The scalar product in the plane : 11th grade math lesson

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The scalar product in the plane in a 11th grade math lesson where we will study the definition and the different properties of the scalar product. In this eleventh grade lesson, we will see the relationship between the scalar product and the concept of orthogonality. Then, we will finish with applications with the cosine of an angle, the Al-Kashi theorem and the median theorem.

I. Different expressions of the scalar product :

1. Colinear vectors :

Definition:

be \,\vec{u} and \vec{v} two non-zero collinear vectors, such that

\,\vec{u}=\vec{OA} and \vec{v}=\vec{OB}.

– If \vec{u} and \vec{v} have the same meaning: \,\vec{u}.\vec{v}=OA\times  \,OB.

– If \vec{u} and \vec{u} are in opposite directions: \,\vec{u}.\vec{v}=OA\times  \,OB.

– If \,\vec{u}=\vec{0} or \vec{v}=\vec{0} then \vec{u}.\vec{v}=0.

\vec{u}.\vec{u}=%7C%7C\vec{u}%7C%7C is the scalar square of the vector \vec{u}

scalar product

2. Any vectors :

Property 1:

Let \vec{u} and \vec{v} be two non-zero vectors such that

\,\vec{u}=\vec{OA} and \vec{v}=\vec{OB}.

So:

\fbox{\vec{u}.\vec{v}=OA'\times  \,OB=OA\times  \,OB'} .

A’ and B’ are respectively the orthogonal projects of A on (OB) and of B on (OA).

orthogonal vectors

3. Properties :

Property 2:

Let (x;y) and (x’;y’) be the respective coordinates of the vectors \vec{u} and \,\vec{v} in any orthonormal reference frame.

\,\fbox{\vec{u}.\vec{v}\,=\,x.x'+y.y'} .

II. Scalar product and orthogonality :

Definition:

To say that \,\vec{u} and \vec{v} are two orthogonal vectors means that :

– Either \vec{u}=\vec{0} or \,\vec{v}=\vec{0};

– Let (OA)\perp(OB), with \,\vec{u}=\vec{OA} and \vec{v}=\vec{OB} nonzero.

Scalar product and orthogonality

2. Property:

Ownership:

\,\fbox{\vec{u}\perp\vec{v}\,\,\Longleftrightarrow\,\,\vec{u}.\vec{v}=0} .

III. Properties of the scalar product :

Properties:

Properties:

Let \vec{u}\,,\,\vec{v},\,\vec{w} be three vectors and k a real number.

\vec{u}.\vec{v}=\vec{v}.\vec{u} (symmetry).

\,(k\vec{u}).\vec{v}=\vec{u}.(k\vec{v})=k(\vec{u}.\vec{v}) (linearity)

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

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

(\vec{u}+\vec{v})^2=\vec{u}^2+2\vec{u}.\vec{v}+\vec{v}^2 (remarkable identity)

(\vec{u}-\vec{v})^2=\vec{u}^2-2\vec{u}.\vec{v}+\vec{v}^2 (remarkable identity)

(\vec{u}-\vec{v})(\vec{u}+\vec{v})=\vec{u}^2-\vec{v}^2 (remarkable identity)

IV. Applications of the scalar product:

1. Scalar product and cosine :

Ownership:

Let \vec{u} and \,\vec{v} be non-zero.

\vec{u}.\vec{v}=%7C%7C\vec{u}%7C%7C\times  \,%7C%7C\vec{v}%7C%7C\times  \,cos(\vec{u},\vec{v})

2. Al-Kashi theorem:

Theorem:

Let ABC be a triangle such that AB=c, AC=b and BC=a.

We have:

  1. a^2\,=\,b^2+c^2-2bc\times  \,cos(\widehat{A})
  2. \,b^2\,=\,a^2+c^2-2ac\times  \,cos(\widehat{B})
  3. c^2\,=\,a^2+b^2-2ab\times  \,cos(\widehat{C}

Al-Kashi's theorem

3. Median theorem:

Theorem:

Let A and B be two distinct points and I the middle of the segment [AB].

For any point M, :

\fbox{\,MA^2\,+MB^2\,=\,2MI^2\,+\,\frac{AB^2}{2}}

Median theorem

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