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. $$c(\mathbf{u} \cdot \mathbf{v})=(c \mathbf{u}) \cdot \mathbf{v}=\mathbf{u} \cdot(c \mathbf{v})$$

Short answer

Question: Prove that for any vectors \(\mathbf{u}, \mathbf{v}\) and scalar \(c\), it is true that \(c(\mathbf{u} \cdot \mathbf{v}) = (c \mathbf{u}) \cdot \mathbf{v} = \mathbf{u} \cdot (c \mathbf{v})\). Answer: We showed that multiplying the dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) by a scalar \(c\) results in the same expression as finding the dot product of one of the vectors multiplied by the scalar and the other vector. We also demonstrated that this property holds regardless of whether we multiply the first vector or the second vector by the scalar. Therefore, we can conclude that \(c(\mathbf{u} \cdot \mathbf{v}) = (c \mathbf{u}) \cdot \mathbf{v} = \mathbf{u} \cdot (c \mathbf{v})\).

Step-by-step solution

Find the dot product of \(\mathbf{u} \cdot \mathbf{v}\)

To start, let's find the dot product of the two vectors \(\mathbf{u}\) and \(\mathbf{v}\). The formula for the dot product of two vectors is given by: $$(\mathbf{u} \cdot \mathbf{v}) = u_1 v_1 + u_2 v_2 + u_3 v_3$$

Multiply dot product by the scalar \(c\)

Now, let's multiply the dot product by the scalar \(c\). This will give us the left side of the equation: $$c(\mathbf{u} \cdot \mathbf{v}) = c(u_1 v_1 + u_2 v_2 + u_3 v_3)$$

Find the dot product after scalar multiplication

Now we will calculate the dot product after multiplying one of the vectors by the scalar \(c\). We can start with multiplying \(\mathbf{u}\) by \(c\). The resulting vector \((c \mathbf{u})\) will look like: $$(c \mathbf{u}) = \left\langle c u_{1}, c u_{2}, c u_{3}\right\rangle$$ Now let's find the dot product of \((c \mathbf{u})\) and \(\mathbf{v}\): $$(c \mathbf{u}) \cdot \mathbf{v} = (c u_1) v_1 + (c u_2) v_2 + (c u_3) v_3$$

Rewrite the dot product after scalar multiplication

We can rewrite the dot product after scalar multiplication by factoring out the scalar from each product term: $$(c \mathbf{u}) \cdot \mathbf{v} = c(u_1 v_1) + c(u_2 v_2) + c(u_3 v_3)$$

Compare the results

Now let's compare the results from Steps 2 and 4: $$c(\mathbf{u} \cdot \mathbf{v}) = c(u_1 v_1 + u_2 v_2 + u_3 v_3)$$ $$(c \mathbf{u}) \cdot \mathbf{v} = c(u_1 v_1) + c(u_2 v_2) + c(u_3 v_3)$$ As we can clearly see, both expressions are equal.

Repeat with the other vector

Now let's repeat the process with the other vector, \(\mathbf{v}\): $$(\mathbf{u} \cdot (c \mathbf{v})) = u_1 (c v_1) + u_2 (c v_2) + u_3 (c v_3)$$ Factoring out the scalar \(c\): $$(\mathbf{u} \cdot (c \mathbf{v})) = c(u_1 v_1) + c(u_2 v_2) + c(u_3 v_3)$$

Conclusion

Comparing the results from Steps 2, 4, and 6, we can conclude that: $$c(\mathbf{u} \cdot \mathbf{v}) = (c \mathbf{u}) \cdot \mathbf{v} = \mathbf{u} \cdot (c \mathbf{v})$$ This confirms the given property for dot products, and we have successfully proved the claim.