Scalar product: answer key to 1st grade math exercises in PDF.

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The answer key to the math exercises in 1st grade in PDF on the scalar product in the plane. Use the bilinearity properties of the scalar product and demonstrate that two vectors are orthogonal or collinear. Apply the Chasles relation to vectors in the first grade.

Exercise 1:

a. AB=3 , AC=5 and $\,(\vec{AB}.\vec{AC})=-\frac{\pi}{6}+2k\pi$ .
$\,\vec{AB}.\vec{AC}=AB\times \,AC\,\times \,cos(-\frac{\pi}{6})=15\times \,\frac{\sqrt{3}}{2}.$

b. AB=1 , AC=4 and $(\vec{AB}.\vec{AC})=-\frac{8\pi}{3}+2k\pi$ .
$\,\vec{AB}.\vec{AC}=AB\times \,AC\,\times \,cos(\frac{8\pi}{6})=4\times \,\frac{-1}{2}=-2.$

c. AB=4 , AC=7 and $\,(\vec{AB}.\vec{AC})=-\frac{\pi}{4}+2k\pi$ .
$\,\vec{AB}.\vec{AC}=AB\times \,AC\,\times \,cos(\frac{-\pi}{4})=28\times \,\frac{\sqrt{2}}{2}=14\sqrt{2}.$

d. AB=2 , AC=2 and $\,(\vec{AB}.\vec{AC})=-\frac{5\pi}{3}+2k\pi$ .
$\,\vec{AB}.\vec{AC}=AB\times \,AC\,\times \,cos(\frac{5\pi}{3})=4\times \,\frac{-1}{2}=-2.$

Exercise 2:

Calculate $\vec{AC}.\vec{AB}\,;\,\vec{CA}.\vec{BA}\,;\,\vec{BA}.\vec{AC}\,\,;$ knowing that :

a. $\vec{AB}.\vec{AC}=-3$
$\vec{AC}.\vec{AB}=\vec{AB}.\vec{AC}=-3\,;\,\vec{CA}.\vec{BA}=\vec{AB}.\vec{AC}=-3\,;\\\,\vec{BA}.\vec{AC}=-\vec{AB}.\vec{AC}=3\,\,;$

Exercise 4:

Let ABCD be a square and I a point of [AB].
Let H be the orthogonal project of A onto [ID].

$\,\vec{IA}.\vec{ID}=\vec{IA}.(\vec{IA}+\vec{AD})=\vec{IA}.\vec{IA}+\vec{IA}.\vec{AD}\\=AI^2$
$\,\vec{IA}.\vec{AD}=0$ because (IA) is perpendicular to (AD).

Exercise 5:

Let ABC be an equilateral triangle of side 1.
Let H be the orthogonal project of A onto (BC).

$\,\vec{BA}.\vec{AC}=1\times 1\times \,\frac{\sqrt{3}}{2}$
and $\,\vec{AB}.\vec{AH}=1\times \,\frac{\sqrt{3}}{2}\times \,\frac{1}{2}=\frac{\sqrt{3}}{4}$ .

Exercise 6:

The purpose of the problem is to show that the lines (CQ) and (PR) are perpendicular.

1. Justify that: $\vec{CQ}.\vec{PR}=\vec{CQ}.(\vec{AR}-\vec{AP})$ .

$\vec{CQ}.\vec{PR}=\vec{CQ}.(\vec{PA}+\vec{AR})$ ( according to the relation of Chasles)

$\vec{CQ}.\vec{PR}=\vec{CQ}.(-\vec{AP}+\vec{AR})$

$\vec{CQ}.\vec{PR}=\vec{CQ}.(\vec{AR}-\vec{AP})$

2. Deduce that the lines (CQ) and (PR) are perpendicular.

Indication:

Show that $\vec{CQ}.\vec{PR}=0$

Exercise 7:

We have $\,\%7C\,\vec{u}\,\,\%7C=2$ and $\,\%7C\,\vec{v}\,\,\%7C=3$ and $\vec{u}$ . $\vec{v}$ = -1

