Content code
m1306
Slug (identifier)
proving-vector-statements
Parent content
Vectors
Grades
Secondary V
Topic
Mathematics
Tags
chasles
justifications
losange
milieu
diagonales
relation
démonstrations
vecteurs
propositions
Content
Contenu
Corps

There is no single procedure to prove statements using vectors.​​

Links
Type
Anchor
Title
Prove that if |\overrightarrow{u} = (a,b)|, then |\mid \mid \overrightarrow{u} \mid \mid = \mid \mid - \overrightarrow{u}\mid\mid|
Anchor
formula-1
Type
Anchor
Title
Prove that |\overrightarrow{u} \cdot \overrightarrow{u}=\mid \mid \overrightarrow{u}\mid \mid ^{2}|
Anchor
formula-2
Type
Anchor
Title
Show that the diagonals of a rhombus are perpendicular
Anchor
diagonals
Type
Anchor
Title
If |ABCD| is a parallelogram, then the diagonals intersect at their midpoints
Anchor
formula-3
Type
Anchor
Title
Prove that the line segment that connects the midpoints of two sides of a triangle is parallel to the 3rd side and equal to half of the 3rd side’s length
Anchor
line-segment
Corps

However, some strategies can be effective for proving these statements.

  1. Set up a plan for proving the statement (mentally or on paper) before starting the calculations.

  2. Work in a structured way. Tables like the ones used in the examples below often make a proof clearer.

  3. Make use of Chasles' relation.

  4. Use the properties of vectors and vector operations.

Title (level 3)
Prove that if |\overrightarrow{u} = (a,b)|, then |\mid \mid \overrightarrow{u}\mid \mid =\mid \mid -\overrightarrow{u}\mid \mid |.
Title slug (identifier)
formula-1
Corps

Statement

Justification

|\mid \mid \overrightarrow{u}\mid \mid =\sqrt{a^{2}+b^{2}}|

By definition

|-\overrightarrow{u}=(-a,-b)|

Opposite vector

|\mid \mid - \overrightarrow{u}\mid \mid =\sqrt{(-a)^{2}+(-b)^{2}}|

By definition

|\mid \mid -\overrightarrow{u}\mid \mid =\sqrt{a^{2}+b^{2}}|

By calculation

The right sides of statements 1 and 4 are the same.

Observation

|\mid \mid \overrightarrow{u}\mid \mid =\mid \mid -\overrightarrow{u}\mid \mid |

By the transitive property
Title (level 3)
Prove that |\overrightarrow{u} \cdot \overrightarrow{u} =\mid \mid \overrightarrow{u}\mid \mid ^{2}|.
Title slug (identifier)
formula-2
Corps

Let |\overrightarrow{u}=(r,s)|.
​​​

Statement

Justification

|\overrightarrow{u} \cdot \overrightarrow{u}=ac+bd|

Definition of the dot product

|\overrightarrow{u} \cdot \overrightarrow{u} =r^{2}+s^{2}|

Substituting in the correct components

|\mid \mid \overrightarrow{u}\mid \mid ^{2}=(\sqrt{a^{2}+b^{2}})^{2}|

By definition

|\mid \mid \overrightarrow{u}\mid \mid ^{2}=(\sqrt{r^{2}+s^{2}})^{2}|

Substituting in the correct components

|\mid \mid \overrightarrow{u}\mid \mid ^{2}=r^{2}+s^{2}|

By calculation

The right sides of statements 3 and 6 are the same.

Observation

|\overrightarrow{u} \cdot \overrightarrow{u}=\mid \mid \overrightarrow{u}\mid \mid ^{2}|

By the transitive property
Title (level 3)
Show that the diagonals of a rhombus are perpendicular
Title slug (identifier)
diagonals
Corps

Hypothesis: |ABCD| is a rhombus.
Conclusion: |\overrightarrow{AC}\perp\overrightarrow{BD}|
So, we need to show that |\overrightarrow{AC}\cdot\overrightarrow{BD}=0| (dot product).
​

Affirmations

Justifications

|\overrightarrow{AC}\cdot\overrightarrow{BD}=(\overrightarrow{AB}+\overrightarrow{BC})\cdot(\overrightarrow{BA}+\overrightarrow{AD})|

