In that test, you will calculate the dot product of two positively or negatively aligned vectors, based on their norms.

You will then prove two imprtant inequalities for two vectors \( \overrightarrow{u}\) and \( \overrightarrow{v}\) in the plane \( \mathbb{P}\) :

1. \( -\left\| \overrightarrow{u} \right\|\left\| \overrightarrow{v} \right\| \le\overrightarrow{u}.\overrightarrow{v}\le \left\| \overrightarrow{u} \right\|\left\| \overrightarrow{v} \right\|\)

2. The triangular inequality: \( \left\| \overrightarrow{u}+\overrightarrow{v} \right\| \leq \left\| \overrightarrow{u} \right\|+\left\| \overrightarrow{v} \right\|\) .

For the whole test, \( \overrightarrow{u}\) and \( \overrightarrow{v}\) are any column vectors with \( 2\) real elements.

They are positively aligned i.e \( k>0\) , so that they have the same direction.

Prove that, in that case, \( \overrightarrow{u}.\overrightarrow{v}=\left\| \overrightarrow{u} \right\|\left\| \overrightarrow{v} \right\|\) .

Use the bilinearity of the dot product and the facts that \( \overrightarrow{u}.\overrightarrow{u}=\left\| \overrightarrow{u} \right\|^2\) and \( \left\| k\overrightarrow{u} \right\| =\left| k \right| \left\| \overrightarrow{u} \right\|\) .

They are negatively aligned i.e \( k<0\) , so that they have the opposite direction.

Prove that, in that case, \( \overrightarrow{u}.\overrightarrow{v}=-\left\| \overrightarrow{u} \right\|\left\| \overrightarrow{v} \right\|\) .

Denote the vectors \( \overrightarrow{u}=\begin{bmatrix}x_1\\ y_1\end{bmatrix}\) and \( \overrightarrow{v}=\begin{bmatrix}x_2\\ y_2\end{bmatrix}\) , with \( x_1\) , \( x_2\) , \( y_1\) and \( y_2\) real numbers.

Then develop \( (\overrightarrow{u}.\overrightarrow{v})^2\) .

Use the remarkable identity \( (a+b)^2=a^2+b^2+2ab\) , the fact that \( (ab)^2=a^2b^2\) and the associativity of the multiplication to get rid of the parentheses.

Denote the vectors \( \overrightarrow{u}=\begin{bmatrix}x_1\\ y_1\end{bmatrix}\) and \( \overrightarrow{v}=\begin{bmatrix}x_2\\ y_2\end{bmatrix}\) , with \( x_1\) , \( x_2\) , \( y_1\) and \( y_2\) real numbers.

Then develop \( \left(\left\| \overrightarrow{u} \right\|\left\| \overrightarrow{v} \right\|\right)^2\) .

Use the fact that \( (ab)^2=a^2b^2\) and the distributivity of the multiplication on the addition.

Denote the vectors \( \overrightarrow{u}=\begin{bmatrix}x_1\\ y_1\end{bmatrix}\) and \( \overrightarrow{v}=\begin{bmatrix}x_2\\ y_2\end{bmatrix}\) , with \( x_1\) , \( x_2\) , \( y_1\) and \( y_2\) real numbers.

Then deduce from questions 2 and 3, that \( \left(\left\| \overrightarrow{u} \right\|\left\| \overrightarrow{v} \right\|\right)^2-(\overrightarrow{u}.\overrightarrow{v})^2\) is the square of a real number.

Eliminate the equal terms, group the factors under the exponent \( 2\) and rearrange the product in order to use the identity \( (a+b)^2=a^2+b^2+2ab\) in the reverse order.

Deduce from the question 4 the double inequality:

\( -\left\| \overrightarrow{u} \right\|\left\| \overrightarrow{v} \right\| \le\overrightarrow{u}.\overrightarrow{v}\le \left\| \overrightarrow{u} \right\|\left\| \overrightarrow{v} \right\|\) .

Consider the cases where \( \overrightarrow{u}.\overrightarrow{v}\ge 0\) and \( \overrightarrow{u}.\overrightarrow{v}<0\) , and use the fact that \( \left\| \overrightarrow{u} \right|\left\| \overrightarrow{v} \right\|\ge 0\) . Don’t forget the two sides of the double inequality.

Set both members of the inequality to the square, use the remarkables identities for the vectors and for the real numbers, and use the fact that both members are positive or zero.