1) Calculate $(\vec{u}+\vec{v})^2$ and $\,\%7C\,(\vec{u}\,-\vec{v})^2\,\%7C$

$(\vec{u}+\vec{v})^2=\,\%7C\,\vec{u}\,\,\%7C^2+\,\%7C\,\vec{v}\,\,\%7C^2+2\vec{u}.\vec{v}=4+9-2=11$

2) Calculate ( $\vec{u}$ + $\vec{v}$ ) . (2 $\vec{u}$ -3 $\vec{v}$ )

$(\vec{u}+\vec{v}).(2\vec{u}-3\vec{v})=2\vec{u}.\vec{u}-3\vec{u}.\vec{v}+2\vec{v}.\vec{u}-3\vec{v}.\vec{v}$

$\\=2\times \,4-3\times \,(-1)+2\times \,(-1)-3\times \,3^2$

$\\=8+3-2-27$

$\\=-18$

Exercise 8:

Prove that whatever the point M of the plane, we have the equality :

$MA^2-MB^2=(\vec{MA}+\vec{MB}).\vec{BA}=2\vec{MI}.\vec{BA}$

$MA^2-MB^2=(\vec{MA}+\vec{MB}).(\vec{MA}-\vec{MB})$

$MA^2-MB^2=(\vec{MA}+\vec{MB}).(\vec{MA}+\vec{BM})$

$MA^2-MB^2=(\vec{MA}+\vec{MB}).(\vec{BM}+\vec{MA})$

${\color{DarkRed}\,MA^2-MB^2=(\vec{MA}+\vec{MB}).\vec{BA}}$

$MA^2-MB^2=(\vec{MA}+\vec{MB}).\vec{BA}$

$MA^2-MB^2=(\vec{MI}+\vec{IA}+\vec{MI}+\vec{IB}).\vec{BA}$

$MA^2-MB^2=(2\vec{MI}+\vec{IA}+\vec{IB}).\vec{BA}$

and I is the middle of [AB] so $\vec{IA}+\vec{IB}=\vec{0}$

$MA^2-MB^2=(2\vec{MI}+\vec{IA}+\vec{IB}).\vec{BA}$

$MA^2-MB^2=(2\vec{MI}+\vec{0}).\vec{BA}$

$MA^2-MB^2=(2\vec{MI}).\vec{BA}$

${\color{DarkBlue}\,MA^2-MB^2=2\vec{MI}.\vec{BA}}$

Exercise 10:

Exercise 11:

Hint: create a marker in the parallelogram.
ABCD is a parallelogram with AB = 4, AD = 5 and AC = 7.

1.Calculate $\vec{AB}.\vec{AD}$ .

2. Deduct BD.

Exercise 12:

MNPQ is a square with MN = 6. I is the center of the square.

Calculate the following scalar products:

1. $\vec{MN}.\vec{QP}=MN\times \,QP\times \,cos(\vec{MN},\vec{QP})\\=6\times \,6\times \,cos(0)=36$

2. $\vec{MN}.\vec{PN}=MN\times \,PN\times \,cos(\vec{MN},\vec{PN})=6\times \,6\times \,cos(90)=0$

3. $\vec{IN}.\vec{IP}=IN\times \,IP\times \,cos(\vec{IN},\vec{IP})=3\sqrt{2}\times 3\sqrt{2}\,\times \,cos(90)=0$

4. $\vec{QI}.\vec{NI}=QI\times \,NI\times \,cos(\vec{QI},\vec{NI})=3\sqrt{2}\times \,3\sqrt{2}\times \,cos(180)=-18$

Exercise 13:
ABC is a triangle in which AB = 2 and AC = 3.

In addition $\vec{AB}.\vec{AC}=4$

Is this triangle rectangular? If yes, specify in which summit.

