Question

Properties of dot products Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle,\) and \(\mathbf{w}=\left\langle w_{1}, w_{2}, w_{3}\right\rangle .\) Prove the following vector properties, where \(c\) is a scalar. $$\mathbf{u} \cdot \mathbf{v}=\mathbf{v} \cdot \mathbf{u}$$

Short answer

Question: Prove the commutative property of the dot product for two 3-dimensional vectors. Answer: The commutative property of the dot product states that \(\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}\). We computed the dot products \(\mathbf{u} \cdot \mathbf{v}\) and \(\mathbf{v} \cdot \mathbf{u}\) for given vectors \(\mathbf{u} = \langle u_1, u_2, u_3 \rangle\) and \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\), and found that both expressions are the same. Thus, we proved that the dot product is commutative for 3-dimensional vectors.

Step-by-step solution

Recall the dot product definition

To start, we need to recall how to compute the dot product of two vectors. The dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by the following formula: $$\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3$$

Calculate \(\mathbf{u} \cdot \mathbf{v}\)

Using the dot product definition, let's compute the dot product \(\mathbf{u} \cdot \mathbf{v}\) using the given vectors \(\mathbf{u} = \langle u_1, u_2, u_3 \rangle\) and \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\): $$\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3$$

Calculate \(\mathbf{v} \cdot \mathbf{u}\)

Now let us compute \(\mathbf{v} \cdot \mathbf{u}\): $$\mathbf{v} \cdot \mathbf{u} = v_1 u_1 + v_2 u_2 + v_3 u_3$$

Compare the results

Now that we have the expressions for \(\mathbf{u} \cdot \mathbf{v}\) and \(\mathbf{v} \cdot \mathbf{u}\), we can compare them: $$\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3$$ $$\mathbf{v} \cdot \mathbf{u} = v_1 u_1 + v_2 u_2 + v_3 u_3$$ As we can observe, the expressions for both dot products are the same because multiplication is commutative for real numbers. Therefore, we can conclude that the dot product is commutative: $$\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$$