$\quad\left(\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right) \cdot\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right)=\ldots$ (A) 0 (B) \(A^{2}+B^{2}\) (C) \(\sqrt{\left(A^{2}+B^{2}\right)}\) (D) \(\mathrm{A}^{2} \mathrm{~B}^{2}\)

We need to find the value of the given expression, which is: \[\left(\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right) \cdot \left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right)\]

Remember that the dot product has the following properties: 1. Distributive: \(\mathrm{A}^{\rightarrow} \cdot (\mathrm{B}^{\rightarrow}+\mathrm{C}^{\rightarrow})= \mathrm{A}^{\rightarrow}\cdot \mathrm{B}^{\rightarrow}+\mathrm{A}^{\rightarrow}\cdot \mathrm{C}^{\rightarrow}\) 2. Commutative: \(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow} = \mathrm{B}^{\rightarrow} \cdot \mathrm{A}^{\rightarrow}\) As for the cross product, we have the following properties: 1. Anticommutative: \(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow} = -\mathrm{B}^{\rightarrow} \times \mathrm{A}^{\rightarrow}\) 2. Distributive: \(\mathrm{A}^{\rightarrow} \times (\mathrm{B}^{\rightarrow}+\mathrm{C}^{\rightarrow}) = \mathrm{A}^{\rightarrow}\times \mathrm{B}^{\rightarrow} + \mathrm{A}^{\rightarrow}\times \mathrm{C}^{\rightarrow}\)

Applying the dot product distributive property, the given expression becomes: \[(\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow})\cdot(\mathrm{A}^{\rightarrow}\times\mathrm{B}^{\rightarrow}) = \mathrm{A}^{\rightarrow}\cdot(\mathrm{A}^{\rightarrow}\times\mathrm{B}^{\rightarrow}) + \mathrm{B}^{\rightarrow}\cdot(\mathrm{A}^{\rightarrow}\times\mathrm{B}^{\rightarrow})\]

For both terms on the right-hand side of the equation, the dot product of a cross product is involved, so we can write: \[\mathrm{A}^{\rightarrow} \cdot (\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}) = - \mathrm{A}^{\rightarrow} \cdot (\mathrm{B}^{\rightarrow} \times \mathrm{A}^{\rightarrow})\] \[\mathrm{B}^{\rightarrow} \cdot (\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}) = -\mathrm{B}^{\rightarrow} \cdot (\mathrm{B}^{\rightarrow} \times \mathrm{A}^{\rightarrow})\] Now, the expression from Step 3 becomes: \[-(\mathrm{A}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow}))+(-\mathrm{B}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow}))\]

The scalar triple product is defined as: \[\mathrm{A}^{\rightarrow} \cdot (\mathrm{B}^{\rightarrow} \times\mathrm{C}^{\rightarrow}) = \mathrm{B}^{\rightarrow} \cdot (\mathrm{C}^{\rightarrow} \times\mathrm{A}^{\rightarrow}) = \mathrm{C}^{\rightarrow} \cdot (\mathrm{A}^{\rightarrow}\times\mathrm{B}^{\rightarrow})\] Now we see that the terms in the expression from Step 4 are scalar triple products: \[-(\mathrm{A}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow}))=-\mathrm{A}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow})\] \[-(\mathrm{B}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow}))=-\mathrm{B}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow})\] Since each term on the right-hand side is equal, their sum will be zero: \[-\mathrm{A}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow})-\mathrm{B}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow})=0\] Therefore, the original expression is equal to 0, which corresponds to choice (A).