$\vec{AB}.\vec{AC}=AB\times \,AC\times \,cos(\vec{AB},\vec{AC})$

$cos(\vec{AB},\vec{AC})=\frac{\vec{AB}.\vec{AC}}{AB\times \,AC}=\frac{4}{2\times \,3}$

$cos(\vec{AB},\vec{AC})=\frac{2}{3}$

Conclusion: the triangle is not rectangular because the cosine is different from zero.

Exercise 14:

ABC is an equilateral triangle of side 5 cm. I is the middle of [BC].
Calculate the following scalar products:

1. $\vec{BA}.\vec{BC}$ .

$\vec{BA}.\vec{BC}\\=BA\times \,BC\times \,cos(\vec{BA},\vec{BC})\,\\=25\times \,cos60^{\circ}=\frac{25}{2}$

2. $\vec{CA}.\vec{CI}.$

$\vec{CA}.\vec{CI}.$
$\vec{CA}.\vec{CI}=CA\times \,CI\times \,cos(\vec{CA},\vec{CI})\\=5\times \,2,5\times \,\frac{1}{2}=6,25$

3. $(\vec{AB}-\vec{AC}).\vec{AI}.$

$(\vec{AB}-\vec{AC}).\vec{AI}=\vec{CB}.\vec{AI}=0$
(Ai) is a median but as the triangle is equilateral, it is also a height
so these two vectors are orthogonal and therefore their scalar product is zero.

Exercise 15:

In an orthonormal reference frame $(O;\vec{i},\vec{j})$
we consider the following points: A (2; 1), B (7; 2) and C (3; 4).

All of the following questions are independent and unrelated.

1. Compute the coordinates of the barycenter G of (A; 3), (B; 2) and (C; – 4).

First of all abrod the barycenter exists because $3+2-4=1\neq\,0$

$3\vec{GA}+2\vec{GB}-4\vec{GC}=\vec{0}$

$\,\{\,3(x_A-x_G)+2(x_G-x_B)-4(x_C-x_G)=0\,\\3(y_A-y_G)+2(y_B-y_G)-4(y_C-y_G)=0\,=0$

$\,\{\,3(2-x_G)+2(7-x_G)-4(3-x_G)=0\,\\3(1-y_G)+2(2-y_G)-4(4-y_G)=0$

$\,\{\,6-3x_G+14-2x_G-12+4x_G=0\,\\3-3y_G+4-2y_G-16+4y_G=0$

$\,\{\,-x_G+8=0\,\\-y_G+9=0$

$\,\{\,x_G=8\,\\y_G=9$

2. Determine a Cartesian equation of the perpendicular bisector of [BC].

a direction vector of [BC] is $\vec{u}\,(3-7\\4-2\,\,)$

$\vec{u}\,(-4\\2\,\,)$

$\vec{u}\,(-2\\1\,\,)$

so a normal vector of the median is :

$\vec{u}\,(-2\\1\,\,)$

so a Cartesian equation is of the type :

and the midpoint I of [BC] belongs to the perpendicular bisector : $I\,(\,\frac{7+3}{2}\,\\\frac{2+4}{2}\,)$ so $I\,(\,5\,\\3\,)$

We obtain:

$-2\times \,5+3+c=0$

Conclusion: a Cartesian equation of the median is .

3. Calculate $\vec{CB}.\vec{CA}$ .

$\vec{CB}.\vec{CA}=\,(\,7-3\\2-4\,\,).\,(\,2-3\\1-4\,\,)=\,(\,4\\-2\,\,).\,(\,-1\\-3\,\,)=4\times \,(-1)+(-2)\times \,(-3)=-4+6=2$

4. Is the angle $\widehat{C}$ right?

No since the scalar product is not zero.

Exercise 16:

Exercise 17:

Exercise 18:

Exercise 19:

Knowing that the vectors $\vec{u}$ and $\vec{v}$ are such that $\,\%7C\,\vec{u}\,\,\%7C=3$ , $\,\%7C\,\vec{v}\,\,\%7C=7$ and $\vec{u}.\vec{v}\,=13$ .

