = \sqrt{1^2 + 1^2} = \sqrt{2} \]

\[ |\vec{B}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} \]

### Step 8: Substitute into the cosine formula

Now substitute these values into the cosine formula:

\[ \cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2} \]

### Step 9: Find \(\theta\)

Finally, take the inverse cosine:

\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Conclusion

The vectors are:
- \(\vec{A} = \hat{i} + \hat{j}\)
- \(\vec{B} = \hat{j} + \hat{k}\)

The angle between \(\vec{A}\) and \(\vec{B}\) is \(60^\circ\).

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