The problem is about vector calculations and the dot product on a line $(\Delta)$ with a coordinate system $(A; \vec{i})$. We are given that $x=8$. We need to find expressions for algebraic measures of vectors, calculate dot products, and prove a relationship between the dot product and the cosine of the angle between vectors.

1. Problem Description

The problem is about vector calculations and the dot product on a line $(\Delta)$ with a coordinate system $(A; \vec{i})$. We are given that $x=8$. We need to find expressions for algebraic measures of vectors, calculate dot products, and prove a relationship between the dot product and the cosine of the angle between vectors.

2. Solution Steps

1. a) Given $\vec{AB} = (x_B - x_A) \vec{i}$, express $AB$ in terms of $x_A$ and $x_B$.

Since $AB$ is the algebraic measure of the vector $\vec{AB}$, we have $AB = x_B - x_A$.

2. b) Determine the algebraic measure of each couple of points $(A, B)$, $(A, S)$, $(S, B')$ and $(B, R)$.

From the figure, we have $x_A = 0$. Also, $x=8$, and the length from $A$ to $R$ is $x$, so $x_R = x = 8$. We also have the distance from $A$ to $S$ being $2x$, so $x_S = 2x = 16$. For $B$, note that $x - 3$ is the height of the rectangle.

$AB = x_B - x_A = x_B - 0 = x_B$. From the diagram, it seems $x_B$ is close to 0, however $AB$ seems negative. Since the diagram seems incorrect, since length can not be negative, we calculate

$AB = x_B - x_A$

From figure, we can say $x_A = 0$ and $x_B$ is negative, thus $x_B = - (x-3) = 3 - x = 3 - 8 = -5$. Thus $AB = -5$

$AS = x_S - x_A = 2x - 0 = 2x = 2(8) = 16$.

$SB' = x_{B'} - x_S$. $x_S = 16$. The distance from $B'$ to $A$ seems to be $x$. So $x_{B'} = - (x-3) + x = 3$. $SB' = 3 - 16 = -13$.

$BR = x_R - x_B = x - (3-x) = 8- (3-8) = 8 - (-5) = 13$.

3. a) Give the orthogonal projection of points $A$ and $C$ onto the line $(AS)$.

The orthogonal projection of $A$ onto the line $AS$ is $A$ itself. The orthogonal projection of $C$ onto the line $AS$ is $B$.

4. b) Calculate $AS \times AB$.

The problem defines $AS \times AB$ as the scalar product of $\vec{AS}$ by the vector $\vec{AC}$, where $A$ and $B$ are respectively the projections of $A$ and $C$ on (AS).

Since the projection of $A$ onto $(AS)$ is $A$ and the projection of $C$ onto $(AS)$ is $B$, we have $AS \times AB = \vec{AS} \cdot \vec{AB} = AS \cdot AB$, where AS and AB denote the algebraic lengths of AS and AB.

$AS = 16$ and $AB = -5$, so $AS \cdot AB = 16 \times (-5) = -80$.

5. c) Calculate $AB \cdot AC$ and $BD \cdot SB$.

Since the orthogonal projection of AC onto AS is AB, the scalar product $AB \cdot AC$ is $AB \cdot AB = AB^2 = (-5)^2 = 25$.

Similarly, we need to calculate $BD \cdot SB$. Since the points $B, D$ have same x-coordinate as $C$ and $A$, $x_B = -5$ and $x_D = x - 5 = 3$, from symmetry. So $\vec{BD}$ is $\vec{D} - \vec{B} = (3 - (-5)) \vec{i} = 8 \vec{i}$.

$\vec{SB} = \vec{B} - \vec{S} = (-5 - 16)\vec{i} = -21\vec{i}$. $BD = 8$, $SB = -21$.

The projection of $\vec{BD}$ on AS has length

0. Then $BD \cdot SB = 0$.

Or $BD \cdot SB = (8)(-21) = -168$. However, BD is perpendicular to SB' or perpendicular to AB. so $SB \cdot BD = 0$.

6. d) Given $\cos(\vec{AS}, \vec{AC}) = \cos(\pi - \alpha)$, calculate $||\vec{AS}|| \times ||\vec{AC}|| \times \cos(\vec{AS}, \vec{AC})$ then prove that $\vec{AS} \cdot \vec{AC} = ||\vec{AS}|| \times ||\vec{AC}|| \times \cos(\vec{AS}, \vec{AC})$.

$||\vec{AS}|| = |AS| = 16$. $||\vec{AC}|| = \sqrt{AB^2 + BC^2} = \sqrt{(-5)^2 + (8-3)^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}$.

$||\vec{AS}|| \times ||\vec{AC}|| \times \cos(\vec{AS}, \vec{AC}) = 16 \times 5\sqrt{2} \times \cos(\pi - \alpha) = 80\sqrt{2} \cos(\pi - \alpha)$.

We know $\vec{AS} \cdot \vec{AC} = AS \times$ (projection of AC onto AS) $= AS \times AB = 16 \times (-5) = -80$.

$\cos(\alpha) = \frac{AB}{AC} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.

So, $\alpha = \pi/4$.

Then $\cos(\pi - \alpha) = - \cos(\alpha) = -\frac{1}{\sqrt{2}}$.

$80\sqrt{2} \times -\frac{1}{\sqrt{2}} = -80$.

So, $\vec{AS} \cdot \vec{AC} = ||\vec{AS}|| \times ||\vec{AC}|| \times \cos(\vec{AS}, \vec{AC}) = -80$.

3. Final Answer

1. a) $AB = x_B - x_A$

2. b) $AB = -5$, $AS = 16$, $SB' = -13$, $BR = 13$

3. a) Projection of $A$ is $A$, projection of $C$ is $B$

4. b) $AS \cdot AB = -80$

5. c) $AB \cdot AC = 25$, $BD \cdot SB = 0$

6. d) $\vec{AS} \cdot \vec{AC} = ||\vec{AS}|| \times ||\vec{AC}|| \times \cos(\vec{AS}, \vec{AC}) = -80$