Vectors and translation: corrected math exercises in 2nd grade.

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The answer key to the math exercises on vectors in 2nd grade. Know how to use the Chasles relation and demonstrate that vectors are collinear in second grade.

Exercise 1:

Are the points P, Q and R aligned?

Yes they are aligned, show that the vectors $\vec{RP}$ and $\vec{RQ}$ are collinear .

Exercise 2:
ABCD is a parallelogram.
I is the middle of [AB].

E is the point such that $\vec{DE}=\frac{2}{3}\vec{DI}$

1. Make the following figure.

2. Determine the coordinates of the points of the figure
in the marker $(A;\vec{AB};\vec{AD})$ .

3. Are the points A, E and C aligned?

Yes they are aligned, show that the vectors $\vec{AE}$ and $\vec{AC}$ are collinear.

Exercise 3:

A and B are two distinct points of the plane.

We define the point M by the following vector relation:

$3\vec{MA}+\vec{MB}=\vec{0}$ .

1. Express $\vec{AM}$ as a function of $\vec{AB}$ .

Let’s use the Chasles relation:

$3\vec{MA}+\vec{MA}+\vec{AB}=\vec{0}$

$4\vec{MA}+\vec{AB}=\vec{0}$

$4\vec{AM}=\vec{AB}$

${\color{DarkRed}\,\vec{AM}=\frac{1}{4}\vec{AB}}$

2. Place the point M .

Exercise 4:

ABCD is a parallelogram of center O. The points M, N, P and Q are such that :

$\vec{AM}=\frac{3}{2}\vec{AB}\,;\,\vec{BN}=\frac{3}{2}\vec{BC}\,;\,\vec{CP}=\frac{3}{2}\vec{CD}\,;\,\vec{DQ}=\frac{3}{2}\vec{DA}$

a. Show that $\vec{MB}=\vec{DP}$ .

We have:

$\vec{AM}=\frac{3}{2}\vec{AB}$

$\vec{AB}+\vec{BM}=\frac{3}{2}\vec{AB}$

$\vec{BM}=\frac{3}{2}\vec{AB}-\vec{AB}$

$\vec{BM}=\frac{1}{2}\vec{AB}$

likewise :

$\vec{CP}=\frac{3}{2}\vec{CD}$

$\vec{CD}+\vec{DP}=\frac{3}{2}\vec{CD}$

$\vec{DP}=\frac{3}{2}\vec{CD}-\vec{CD}$

$\vec{DP}=\frac{1}{2}\vec{CD}$

Now ABCD is a parallelogram so $\vec{AB}=\vec{DC}$

so $\vec{BM}=\frac{1}{2}\vec{DC}$

$\vec{BM}=\frac{1}{2}\vec{DC}=\vec{PD}$

thus:

$\vec{MB}=\vec{DP}$

We deduce that the quadrilateral MBPD is a parallelogram.

b. Deduce that O is the midpoint of [MP].

Property : the diagonals of a parallelogram intersect in their middle.

Conclusion:

The point O is the middle of [MP].

Exercise 5:
ABCD is a parallelogram of center O.
Give the set of possible vector relations on this figure.

$\vec{DC}=\vec{AB}$

$\vec{AD}=\vec{BC}$

there are others on the identity of the parallelogram
that you will have the opportunity to meet in other exercises of the site.

Exercise 6:

(O,I,J) is an orthonormal reference frame with OI=OJ=1 cm.
a. Place the points.

b. We have:
$\vec{OA}=-4\vec{OI}+6\vec{OJ}=-2\vec{OC}+2\vec{OD}$
because $\vec{OC}=2\vec{OI}\,,\,\vec{OD}=3\vec{OJ}$
The coordinates of A(-2;2) in (O,C,D)
similarly, show that the coordinates are B(-1;-1) in (O,C,D).
In the (O,D,C) frame of reference, simply invert the abscissa and ordinate.

c. THE point O has coordinate O(1;1) in the frame (E,C,D).

Exercise 7:

The figure below represents regular hexagons of centers a, b, c, d.

1. Determine the images of each of the points C,E,A,M by the translation of vector :
a. The translation of vector $\vec{AB}$ sends C to L, E to D, A to B and M to N.
b. The translation of vector $\vec{BC}$ sends C to d, A to a and M to c.
c. The translation of vector $\vec{AC}$ sends C to K, E to W, A to C and M to I.

2. Use vector translation $\vec{AC}$

Exercise 8:

Prove that for all points A, B, C, D.

