Add and Subtract Vectors

Problem 1.

Given:

ABCD is a quadrilateral

M, N, K, E are the midpoints of the sides AB, BC, DC, AD

Prove:

The quadrilateral MNKE is a parallelogram

Proof:

Join a point A to a point C. We obtain the triangle ∆ ABC, where MN is the midline of the triangle ∆ ABC, and the triangle ∆ ADC, where EK is the middle line of the triangle ∆ ADC.

By the property of the midline of a triangle ∆ it follows that

MN || AC are parallel and MN= AC,

EK || AC are parallel and EK= AC.

Then MN || EK are parallel and MN=EK, therefore

MNKE is the parallelogram (by the first theorem of a parallelogram).

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Problem 2.

Given:

∆ ABC is a triangle

The length of the triangle side AB = 8.5 cm

The length of the triangle side AC = 5 cm

The height AH = 4 cm, i.e. the segment AH is perpendicular to the side BC

HBC, i.e. the point H lies on the side BC

Solve:

The area of the triangle = Area _∆ABC = ?

Solution:

Answer: Area _∆ABC = 21 cm²

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Problem 3.

Prove that the segments connecting the midpoints of the opposite sides of an isosceles trapezoid are perpendicular.

Given:

ABCD is an isosceles trapezoid

Prove: NE KM =

Proof:

Draw perpendiculars BH and CH₁, i.e. BHAD are perpendicular; also CH₁AD are perpendicular.

But BH and CH₁ cross NE then BRNE and CR₁NE are perpendicular.

The sides BH = CH₁ and BH || CH₁ are parallel.

Therefore BH = KM = CH₁ BH KM CH₁ are parallel as segments between parallel lines.

Therefore the angles are equal KON = NR₁C = 90º as corresponding angles.

Then KON = EOM = 90º, as vertical angles.

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Problem 4.

Given:

AB is a segment

AC = CB

O is an arbitrary point

Prove:

The OC vector is half of the sum of the other two vectors OA and OB going out of the same point O

Proof: By the rule of a triangle

(1)

+

(2)

Adding equalities (1) and (2), we get

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Problem 5.

Given:

a, b, c are vectors

Three vectors and are non-collinear vectors.

Build:

The sums and differences of vectors.

Construction:

By the rule of a polygon

a)

b)

=

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Problem 6.

Prove that the segments connecting the midpoints of the opposite sides of an isosceles trapezoid are perpendicular.

Given:

The quadrilateral ABCD is an isosceles trapezoid

Prove: EF NM = , i.e. the angle of intersection of two segments in an isosceles trapezoid is 90º.

Proof:

Draw parallel lines

MK || AB

MR || CD

We get an isosceles triangle ∆MKR

AB=MK, since the trapezoid is isosceles,

CD=MR, since the trapezoid is isosceles.

Therefore, EF is the midline of the triangle ∆MKR, hence

MH=HR and OK=MO.

BM=MC=AK=RD, since ABMK and MCDR are parallelograms.

Therefore HR=KO.

Then MN is the median, the bisector and the height of the isosceles triangle ∆MKR.

Since MN is the height, we see that the segments MN AD are perpendicular.

By the property of the midline of a triangle ∆ it follows that

EF || KR.

Then EF NM =

***

Problem 7.

Prove that the center of a circle inscribed in an isosceles triangle lies on the median drawn to the base.

Given:

an inscribed circle in an isosceles triangle

∆ABC is an isosceles triangle

BH₂ is a median

Prove: O BH₂, i.e. the center of the inscribed circle lies on the median of an isosceles triangle

Proof:

Draw perpendiculars OH₁ ; OH₂ ; OH₃ to the sides of BC, AC, AB.

Here, from the two points, the same perpendicular to the AC side is drawn, but in the triangle, only one perpendicular can be drawn to the side and only from one point.

Therefore that O BH₂

***

Problem 8.

Prove that the center of the circle circumscribed around an isosceles triangle lies on the median drawn to the base or to its continuation.

Given:

The circumcircle around an isosceles triangle

∆ ABC is an inscribed isosceles triangle

BH₃ is a median

Prove: O BH₃

Proof:

Draw perpendiculars from the center of the circle

OH₁ ; OH₂ ; OH₃ to the sides of BC, AC, AB.

Here, a perpendicular to the AC side is drawn from two points, but only one perpendicular can be drawn to the side and only from one point in the triangle.

Therefore, O BH₃

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Collinear vectors

The lemma is a theorem that is subsidiary for the proof of the following theorem.

