91Ó°ÊÓ

The commutativity of the dot product in \(\text{\mathbb{R}^{3}}\) indicates that the order of multiplication does not affect the result. For any vectors \(\vec{u}\) and \(\vec{v}\), we need to show that \(\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}\). When we compute \(\vec{u} \cdot \vec{v}\) using the given definition \(\vec{u} \cdot \vec{v} = u_{1} v_{1} + 2 u_{2} v_{2} + 3 u_{3} v_{3}\), we get a scalar value. By properties of scalar multiplication, we also know that \(a \cdot b = b \cdot a\). Therefore, \(\vec{u} \cdot \vec{v}\) equals \(\vec{v} \cdot \vec{u}\) because \(u_{1} v_{1} + 2 u_{2} v_{2} + 3 u_{3} v_{3} = v_{1} u_{1} + 2 v_{2} u_{2} + 3 v_{3} u_{3}\). This demonstrates that the dot product is commutative.

Linearity in the context of the dot product implies distributivity over vector addition. For vectors \(\vec{u}, \vec{v}, \text{and} \vec{w} \in \text{\mathbb{R}^{3}}\), we want to show \(\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}\).
When adding vectors \(\vec{v}\text{ and }\vec{w}\), their components sum individually: \(\vec{v} + \vec{w} = (v_{1} + w_{1}, v_{2} + w_{2}, v_{3} + w_{3})\).
The dot product with \(\vec{u}\) thus becomes: \(\vec{u} \cdot (\vec{v} + \vec{w}) = u_{1}(v_{1} + w_{1}) + 2 u_{2}(v_{2} + w_{2}) + 3 u_{3}(v_{3} + w_{3})\)
Which expands to: \(u_{1} v_{1} + u_{1} w_{1} + 2 u_{2} v_{2} + 2 u_{2} w_{2} + 3 u_{3} v_{3} + 3 u_{3} w_{3}\)
Combining like terms results in: \(\vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w} = (u_{1} v_{1} + 2 u_{2} v_{2} + 3 u_{3} v_{3}) + (u_{1} w_{1} + 2 u_{2} w_{2} + 3 u_{3} w_{3})\)
This confirms the linearity of the dot product.

Positive definiteness means the dot product of a vector with itself should be non-negative and zero only for the zero vector.
This ensures the dot product is a valid measurement of length in vector space.
For any vector \(\vec{u} = (u_{1}, u_{2}, u_{3})\) in \(\text{\mathbb{R}^{3}}\), we compute \((\vec{u} * \vec{u})\) using the definition: \(\vec{u} * \vec{u} = u_{1}^2 + 2u_{2}^2 + 3u_{3}^2\).
Since the squares of the components \(u_{1}^2, 2u_{2}^2, \text{and } 3u_{3}^2\) are always non-negative and zero only if all \(u_{i}\) are zero, it follows that \(\vec{u} * \vec{u} \geq 0\).
This is zero only when \(\vec{u} = \vec{0}\).
Hence, positive definiteness holds.

The Cauchy-Schwarz inequality is crucial in vector mathematics as it bounds the absolute value of the dot product of two vectors.
Specifically, it states: \(\left| \vec{u} \cdot \vec{v} \right| \leq \| \vec{u} \| \| \vec{v} \|\)
where \(\left| \vec{u} \cdot \vec{v} \right|\) is the absolute value of the dot product and \(\| \vec{u} \|, \| \vec{v} \|\) are the magnitudes of \(\vec{u}\) and \(\vec{v}\). For our defined product, we aim to show: \(\left| u_{1} v_{1} + 2 u_{2} v_{2} + 3 u_{3} v_{3} \right| \leq \sqrt{(\vec{u} * \vec{u})} \sqrt{(\vec{v} * \vec{v})}\).
Utilizing the definition again: \(\vec{u} * \vec{u} = u_{1}^2 + 2u_{2}^2 + 3u_{3}^2\) and \(\vec{v} * \vec{v} = v_{1}^2 + 2v_{2}^2 + 3v_{3}^2\), we apply the generalized Cauchy-Schwarz inequality.
As a result: \(\left| u_{1} v_{1} + 2 u_{2} v_{2} + 3 u_{3} v_{3} \right| \leq \sqrt{(u_{1}^2 + 2u_{2}^2 + 3u_{3}^2)} \sqrt{(v_{1}^2 + 2v_{2}^2 + 3u_{3}^2)}\).
Thus, the Cauchy-Schwarz inequality is satisfied.