$\vec{AD}+\vec{BC}=\vec{AC}+\vec{CD}+\vec{BC}\\=\vec{AC}+\vec{BD}$ .
According to the relation of Chasles.

Exercise 9:

In a reference frame, consider the points A(-5;3), B(2;-1), C(0;4).

a. Place the points A,B,C.
b. $\vec{AB}(2-(-5);-1-3)=(7;-4)\\\vec{AC}(0-(-5);4-3)=(5;1),\vec{BC}=(0-2;4-(-1))=(-2;5).$ .

e. $AB=\sqrt{7^2+4^2}=\sqrt{49+16}=\sqrt{65}$

Exercise 11:

Exercise 12:

Analysis of the figure and preliminary remarks.

(ME) is the line of the middles so (ME) and (BC) are parallel and 2.ME = BC. We deduce that ME = BF = FC

(AF) is the axis of symmetry of the isosceles triangle BAC so (AF) is perpendicular to [BC] and [ME]

In the triangle AMC, [ME] is a median. The center of gravity G of the triangle AMC is at [ME] and we have : MG = 2.GE.

Demonstration

Consider the MECF quadrilateral. Opposite sides [ME] and[CF] are parallel and isometic. MECF is therefore a parallelogram; the diagonals [MC] and [EF] have the same middle I.

In the triangle MEF, [MI] is a median, another median is carried by the line (AF), it follows that the point D , intersection of two medians is the center of gravity of the triangle MEF. Thus we have : MD = 2.DI

Consider then the triangle MEI, on the one hand the points M , G , E are aligned in this order, as well as the points M , D , I, and on the other hand we have the relations :

MD = 2.MI and MG = 2.GE

It follows from the reciprocal of Thales’ theorem that the lines (GD) and (IE) are parallel.

Since F is the middle of [BC] and E is the middle of [AC] then [EF] is parallel to [AB]. Thus the line (MO) bisector of [AB] is perpendicular to [GD](because (MO) is perpendicular to [EF] and therefore to [EI] and therefore to [GD])

Finally, consider the triangle MGD

(DO) is perpendicular to [ME] and (MO) is perpendicular to (GD), it follows that O appears as the orthocenter of the triangle MGD. The third height is carried by the right (OG).

Conclusion:

The lines (OG) and (MC) are perpendicular.

Exercise 13:

Let us designate by A’ the point diametrically opposite to the point A. It follows that the two angles $\hat{ABA'}\,\,et\;\,\hat{ACA'}$ are right angles (angles inscribed in a semicircle)

Since the lines (BH) and (A’C) are perpendicular to the line (AC) then they are parallel. The same is true of the lines (BA’) and (CH) which are perpendicular to the line (AB).

Thus the quadrilateral BHCA’ is a parallelogram since its opposite sides are parallel two by two.

We know that the diagonals of a parallelogram intersect in their middle, so [HA’] $\cap$ [BC] = {M}. Thus the middle M of [BC] also belongs to the line (A’H).

Consequences:

The lines (A’H) and (MH) are merged. The points A’ , M , H , I are therefore aligned. It follows that the angle $\hat{A'IA}$ is right since it is inscribed in a semi-circle.

The lines (MH) and (AI) are therefore perpendicular

Exercise 14:

In a reference frame $(O;\vec{i};\vec{j})$ , we give K ( – 3 ; 5) and L(4 ; 2).
Determine the abscissa of the point M of ordinate – 2 such that K, L and M are aligned.

For the points to be aligned, the vectors $\vec{KL}$ and $\vec{LM}$ must be collinear with .

Let’s calculate the coordinates of these two vectors:

$\vec{KL}(4+3;2-5)$

$\vec{KL}(7;-3)$

and

$\vec{LM}(x-4;-2-2)$

$\vec{LM}(x-4;-4)$

These two vectors are collinear if and only if :

$7\times \,(-4)-(-3)(x-4)=0$

$x=\frac{40}{3}$

Conclusion: $M(\frac{40}{3};-2).$

Exercise 15:

In a reference frame $(O;\vec{i},\vec{j})$ , we give A(2 ; – 3) B(0 ; – 3) C( – 3 ; 0).

1. Determine by calculation the coordinates of the point E such that $\vec{CE}=\frac{1}{2}\vec{AB}$ .

Let’s solve a system:

$\,(\,x+3\\y-0\,\,)=\,(\,\frac{0-2}{2}\\\frac{-3+3}{2}\,\,)$

$\,(\,x+3\\y-0\,\,)=\,(\,-1\\0\,\,)$

We obtain the two equations :

$x+3=-1\,et\,y-0=0$

$x=-4\,et\,y=0$

Conclusion : the coordinates of point E are E( – 4 ; 0 ) .