Calculate the following scalar products:

1. $\vec{u}.\,(\vec{u}+3\vec{v}\,\,)$

$\vec{u}.\,(\vec{u}+3\vec{v}\,\,)\,\\=\vec{u}.\vec{u}+3\vec{u}.\vec{v}\\=\,\%7C\,\vec{u}\,\,\%7C^2+3\vec{u}.\vec{v}\\=9+39=48$ $\,(\vec{u}-2\vec{v}\,\,)\,^2=\,\%7C\,\vec{u}\,\,\%7C^2-4\vec{u}.\vec{v}+4\,\%7C\,\vec{v}\,\,\%7C^2\\=9-52+196\\=153$

2. $\,(\vec{u}-2\vec{v}\,\,)\,^2$

$\,(\vec{u}-2\vec{v}\,\,)\,^2=\,\%7C\,\vec{u}\,\,\%7C^2-4\vec{u}.\vec{v}+4\,\%7C\,\vec{v}\,\,\%7C^2\\=9-52+28\\=-15$

Exercise 20:

Under which condition on points A, B and C do we have :

$(\vec{AB}+\vec{AC})^2=(AB+AC)^2$

We have:

$(\vec{AB}+\vec{AC})^2=AB^2+AC^2+2\vec{AB}.\vec{AC}=AB^2+AC^2+2AB\times \,AC\times \,cos(\vec{AB};\vec{AC})$
and
$(AB+AC)^2=AB^2+AC^3+2\times \,AB\times \,AC$

Therefore, it is necessary that :

$cos(\vec{AB};\vec{AC})=1$
$(\vec{AB};\vec{AC})=2k\pi$

so
that the point A belongs to the line (BC) deprived of the segment [BC].

Exercise 21:
1. $\vec{MA}.\vec{MB}=1.$

Let I be the middle of [AB] and therefore the isobarycenter of [AB].
$\vec{MA}.\vec{MB}=1$

$\,(\vec{MI}+\vec{IA}\,\,).\,(\vec{MI}\,+\vec{IB}\,\,)=1$

using the properties of the scalar product :

$\vec{MI}.\vec{MI}+\vec{MI}.\vec{IB}+\vec{IA}.\vec{MI}+\vec{IA}.\vec{IB}=1$

$MI^2+\vec{MI}.\,(\vec{IA}+\vec{IB}\,\,)+\vec{IA}.\vec{IB}=1$

$MI^2+\vec{MI}.\vec{0}\,+\vec{IA}.\vec{IB}=1$

$MI^2+0\,+\vec{IA}.\vec{IB}=1$

$MI^2+IA\times \,IB\times \,cos\,\pi=1$

$MI^2-IA\times \,IB=1$

$MI^2=1+IA\times \,IB$ > 1

$MI^2=1+0,5\times \,0,5$

$MI=\sqrt{1,25}$

Conclusion: this is the circle with center I and radius $\sqrt{1,25}$ dm .

Exercise 22:

[AB] is a segment of middle I and AB = 2 cm.

1. Show that for any point M in the plane :

$MA^2-MB^2=2\vec{IM}.\vec{AB}$

We have:

$MA^2-MB^2=(\vec{MA}+\vec{MB}).(\vec{MA}-\vec{MB})$

$MA^2-MB^2=(\vec{MA}+\vec{MB}).(\vec{MA}+\vec{BM})$

$MA^2-MB^2=(\vec{MA}+\vec{MB}).\vec{BA}$

$MA^2-MB^2=(\vec{MI}+\vec{IA}+\vec{MI}+\vec{IB}).\vec{BA}$

$MA^2-MB^2=(2\vec{MI}+\vec{IA}+\vec{IB}).\vec{BA}$

$MA^2-MB^2=2\vec{MI}.\vec{BA}$

because I is in the middle of [AB] so $\vec{IA}+\vec{IB}=\vec{0}$

Exercise 23:

Prove that :

1. $\,\%7C\,\vec{u}+\vec{v}\,\,\%7C^2-\,\%7C\,\vec{u}-\vec{v}\,\,\%7C^2=4\vec{u}.\vec{v}$ .

