Question

Prove that →A.(→A×→B)=0.

Answer 1

Explanation:

assume you are working with real inner products. Since ∥a−b∥=∥a+b∥‖a−b‖=‖a+b‖, we have ⟨a−b,a−b⟩=⟨a+b,a+b⟩⟨a−b,a−b⟩=⟨a+b,a+b⟩. But

⟨a−b,a−b⟩=⟨a,a−b⟩−⟨b,a−b⟩=⟨a,a⟩−⟨a,b⟩−⟨b,a⟩+⟨b,b⟩⟨a−b,a−b⟩=⟨a,a−b⟩−⟨b,a−b⟩=⟨a,a⟩−⟨a,b⟩−⟨b,a⟩+⟨b,b⟩

and

⟨a+b,a+b⟩=⟨a,a+b⟩+⟨b,a+b⟩=⟨a,a⟩+⟨a,b⟩+⟨b,a⟩+⟨b,b⟩⟨a+b,a+b⟩=⟨a,a+b⟩+⟨b,a+b⟩=⟨a,a⟩+⟨a,b⟩+⟨b,a⟩+⟨b,b⟩

so subtracting the two right-hand sides gives us ⟨a,b⟩+⟨b,a⟩=0⟨a,b⟩+⟨b,a⟩=0. What can you conclude from this?

Answer 2

Explanation:

Step 1:

It is to be proven that $A \rightarrow A \rightarrow \times B \rightarrow=0$

Vector product is shown as $A \rightarrow \times B \rightarrow=A \rightarrow B \rightarrow \sin n^{\wedge}$

To the plane comprising $A \rightarrow$ and $B \rightarrow$, $A \rightarrow B \rightarrow \sin n^{\wedge}$ is a vector.

Step 2:

This shows that to $A \rightarrow$, it is perpendicular. It is to be known that the two perpendicular vectors have the dot products, which is zero.

Therefore, $A \rightarrow A \rightarrow \times B \rightarrow=0$.

Hence, this is proven.

It is to be known that the vector product, cross product or the area product of the two vectors is a binary operation.