2. What can we say about the lines (CE) and (AB) ? Justify.

The vectors $\vec{CE}$ and $\vec{AB}$ being collinear, the lines (CE) and (AB) are parallel.

Exercise 16:

In a reference frame $(O;\vec{i},\vec{j})$ , we give :

E(3; – 1) F(7; – 7) G(5; – 4).

Determine if the three points E, F and G are aligned.

Let’s check if the vectors $\vec{EF}$ and $\vec{EG}$ are collinear then the three points will be aligned.

$\vec{EF}(7-3;-7+1)$ so $\vec{EF}(4;-6)$

$\vec{EG}(5-3;-4+1)$ so $\vec{EG}(2;-3)$

We note that $\vec{EF}=2\vec{EG}$ so the vectors are collinear

and the points E, F and G are aligned.

Exercise 17:

1. Are the vectors $\vec{u}(1+\sqrt{3};4)$ and $\vec{v}(\frac{1}{2};\sqrt{3}-1)$ collinear ?

Let’s calculate the determinant:

$(1+\sqrt{3})(\sqrt{3}-1)-4\times \,\frac{1}{2}=(\sqrt{3}^2-1^2)-2=2-2=0$

Conclusion: these two vectors are collinear.

2. Determine such that the vectors $\vec{u}(2;m)$ and $\vec{v}(5;-1)$ are collinear.

The determinant must be zero:

$2\times \,(-1)-5m=0$

${\color{DarkRed}\,m=-\frac{2}{5}}$

Exercise 18:

Let A, B, C and D be any four points in the plane.

Show that :

$3\vec{DA}-\vec{DB}-2\vec{DC}=3\vec{BA}-2\vec{BC}$

We have

$3\vec{DA}-\vec{DB}-2\vec{DC}$

$=3\vec{DB}+3\vec{BA}-\vec{DB}-2\vec{DB}-2\vec{BC}$

$=3\vec{DB}-\vec{DB}-2\vec{DB}+3\vec{BA}-2\vec{BC}$

$=\vec{0}+3\vec{BA}-2\vec{BC}$

${\color{DarkRed}\,=3\vec{BA}-2\vec{BC}}$

Exercise 19:

$\vec{BA}+\vec{CB}+\vec{DC}=\vec{CA}+\vec{DB}-\vec{CD}$

$\vec{BA}+\vec{CB}+\vec{DC}+\vec{CD}=\vec{CA}+\vec{DB}$

$\vec{BA}+\vec{CB}+\vec{0}=\vec{CA}+\vec{DB}$

$\vec{CA}=\vec{CA}+\vec{DB}$

$\vec{CA}-\vec{CA}=\vec{DB}$

$\vec{0}=\vec{DB}$

Conclusion: points B and D are merged.

Exercise 20:
A and B are two distinct points.

We try to construct the point M such that :

$3\vec{MA}+4\vec{MB}=\vec{0}$

1. Are the vectors $\vec{MA}$ and $\vec{MB}$ collinear? Do they have the same direction?

$\vec{MA}=-\frac{4}{3}\vec{MB}$
$MA=\frac{4}{3}MB$

Conclusion: these two vectors are collinear in opposite directions and do not have the same norm.

2. Using the Chasles relation, show that we have the equality :

$7\vec{MA}+4\vec{AB}=\vec{0}$

We know that:

$3\vec{MA}+4\vec{MB}=\vec{0}$

$3\vec{MA}+4\vec{MA}+4\vec{AB}=\vec{0}$

$7\vec{MA}+4\vec{AB}=\vec{0}$

3. Deduce $\vec{AM}$ as a function of $\vec{AB}$ .

$7\vec{AM}=4\vec{AB}$

$\vec{AM}=\frac{4}{7}\vec{AB}$

Construct the point M.

Exercise 21:

Are the vectors $\vec{u}(\sqrt{2};1-\sqrt{3})$ and $\vec{v}(1+\sqrt{3};-\sqrt{2})$ collinear?

Let’s calculate the determinant:

$\sqrt{2}\times \,(-\sqrt{2})-(1-\sqrt{3})(1+\sqrt{3})$

$=-2-(1^2-\sqrt{3}^2)$

Conclusion: these two vectors are indeed collinear.