The lemma about collinear vectors:

If the vectors and are collinear (where ), then we can find a such number k that the equality is true (the vector is equal to the product of the number k by the vector )

Given: vector a, vector b

The vectors and are collinear, i.e. the vector b is collinear with the vector a.

Prove: there is a such number k that the following equality is true

Proof:

Case 1. Let the vectors b and a be co-directional vectors, i.e.

, where k>0, since . Then and are the co-directional vectors.

Therefore

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Case 2.

Let a, b vectors be opposite vectors, i.e.

Let us take , where k<0

Consequently,

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Problem 9.

Given:

vector m, vector n

1) are oppositely directed vectors,

= 0.5 cm, = 2 cm

2) are co-directional vectors,

= 12 cm, = 240 cm

Solve: k – ?

Solution:

Answer: k = – 4.

Solution:

Answer: k = 20.

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Problem 10.

Given:

ABCD is a parallelogram

BD AC = O

M is the midpoint of the segment AO

1)

2)

Solve: k – ?

Solution:

Answer: k=

2) Since , we see that k<0. , are collinear, since they lie on one straight line. Find the midpoint of OC and call it as the point N.

We see that AM=MO=ON=NC

Since k<0, we see that

Answer: k=

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Problem 11.

Given:

1) are oppositely directed vectors,

= 400 mm, = 4dm = 400mm

2) are co-directional vectors, = , =

Solve: k – ?

Solution:

Answer: k = –1.

Solution:

Answer: k = 5.

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Problem 12.

Solve the equation: find the values of x, y.

Solution:

Answer: x=0, y=3

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Solve the equation: find the values of x, y.

Solution:

Answer: x= – 1, y=

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The resolution of a vector into two non-collinear vectors

Definition: If , where and are the given vectors, x and y are some numbers, then we say that the vector is decomposed into vectors and , where x and y are the resolution coefficients.

Express the vector:

through vectors and

through and

through and

through and

Solution:

Answer:

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Problem 14.

Given: vectors and are non-collinear.

a)

b)

Solve: the resolution coefficients of the x, y – ?

Solution:

Answer: a) x= –1, y= 3 б) x = 4, y= –5

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Problem 15.

Given: a figure ABCD is a trapezoid

EF is a middle line of the trapezoid

Prove: EFAD, i.e. the midline of the trapezoid is parallel to its base,

i.e. the length of the midline of the trapezoid is a half of the sum of the lengths of the trapezoid bases.

Proof:

By the rule of a polygon

+

Adding both expressions, we get

Since E and F are the midpoints of the sides AB and CD, we see that

Since , we see that , and

Therefore EF || AD and

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Theorem: Any vector can be decomposed into two non-collinear vectors, and the coefficients of the resolution are determined uniquely.

Given:

vector a, vector b

and are non-collinear vectors.

Prove:

Proof:

Through the point A and the point B we draw lines parallel to the straight lines containing the vectors and . Find the point C.

Then by the rule of the triangle

Let us remark that the vectors and are collinear, and the vectors are are collinear too.

By the lemma about collinear vectors

,

Then

Uniqueness of resolution

Proof:

We know that (1)

Let there be (2)

As a result of the difference between expressions (1) and (2) we get

This equality is possible

;

I.e. ;

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Coordinates of a vector

Definition: An unit vector is a vector of length is equal to 1.

i and j are coordinate vectors

Since and are non-collinear vectors, we see that any vector can be decomposed into the vectors and .

I.e. , where x and y are the vector coordinates.

{1:2}

{2:–3}

{0;0}

If and ,

then if and

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Problem 16.

Find the coordinates of the vectors.

Solution:

Answer: {0;6}

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Problem 19.

Find the coordinates of a vector if given are the vectors with coordinates

{–7;–1}; {–1;7}; {4;–6}

Solution:

Answer: {–21;–14}

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Problem 20.

Given:

1)

2)

Solve: resolution coefficients of x, y – ?

Solution:

1)

By the theorem about the resolution of a vector into two non-collinear vectors:

x=–3, y=7

2)

By the theorem about the resolution of a vector into two non-collinear vectors:

x= –4, y=0

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Problem 21.

Given: the coordinates of the vectors

1) {3;6}; {4;–3}

2) {–5;–6}; {2;–4}

Solve: the difference of vectors –

Solution:

1) –= = {–1;9}

–{–1;9}

2) –= ={–7;–2}

–{–7;–2}

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Problem 22.

Given: the coordinates of the vectors

{–2;–3}; {2;–3}; {0;5}

Solve: the coordinates of the vectors are opposite to the given coordinates.

Solution:

{–2;–3} {2;3}

{2;–3} {–2;3}

{0;5} {0;–5}

***