The properties of the scalar product

Commutative law

Let \(\vec{u} \) and \(\vec{v} \) be two vectors.

Performing both scalar products \( \vec{u}\cdot\vec{v} \) and \( \vec{u}\cdot\vec{u} \), we do have:

Let us call \(\alpha \) = \( (\vec{u}, \vec{v}) \), the forming angle by \( \vec{u} \) and \(\vec{v} \), then:

But, the cosine function is pair and: \( \forall \alpha \in \mathbb{R}, \ \cos(\alpha ) = \cos(-\alpha ) \).

From this we can deduce that: \( \cos(\vec{u}, \vec{v}) = \cos(\vec{v}, \vec{u}) \).

And finally,

Orthogonal vectors

Let \(\vec{u} \) and \(\vec{v} \) be two vectors, different from the zero vector.

1. From left to right implication

   Let us assume that \(\vec{u} \) and \(\vec{v} \) are orthogonal.

   Then, by the definition of the scalar product,

   $$ \vec{u}\cdot\vec{v} = ||\vec{u}|| \times ||\vec{v}|| \times \cos(\vec{u}, \vec{v}) $$

   But, if \(\vec{u} \) ans \(\vec{v} \) are orhtogonal, then \(\cos(\vec{u}, \vec{v}) = \cos \left(\frac{\pi}{2}\right) = 0 \).

   Thus, their scalar product will be zero, and:

   $$ \forall (\vec{u}, \vec{v}) \neq \vec{0},$$
   $$ \vec{u} \text{ and } \vec{v} \text{ orthogonal vectors }\Longrightarrow \vec{u}\cdot \vec{v} = 0 $$

2. Reciprocal

   Reciprocally, being said that the two vectors \(\vec{u} \) and \(\vec{v} \) are non-zero vectors, so:

   $$ \vec{u}\cdot \vec{v} = 0 \Longrightarrow \cos(\vec{u}, \vec{v}) = 0 $$

   And,

   $$ \forall k \in \mathbb{N}, \ \cos(\vec{u}, \vec{v}) = 0 \Longrightarrow \Biggl\{ (\vec{u}, \vec{v}) = \frac{k\pi}{2} \Biggr \} $$

   And as a result, \(\vec{u} \) and \(\vec{v} \) are necessarily orthogonal vectors.

   $$\vec{u}\cdot \vec{v} = 0 \Longrightarrow \vec{u} \text{ and } \vec{v} \text{ orthogonal vectors }$$

3. Conclusion

   $$ \forall (\vec{u}, \vec{v}) \neq \vec{0},$$
   $$ \vec{u} \text{ and } \vec{v} \text{ orthogonal vectors }\Longleftrightarrow \vec{u}\cdot \vec{v} = 0 $$

Squared vector

Let \(\vec{u} \) be a vector.

Then, by the definition of the scalar product,

But \(\cos(\vec{u}, \vec{u}) = \cos \left(0\right) = 1 \).

Thus, only magnitudes remains in the product.

And finally:

Scalar product by the vectors coordinates

Let \(\vec{u}\begin{pmatrix} x\\ y\\z \end{pmatrix}\) and \(\vec{v}\begin{pmatrix} x'\\ y'\\z' \end{pmatrix}\) be two vectors represented in an orthonormal coordinate system \( (O, \vec{i}, \vec{j}, \vec{k})\) and such as the following figure:

So, we can express \(\vec{u} \) and \(\vec{v} \) in the following formula:

Then, the scalar product \(\vec{u}\cdot\vec{v} \) is worth:

The three vectors \(\vec{i} \cdot \vec{j}, \vec{k} \) are by hypothesis our unit vectors, then:

Furthermore, being in an orthognal reference frame, the three vectors \(\vec{i} \cdot \vec{j}, \vec{k} \) are orthogonal , so their scalar product is zero:

And by the commutative law of the scalar product , we do also have:

Now, rewriting \((1) \),

Et finalement,

Bilinearity

Let \((\lambda, \mu) \in \hspace{0.04em} \mathbb{R}^2\) be two real numbers, and \(\vec{u}\begin{pmatrix} x\\ y\\z \end{pmatrix}, \ \vec{v}\begin{pmatrix} x'\\ y'\\z' \end{pmatrix}\) two vectors.