${\color{DarkBlue}\,\,\%7C\,\vec{u}\,\,\%7C^2+2\vec{u}.\vec{v}+\,\%7C\,\vec{v}\,\,\%7C^2-\,\%7C\,\vec{u}\,\,\%7C^2+2\vec{u}.\vec{v}-\,\%7C\,\vec{v}\,\,\%7C^2=4\vec{u}.\vec{v}}$

2. $\,\%7C\,\vec{u}+\vec{v}\,\,\%7C^2+\,\%7C\,\vec{u}-\vec{v}\,\,\%7C^2=2(\,\%7C\vec{u}\,\,\%7C^2+\,\%7C\,\vec{v}\,\,\%7C^2)$ .

${\color{DarkBlue}\,\,\%7C\,\vec{u}\,\,\%7C^2+2\vec{u}.\vec{v}+\,\%7C\,\vec{v}\,\,\%7C^2+\,\%7C\,\vec{u}\,\,\%7C^2-2\vec{u}.\vec{v}+\,\%7C\,\vec{v}\,\,\%7C^2=2\,\%7C\,\vec{u}\,\,\%7C^2+2\,\%7C\,\vec{v}\,\,\%7C^2=2(\,\%7C\,\vec{u}\,\,\%7C^2+\,\%7C\,\vec{v}\,\,\%7C^2)}$

3. What is the link with the rhombus, the parallelogram?

Make a figure…

4. Show that :

$(\vec{u}+\vec{v}).(\vec{u}-\vec{v})=\,\%7C\,\vec{u}\,\,\%7C^2-\,\%7C\,\vec{v}\,\,\%7C^2$

$(\vec{u}+\vec{v}).(\vec{u}-\vec{v})=\vec{u}.\vec{u}-\vec{u}.\vec{v}+\vec{v}.\vec{u}-\vec{v}.\vec{v}$

$(\vec{u}+\vec{v}).(\vec{u}-\vec{v})=\vec{u}.\vec{u}-\vec{u}.\vec{v}+\vec{u}.\vec{v}-\vec{v}.\vec{v}$ (because the scalar product is symmetrical)

$(\vec{u}+\vec{v}).(\vec{u}-\vec{v})=\vec{u}.\vec{u}-\vec{v}.\vec{v}$

${\color{DarkBlue}\,(\vec{u}+\vec{v}).(\vec{u}-\vec{v})=\,\%7C\,\vec{u}\,\,\%7C^2-\,\%7C\,\vec{v}\,\,\%7C\,^2}$

5. Deduce that a parallelogram has its diagonals perpendicular if and only if its sides are equal.

Make a figure….

Exercise 24:

Indication:

The equation of a circle is

with $\Omega\,(x_0;y_0)$ center of the circle of radius R.
In an orthonormal reference frame $(O;\vec{i},\vec{j})$ , we give a point $\Omega\,(2;-3)$ .

1. Determine the equation of the circle (C) with center $\Omega$ and radius R = 5.

2. Prove that the point A( – 2 ; 0) is a point of the circle (C).

3. Determine a Cartesian equation of the tangent at A to the circle (C).

Exercise 26:

We place ourselves in an orthonormal reference frame $(O;\vec{i},\vec{j}).$
Consider a triangle ABC with A (-1; 2), B (3; 1) and C (2; 4).

1. Determine an equation of the perpendicular bisector of the segment [AB].

Let’s determine the coordinates of the vector $\vec{AB}$ .

