If `A.B=AxxB`, then angle between `A` and `B` is

Options

A. `45^(@)`
B. `30^(@)`
C. `60^(@)`
D. `90^(@)`

Explanation

To solve the problem, we start with the given equation:

**Step 1: Write down the given equation.**
We have:
\[ A \cdot B = A \times B \]

**Step 2: Use the formulas for dot and cross products.**
The dot product \( A \cdot B \) is given by:
\[ A \cdot B = |A| |B| \cos \theta \]
The cross product \( A \times B \) is given by:
\[ A \times B = |A| |B| \sin \theta \]

**Step 3: Set the two expressions equal to each other.**
From the given equation, we can write:
\[ |A| |B| \cos \theta = |A| |B| \sin \theta \]

**Step 4: Simplify the equation.**
Assuming \( |A| \) and \( |B| \) are not zero, we can divide both sides by \( |A| |B| \):
\[ \cos \theta = \sin \theta \]

**Step 5: Use the identity of tangent.**
We can rewrite the equation as:
\[ \frac{\sin \theta}{\cos \theta} = 1 \]
This implies:
\[ \tan \theta = 1 \]

**Step 6: Find the angle \( \theta \).**
The angle \( \theta \) for which \( \tan \theta = 1 \) is:
\[ \theta = 45^\circ \]

Thus, the angle between vectors \( A \) and \( B \) is:
\[ \theta = 45^\circ \]

**Final Answer:**
The angle between vectors \( A \) and \( B \) is \( 45^\circ \).

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