If $ heta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}.\vec{b}|=|\vec{a} imes\vec{b}|$, then $ heta$ is equal to

Q: If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}.\vec{b}|=|\vec{a}\times\vec{b}|$, then $\theta$ is equal to

If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}.\vec{b}|=|\vec{a}\times\vec{b}|$, then $\theta$ is equal to

Question

If $	heta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}.\vec{b}|=|\vec{a}	imes\vec{b}|$, then $	heta$ is equal to

Accepted Answer

To determine why Option 2 is the correct answer, we need to analyze the condition given in the problem: $ |\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}| $.

### Explanation:

1. **Dot Product and Cross Product:**
   - The dot product of two vectors $\vec{a}$ and $\vec{b}$ is given by:
     $$
     \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta
     $$
   - The magnitude of the cross product of two vectors $\vec{a}$ and $\vec{b}$ is given by:
     $$
     |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta
     $$

2. **Given Condition:**
   - According to the problem, $ |\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}| $.
   - Substituting the expressions for the dot and cross products, we have:
     $$
     |\vec{a}||\vec{b}||\cos\theta| = |\vec{a}||\vec{b}||\sin\theta|
     $$
   - Assuming $ |\vec{a}| $ and $ |\vec{b}| $ are non-zero, we can divide both sides by $ |\vec{a}||\vec{b}| $, yielding:
     $$
     |\cos\theta| = |\sin\theta|
     $$

3. **Solving the Equation:**
   - The equation $ |\cos\theta| = |\sin\theta| $ implies that:
     $$
     \cos^2\theta = \sin^2\theta
     $$
   - This simplifies to:
     $$
     \cos^2\theta - \sin^2\theta = 0
     $$
   - Using the identity $\cos^2\theta = 1 - \sin^2\theta$, we get:
     $$
     1 - 2\sin^2\theta = 0
     $$
   - Solving for $\sin^2\theta$, we find:
     $$
     \sin^2\theta = \frac{1}{2}
     $$
   - Therefore, $\sin\theta = \pm\frac{\sqrt{2}}{2}$ and $\cos\theta = \pm\frac{\sqrt{2}}{2}$.

4. **Determining $\theta$:**
   - The angles that satisfy $\sin\theta = \cos\theta$ are $\theta = \frac{\pi}{4}$ and $\theta = \frac{3\pi}{4}$.
   - However, since we are looking for the angle between two vectors, the principal angle is typically considered, which is $\theta = \frac{\pi}{4}$.

### Why Other Options are Incorrect:

1. **Option 1 (0):**
   - If $\theta = 0$, then $\sin\theta = 0$ and $\cos\theta = 1$. This would imply $ |\vec{a} \times \vec{b}| = 0 $, which does not satisfy the given condition.

2. **Option 3 ($\frac{\pi}{2}$):**
   - If $\theta = \frac{\pi}{2}$, then $\sin\theta = 1$ and $\cos\theta = 0$. This would imply $ |\vec{a} \cdot \vec{b}| = 0 $, which does not satisfy the given condition.

3. **Option 4 ($\pi$):**
   - If $\theta = \pi$, then $\sin\theta = 0$ and $\cos\theta = -1$. This would imply $ |\vec{a} \times \vec{b}| = 0 $, which does not satisfy the given condition.

Thus, the correct answer is indeed Option 2, $\theta = \frac{\pi}{4}$, as it satisfies the condition $ |\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}| $.