Performing the scalar product \( (\lambda\vec{u})\cdot\vec{v}\), we do have:

Using the coordinates expression, we do have this:

But, in the same time:

And we then obtain bilinearity:

Furthermore, performing the scalar product \( (\lambda\vec{u})\cdot(\mu\vec{v})\), we do obtain:

Distributive law in relation to the addition

Let \(\vec{u}\), \(\vec{v}\) and \(\vec{w}\) be three vectors, also expressed \(\overrightarrow{AB}\), \(\overrightarrow{AC}\) and \(\overrightarrow{CD}\) such as the following figure:

We also note in this figure the sum \((\vec{v} + \vec{w})\) giving \(\overrightarrow{AD}\).

1. Right-hand distributivity

   So the scalar product \(\vec{u}\cdot( \vec{v} + \vec{w}) \) is worth :

   $$ \vec{u}\cdot( \vec{v} + \vec{w}) = || u || \times || v + w || \times \cos(\vec{u}, \vec{v} + \vec{w})$$
   $$ \vec{u}\cdot( \vec{v} + \vec{w}) = || \overrightarrow{AB} || \times || \overrightarrow{AD} || \times \cos(\overrightarrow{AB}, \overrightarrow{AD}) \qquad (1) $$

   Let us now calculate what the scalar products are independently worth \(\vec{u}\cdot\vec{v}\) and \(\vec{u}\cdot\vec{w}\):

   $$ \left \{ \begin{gather*} \vec{u}.\vec{v} = || \overrightarrow{AB} || \times || \overrightarrow{AC} || \times \cos(\overrightarrow{AB}, \overrightarrow{AC}) \\ \vec{u} \cdot \vec{w} = || \overrightarrow{AB} || \times || \overrightarrow{CD} || \times \cos(\overrightarrow{AB}, \overrightarrow{CD}) \end{gather*} \right \} $$
   

   But \(\left[|| \overrightarrow{AC} || \times \cos(\overrightarrow{AB}, \overrightarrow{AC})\right]\) and \(\left[|| \overrightarrow{CD} || \times \cos(\overrightarrow{AC}, \overrightarrow{CD})\right]\) are the respective orthogonal projections of vectors \(\overrightarrow{AC}\) and \(\overrightarrow{CD}\) on vector \(\overrightarrow{AB}\).

   $$ \left \{ \begin{gather*} \vec{u}.\vec{v} = || \overrightarrow{AB} || \times || \overrightarrow{AC'} || \qquad(2) \\ \vec{u} \cdot \vec{w} = || \overrightarrow{AB} || \times || \overrightarrow{C'D'} || \qquad(3) \end{gather*} \right \} $$

   So, the sum of the two expressions \((2)\) and \((3)\) is worth:

   $$ \vec{u}\cdot\vec{v} + \vec{u} \cdot \vec{w} = || \overrightarrow{AB} || \times || \overrightarrow{AC'} || + || \overrightarrow{AB} || \times || \overrightarrow{C'D'} || $$
   $$ \vec{u}\cdot\vec{v} + \vec{u} \cdot \vec{w} = || \overrightarrow{AB} || \times \left[ || \overrightarrow{AC'} || + || \overrightarrow{C'D'} || \right] $$

   Moreover, both vectors \(|| \overrightarrow{AC'} ||\) and \(|| \overrightarrow{C'D'} ||\) being aligned, we have:

   $$ \vec{u}\cdot\vec{v} + \vec{u} \cdot \vec{w} = || \overrightarrow{AB} || \times || \overrightarrow{AD'} || $$

   And \(|| \overrightarrow{AD'} ||\) being the orthogonal projection of the vector \(\overrightarrow{AD}\) on \(\overrightarrow{AB}\), we now have:

   $$ \vec{u}\cdot\vec{v} + \vec{u} \cdot \vec{w} = || \overrightarrow{AB} || \times || \overrightarrow{AD} || \times \cos(\overrightarrow{AB}, \overrightarrow{AD}) $$