$\vec{AB}\,(\,3+1\\1-2\,\,)=\,(\,4\\-1\,\,)$

Let’s determine the coordinates of I, the middle of the segment [AB]:

$I\,(\,\frac{-1+3}{2}\\\frac{2+1}{2}\,\,)=\,(\,1\\\frac{3}{2}\,\,)$

The coordinates of a director vector of the median is a normal vector to the vector $\vec{AB}$
so $\vec{u}\,(\,1\\4\,\,)$

Any point M(x,y) belongs to the bisector of [AB]

if and only if :

$\vec{IM}=k\vec{u}$

(k non-zero)

$\,(\,x-1\\y-\frac{3}{2}\,\,)\,=\,(\,k\\4k\,\,)$
so
$\,\{\,k=x-1\\y=4k+\frac{3}{2}\,.$

$\,\{\,k=x-1\\y=4\,(x-1\,\,)+\frac{3}{2}\,.$

$\,\{\,k=x-1\\y=4x-4+\frac{3}{2}\,.$

so the reduced equation of the perpendicular bisector of [AB] is :

${\color{DarkRed}\,y=4x-\frac{5}{2}}$

2. Determine an equation of the height from A in triangle ABC.

Let’s determine the coordinates of the vector $\vec{BC}$ :

$\vec{BC}\,\,(\,2-3\\4-1\,\,)=\,(\,-1\\3\,\,)$

Let H be the orthogonal project of A onto (BC).

The line (AH) is therefore the height from the vertex A.

Let M(x,y) belong to (AH)

if and only if :

$\vec{BC}.\vec{AM}=0$

$\,(\,-1\\3\,\,).\,(x+1\\y-2\,\,)=0$

$-x\,+3y-7=0$

$y=\frac{x+7}{3}$

${\color{DarkRed}\,y=\frac{1}{3}x+\frac{\,7}{3}}$ is a reduced equation of the height from A.

Exercise:

We place ourselves in an orthonormal reference frame $(O;\vec{i},\vec{j})$ .

1. Determine the equation of the circle with center $\Omega\,(5;1)$ tangent to the line (D) of equation :

$x\,+\,y\,-\,4\,=\,0.$

Indication:
we recall that the distance between a point $A(\alpha\,;\beta\,)$ and a line (D) of equation ax + by + c = 0 is
given by the formula :

$d(A,D)=\frac{\,%7C\,a\alpha\,+b\beta\,+c\,\,%7C}{\sqrt{a^2+b^2}}$

Let’s determine the distance between the tangent and the center of the circle, this will be the radius of the circle.

$d(\Omega\,,D)=\frac{\,%7C\,1\times \,5\,+1\times \,1\,-4\,\,%7C}{\sqrt{1^2+1^2}}$

$d(\Omega\,,D)=\frac{2}{\sqrt{2}}=\frac{2\sqrt{2}}{2}=\sqrt{2}$

The equation of the circle is :

$(x-5)^2+(y-1)^2=\sqrt{2}^2$

${\color{DarkRed}\,(x-5)^2+(y-1)^2=2}$

Exercise 27:
We place ourselves in an orthonormal reference frame $(O;\vec{i},\vec{j})$ .

Consider whether the following equations are equations of a circle and, if so, specify the center and radius of the circle.

1. $x^2\,+\,y^2\,-\,2x\,-\,6y\,+\,5\,=\,0.$

It is a circle with center point I ( 1 ; 3 ) and radius $\sqrt{5}$ .

2. $x^2\,+\,y^2\,-\,x\,-\,3y\,+\,3\,=\,0.$

$(x-\frac{1}{2})^2-\frac{1}{4}+(y-\frac{3}{2})^2+\frac{9}{4}+3=0$

$(x-\frac{1}{2})^2+(y-\frac{3}{2})^2+\frac{8}{4}+\frac{12}{4}=0$

$(x-\frac{1}{2})^2+(y-\frac{3}{2})^2=-5$

This is not the equation of a circle.

Exercise 28:

Hint: use the generalized Pythagorean formula in any triangle.
ABC is a triangle and I is the middle of [BC].
We give: BC = 4, AI = 3 and $(\vec{IA},\vec{IB})=\frac{\pi}{3}$ .

Calculate:

1. $\vec{AB}.\vec{AC}.$

4. $AB\,et\,AC.$

The answers to the exercises on the scalar product in the plane in 1ère.

After having consulted the answers to these exercises on the scalar product in the plane in 1st grade, you can return to the exercises in 1st grade.

Exercises in first grade.

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