$a^{2}b^{2}-(\vec{a} imes\vec{b})^{2}=$

Q: $a^{2}b^{2}-(\vec{a}\times\vec{b})^{2}=$

$a^{2}b^{2}-(\vec{a}\times\vec{b})^{2}=$

Question

$a^{2}b^{2}-(\vec{a}	imes\vec{b})^{2}=$

Accepted Answer

To solve the problem and justify why Option 1 is the correct answer, we need to delve into vector algebra and the properties of dot and cross products.

### Explanation for Option 1: $(\vec{a}.\vec{b})^{2}$

The expression given in the question is $a^{2}b^{2}-(\vec{a}	imes\vec{b})^{2}$. Here, $a$ and $b$ are the magnitudes of vectors $\vec{a}$ and $\vec{b}$, respectively. The cross product $\vec{a} 	imes \vec{b}$ results in a vector whose magnitude is given by $|\vec{a} 	imes \vec{b}| = ab \sin 	heta$, where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$.

The square of the magnitude of the cross product is:
$$(\vec{a} 	imes \vec{b})^2 = (ab \sin 	heta)^2 = a^2b^2 \sin^2 	heta.$$

Now, consider the dot product $\vec{a} \cdot \vec{b}$, which is given by:
$$\vec{a} \cdot \vec{b} = ab \cos 	heta.$$

The square of the dot product is:
$$(\vec{a} \cdot \vec{b})^2 = (ab \cos 	heta)^2 = a^2b^2 \cos^2 	heta.$$

Using the Pythagorean identity $\sin^2 	heta + \cos^2 	heta = 1$, we can express $a^2b^2$ as:
$$a^2b^2 = a^2b^2 \sin^2 	heta + a^2b^2 \cos^2 	heta.$$

Therefore, the expression $a^{2}b^{2}-(\vec{a}	imes\vec{b})^{2}$ becomes:
$$a^2b^2 - a^2b^2 \sin^2 	heta = a^2b^2 \cos^2 	heta = (\vec{a} \cdot \vec{b})^2.$$

This confirms that Option 1, $(\vec{a}.\vec{b})^{2}$, is indeed the correct answer.

### Explanation for Why Other Options are Incorrect:

- **Option 2: $(\vec{a}+\vec{b}).(\vec{a}-\vec{b})$**

This expression simplifies to $\vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{b}$, which is $a^2 - b^2$. This does not relate to the expression $a^{2}b^{2}-(\vec{a}	imes\vec{b})^{2}$, which involves both $a^2b^2$ and the angle between the vectors.

- **Option 3: $(\vec{a}-\vec{b})^{2}$**

This expression expands to $(\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) = a^2 - 2\vec{a} \cdot \vec{b} + b^2$. This is not equivalent to $a^{2}b^{2}-(\vec{a}	imes\vec{b})^{2}$, as it does not involve the product of the magnitudes $a^2b^2$ or the sine of the angle between the vectors.

- **Option 4: $(\vec{a}+\vec{b})^{2}$**

This expression expands to $(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = a^2 + 2\vec{a} \cdot \vec{b} + b^2$. Similar to Option 3, this does not match the form of $a^{2}b^{2}-(\vec{a}	imes\vec{b})^{2}$, as it lacks the necessary terms involving the cross product.

In conclusion, Option 1 is the only expression that correctly represents the given equation, based on the properties of dot and cross products in vector algebra.