Exercise 22:

Consider a triangle ABC and the points I and J such that :

$\vec{AI}=\frac{1}{3}\vec{AB}$

$\vec{AJ}=3\vec{AC}$

1. Show using the Chasles relation that $\vec{BJ}=3\vec{IC}$ .

$\vec{BJ}=\vec{BA}+\vec{AJ}=-3\vec{AI}+3\vec{AC}=3\vec{IA}+3\vec{AC}=3\vec{IC}$

2. What can we deduce for the lines (BJ) and (IC)?

the vectors $\vec{BJ}$ and $\vec{IC}$ are thus collinear and we deduce that the lines (BJ) and (IC) are parallel.

Exercise 23:

In each of the following cases, show that the vectors $\vec{AB}$ and $\vec{CD}$ are collinear.

1. $\vec{AC}+\vec{DC}=\vec{BD}$ .

Let’s use the Chasles relation:

$\vec{AB}+\vec{BC}+\vec{DC}=\vec{BC}+\vec{CD}$

$\vec{AB}+\vec{DC}=\vec{CD}$

$\vec{AB}=-\vec{DC}+\vec{CD}$

$\vec{AB}=2\vec{CD}$

2. $2\vec{CB}-9\vec{CA}-7\vec{AD}=\vec{0}$

Let’s use the Chasles relation

$2\vec{CB}-2\vec{CA}-7\vec{CA}-7\vec{AD}=\vec{0}$

$2\vec{CB}+2\vec{AC}-7\vec{CD}=\vec{0}$

$2\vec{AB}-7\vec{CD}=\vec{0}$

$2\vec{AB}=7\vec{CD}$

$\vec{AB}=\frac{7}{2}\vec{CD}$

Exercise 24:

$\vec{AC}+\vec{AD}-\vec{BC}=\vec{AB}$ .

Let’s use the Chasles relation:

$\vec{AB}+\vec{BC}+\vec{AD}-\vec{BC}=\vec{AB}$

$\vec{AB}+\vec{AD}=\vec{AB}$

$\vec{AD}=\vec{AB}-\vec{AB}$

$\vec{AD}=\vec{0}$

Conclusion: points A and D are merged.

Exercise 25:

1. Place the point E such that $\vec{BE}=\vec{AC}$ .

2. Place the point F such that $\vec{BF}=-\vec{AC}$ .

3. Place the point G such that $\vec{BG}=\vec{AC}+\vec{BA}$ .

Exercise 30:

In the plane provided with an orthonormal reference frame, we note E the set of points whose coordinates (x;y) verify the relation :
$\frac{x^2}{25}+\frac{y^2}{9}=1$ .

Consider also the points F(4;0) and F'(-4;0).

1. Calculate the coordinates of the intersection points of E with the axes of the benchmark.

When x=0 , y=3 and y= -3.

When y=0, x=5 and x= – 5 .

2. Using the geogebra software, visualize the set E and make a conjecture about the sum of the distances MF + MF’ when M is a point of E.

The distance MF+MF’ is constant.

3. Let M(x;y) be a point of E.
a) Express as a function of and deduce that $x^2\leq\,\,25$ .

$y^2=9(1-\frac{x^2}{25})$

or which is equivalent to saying that $1-\frac{x^2}{25}\geq\,\,0$ which is equivalent to $x^2\leq\,\,25$

b) Show that $MF^2=(\frac{4}{5}x-5)^2$ .

c) Knowing that $x\leq\,\,5$ , show that $\frac{4}{5}x-5\leq\,\,0$

then deduce that $MF=5-\frac{4}{5}x$ .

d) Validate the conjecture.

Exercise 31:

Let ABCD be a parallelogram.

1) Place the points M and N defined by the following equalities:

$\vec{AM}=\vec{AD}+\frac{2}{5}\times \,\vec{DB}$

$\vec{CN}=-\vec{CB}-\frac{1}{3}\times \,\vec{BA}$

2) Show using the chasles relation that $\vec{DN}=-\vec{CB}-\frac{2}{3}\times \,\vec{BA}$ .

3) Express the vector $\vec{DN}$ in terms of the vectors $\vec{AD}$ and $\vec{DB}$ .

<strong”> Indications:

Make a figure and use the Chasles relationship.

Exercise 32:

Exercise 33:

The answer key to the exercises on vectors in 2nd grade.

After having consulted the answers to these exercises on vectors in 2de, you can return to the exercises in seconde.

Second grade exercises.

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Vectors and translation : corrected 2nd grade math exercises in PDF.

The answer key to the exercises on vectors in 2nd grade.

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