By Chasles’ relation

|\overrightarrow{AC}\cdot\overrightarrow{BD}=(\overrightarrow{AB}+\overrightarrow{AD})\cdot(\overrightarrow{BA}+\overrightarrow{AD})|

These are congruent sides of the rhombus

|\overrightarrow{AC}\cdot\overrightarrow{BD}=(\overrightarrow{AD}+\overrightarrow{AB})\cdot(\overrightarrow{AD}+\overrightarrow{BA})|

By the commutative property

|\overrightarrow{AC}\cdot\overrightarrow{BD}=(\overrightarrow{AD}+\overrightarrow{AB})\cdot(\overrightarrow{AD}-\overrightarrow{AB})|

Opposite vector

|\overrightarrow{AC}\cdot\overrightarrow{BD}=\overrightarrow{AD}^2-\overrightarrow{AB}^2|

Difference of squares

|\overrightarrow{AC}\cdot\overrightarrow{BD}=\mid \mid \overrightarrow{AD}\mid \mid ^{2}-\mid \mid \overrightarrow{AB}\mid \mid ^{2}|

|\overrightarrow{u}^{2}=\mid \mid \overrightarrow{u}\mid \mid ^{2}|

|\overrightarrow{AC}\cdot\overrightarrow{BD}=\mid \mid \overrightarrow{AB}\mid \mid ^{2}-\mid \mid \overrightarrow{AB}\mid \mid ^{2}|

Congruent sides of the rhombus

|\overrightarrow{AC}\cdot\overrightarrow{BD}=0|

By subtraction
Title (level 3)
If |ABCD| is a parallelogram, then the diagonals intersect at their midpoints.
Title slug (identifier)
formula-3
Image
Parallelogram
Corps

Hypotheses: |ABCD| is a parallelogram and |M| is the midpoint of the diagonal |AC|.
Conclusion: |M| is the midpoint of the diagonal |DB|.

So, we need to show that |\overrightarrow{MB}=\overrightarrow{DM}|.
​​​

Statement

Justification

|\overrightarrow{DM}=\overrightarrow{DA}+\overrightarrow{AM}|

Chasles’ relation

|\overrightarrow{DM}=\overrightarrow{CB}+\overrightarrow{AM}|

Opposite sides of a parallelogram

|\overrightarrow{DM}=\overrightarrow{CB}+\overrightarrow{MC}|

By the hypothesis

|\overrightarrow{DM}=\overrightarrow{MC}+\overrightarrow{CB}|

Commutative property of addition

|\overrightarrow{DM}=\overrightarrow{MB}|

Chasles’ relation
Title (level 3)
Show that the line segment that connects the midpoints of two sides of a triangle is parallel to the 3rd side and equal to half of the 3rd side’s length.
Title slug (identifier)
line-segment
Image
Segment
Corps

Hypotheses: |ABC| is a triangle and |\overrightarrow{AM}=\overrightarrow{MB}|, so |\overrightarrow{BN}=\overrightarrow{NC}|.
Conclusion: |\displaystyle \overrightarrow{MN}=\frac{1}{2} \overrightarrow{AC}|​​​

Statement

Justification

|\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}|

Chasles’ relation

|\overrightarrow{AC}=(\overrightarrow{AM}+\overrightarrow{MB})+(\overrightarrow{BN}+\overrightarrow{NC})|

Chasles’ relation

|\overrightarrow{AC}=\overrightarrow{MB}+\overrightarrow{MB}+\overrightarrow{BN}+\overrightarrow{BN}|

By the hypothesis

|\overrightarrow{AC}=2\overrightarrow{MB}+2\overrightarrow{BN}|

Vector addition

|\overrightarrow{AC}=2(\overrightarrow{MB}+\overrightarrow{BN})|

Common factor

|\overrightarrow{AC}=2\overrightarrow{MN}|

Chasles’ relation

|\displaystyle \frac{1}{2} \overrightarrow{AC} = \overrightarrow{MN}|

Divide by 2
Contenu
Corps

Pour valider ta compréhension à propos des démonstrations de façon interactive, consulte la MiniRécup suivante :

MiniRécup

Pour valider ta compréhension à propos des vecteurs de façon interactive, consulte la MiniRécup suivante :

MiniRécup
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