   So, thanks to the previous expression \((1)\), we finally have that:

   $$ \forall (\vec{u}, \vec{v}, \vec{w}), $$
   $$ \vec{u}\cdot( \vec{v} + \vec{w}) = \vec{u}.\vec{v} + \vec{u} \cdot \vec{w} $$

2. Left-handed distributive law

   And thanks to the commutative law of the scalar product , we do have the same thing to the left:

   $$ (\vec{u} + \vec{v}) \cdot \vec{w} = \vec{w} . (\vec{u} + \vec{v}) $$

   Devlopping the expression with the right-handed distributive law:

   $$ (\vec{u} + \vec{v}) \cdot \vec{w} = \vec{w}.\vec{u} + \vec{w} \cdot \vec{v} $$

   And still with the commutative law :

   $$ (\vec{u} + \vec{v}) \cdot \vec{w} = \vec{u} \cdot \vec{w} + \vec{v} \cdot \vec{w} $$

   And finally,

   $$ \forall (\vec{u}, \vec{v}, \vec{w}), $$
   $$ (\vec{u} + \vec{v}) \cdot \vec{w} = \vec{u} \cdot \vec{w} + \vec{v} \cdot \vec{w} $$

Remarkable identities

We saw above that the scalar product is distributive , that is the property that will be used in this part.

1. Calculation of \( (\vec{u} + \vec{v})^2 \)

   $$ (\vec{u} + \vec{v})^2 = (\vec{u} + \vec{v})\cdot(\vec{u} + \vec{v}) $$
   $$ (\vec{u} + \vec{v})^2 =\vec{u}\cdot \vec{u} + \vec{u}.\vec{v} + \vec{u} \cdot \vec{v} - \vec{v}. \vec{v} $$
   $$ (\vec{u} + \vec{v})^2 = {|| \vec{u} ||}^2 + 2 \vec{u}\cdot\vec{v} + {|| \vec{v} ||}^2 $$

2. Calculation of \( (\vec{u} - \vec{v})^2 \)

   This is the same calculation but with a \((-)\) sign in the double product.

   $$ (\vec{u} - \vec{v})^2 = (\vec{u} - \vec{v})\cdot(\vec{u} - \vec{v}) $$
   $$ (\vec{u} - \vec{v})^2 =\vec{u}\cdot \vec{u} - \vec{u}.\vec{v} - \vec{u} \cdot \vec{v} - \vec{v}. \vec{v} $$
   $$ (\vec{u} - \vec{v})^2 = {|| \vec{u} ||}^2 - 2 \vec{u}\cdot\vec{v} + {|| \vec{v} ||}^2 $$

3. Calculation of \( (\vec{u} + \vec{v})\cdot (\vec{u} - \vec{v}) \)

   $$ (\vec{u} + \vec{v})\cdot (\vec{u} - \vec{v}) = \vec{u}. \vec{u} - \vec{u} \cdot \vec{v} + \vec{v}. \vec{u} - \vec{v}. \vec{v} $$

   We saw above that the scalar product is commutative , so \( \vec{u}\cdot \vec{v} = \vec{v}. \vec{u} \), and:

   $$ (\vec{u} + \vec{v}). (\vec{u} - \vec{v}) = {|| \vec{u} ||}^2 \hspace{0.1em} \underbrace{ - \ \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{v} } _{ = \hspace{0.1em} 0 } \hspace{0.1em} - {|| \vec{v} ||}^2$$
   $$ {|| \vec{u} ||}^2 - {|| \vec{v} ||}^2= (\vec{u} + \vec{v}) \cdot (\vec{u} - \vec{v}) $$

As a result, we obtain the same formulas as the classical remarkable identities .

Expression depending on norms

We saw above that:

Working with the expression \( (4) \), we do have:

Then \( (4) \) becomes now:

And finally,

In the same way, working with the expression \( (5) \):

Projection of a vector onto a vector

The scalar projection of \(\vec{v}\) onto \(\vec{u}\) is worth:

Thus, having its standard we can attribute to it the direction and the sens of \(\vec{u}\) via the operation:

And with the above, we have:

The vectorial projection of \(\vec{u}\) onto \(\vec{v}\) is worth:

Recap table of the properties of the scalar product