SLC S23 Week5 || Coordinates&Vectors
Assalamualaikum my fellows I hope you will be fine by the grace of Allah. Today I am going to participate in the steemit learning challenge season 23 week 5 by @sergeyk under the umbrella of steemit team. It is about Coordinates&Vectors. Let us start exploring this week's teaching course.
Tell about the number line and the coordinate plane. When did you first learn about coordinates? Was it difficult?
The number line is a straight and horizontal line. It is used represent numbers. It has a central point which is called zero.
On this horizontal line there are two types of numbers. The number on the right side from the zero point are the positive numbers and the numbers on the left side of the zero point are the negative numbers. This helps us to understand the basic mathematical concepts like addition, subtraction and comparing the numbers.
The coordinate plane is a two-dimensional plane. It is formed by the two perpendicular lines. One is a horizontal axis which is OX and the other is a vertical axis which is OY. These two axis intersect each other at one point which is the origin point denoted by (0,0). The coordinate plane is divided into four quadrants. In the quadrants we can plot points by using the ordered pairs in the format (x,y). The position of each point is determined by the distance from the x-axis and y-axis.
As far my experience I first learned about the coordinates in the elementary school. It was really a new thing for me and it was interesting to learn. I learnt it from my maths teacher many years ago. It was tricky and was not easy to remember the correct order of x and y. In the start I was confused about plotting the points on the negative coordinates. But with the practice I learnt it and it became easier than the start.
Connect the pairs of points with vectors. Which vectors are formed? Show these vectors on the plane. A(-3, 11), B(4,7), C(0,4), D(4,0), E(-4,-7), F(11,3).
To complete this task I first drew all the points on the coordinate plane.
- A(-3, 11)
- B(4,7)
- C(0,4)
- D(4,0)
- E(-4,-7)
- F(11,3)
All the points have been placed according to their required place on the coordinate plane. Most of the points are in the first quadrant. The quadrant II and III have one point other than common and similarly IV has no any point.
After carefully placing all the points I used vector tool to combine all the points with each other. I formed 5 vectors to avoid overcrowding of the vectors. This is the details of all the vectors which I formed.
Vector AB: Start at A and end at B
Coordinates: (4 - (-3), 7 - 11) = (7, -4)Vector CD: Start at C and end at D
Coordinates: (4 - 0, 0 - 4) = (4, -4)Vector EF: Start at E and end at F
Coordinates: (11 - (-4), 3 - (-7)) = (15, 10)Vector AF: Start at A and end at F
Coordinates: (11 - (-3), 3 - 11) = (14, -8)Vector BD: Start at B and end at D
Coordinates: (4 - 4, 0 - 7) = (0, -7)
The vector AF passes over the vector AB. Actually the points A, B and F are in the same way. To distinguish the vector AF from AB I have increased the thickness of the vector AB and changed the colour and style of the vector AF. I have labelled all the vectors with their names and with their coordinates.
Task 3. Place the vectors a(3,7), b(-1,-3), and c(1,5) on the plane. Construct the vector a + b + c.
I have placed all the three vectors a, b, and c on the plane keeping in mind their coordinates. All the vectors are starting from the origin point A. The vector c has negative coordinates and this is the reason it is lying in the III quadrant.
For the addition of the vectors I have used triangle method. To find the sum vector a+b+c I have placed the first vector a(3,7) on the plane as usual and it is a starting vector. Then I have placed the next vector starting from the head of the vector a. I calculated the coordinates of the vector b by summing up the x coordinates and y coordinates (3-1, 7-3) = (2,4). So I started the vector b from the ending of vector a and placed it at (2,4). Then starting from the ending of the vector b I placed the vector c by calculating the coordinates (1+2, 5+4) = (3, 9). Then to find the resultant or the sum vector I have connected the point A or origin with the head of the last vector head with a vector. This new vector is the sum of these three vectors a,b, and c.
If we see we can calculate the resultant in this way:
So it has been proved that our resultant vector is completely fine as it satisfy the manual calculation. The vectors a and c are overlapping with each other.
Place two random points and determine their coordinates.Create a vector from these points and write its coordinates.
I have taken two random points on the coordinate plane. Both the points are in the first quadrant.
I have found out the coordinates of both the points. The point has (2,3) coordinates and the point B has (5,7) coordinates.
I have used the tool to create the vector from point A to B and a vector AB has formed.
I have found the coordinates of the vector AB as shown in the image. The coordinates are given by AB=(5−2,7−3)=(3,4) so the vector AB(3,4).
To build the twice vector of the vector AB I have multiplied the coordinates of the vector AB with 2 so 2×(3,4)=(6,8) so the twice large vector will have these coordinates (6,8).
I have drawn the vector PQ which is the twice large of the vector AB.
I have removed all the helping lines and points from the coordinate plane. I also came to know about the value of the points in the coordinate so I used that and before I was writing it manually by writing the coordinates in the name. So I learned something new. I measured the distance using the distance tool to show the new vector PQ is the twice of the vector AB. Vector AB has distance of 5 and the vector PQ has distance 10.
Construct three arbitrary vectors: a, b, k.
I have constructed 3 arbitrary vectors a,b, and k. I have tried to construct the three arbitrary vectors in different coordinates. These three vectors are a(2, 3), b(4, -1), and k(-3, 2).
Build the vectors u = a + b and v = b - k
Vector u is the sum of the vector a and b like given by the teacher u = a + b. To calculate the vector u = a+b = (2+4,3+(−1)) = (6,2) so the vector u must have these coordinates.
I have used triangle method to find the vector sum. By using the vector sum method I successfully found the vector u which is the sum of the vectors a and b. Red vector is representing the vector u as the sum of vectors a and b.
Vector v is the difference of the vectors b and k like v = b-k. By calculation the coordinates of v are v = b−k = (4−(−3),−1−2) = (4+3,−3) = (7,−3). The new vector v must have the coordinates (7,-3).
I have again used triangle method to find the vector v with the difference of the vectors b and k. The vector v with the coordinates (7, -3) is represented by the blue colour.
Then construct u + v and u - v.
The sum of the vector u ad v is u + v = (6+7,2 + (−3)) = (13,−1). Now I will find this sum vector of u and v.
I have calculated the sum of the vectors u and v and in the vector representation I used triangle method. u+v=d(13,-1) is represented by the red dotted vector starting from the origin A and ending at the point G.
The difference of the vector u ad v is u - v = (6-7, 2 - (−3)) = (−1, 5 ). Now I will find this difference vector of u and v.
In the vector representation I have again used triangle method to find the difference of the vectors u and v. I have hidden the arbitrary vectors which I took in the start to make the screen clear free from the rush. I have represented the difference vector with the colour of blue to make it distinct from the sum. u-v=f(-1, 5)
This is all about the coordinates and vectors. I invite @wuddi @chant and @suboohi to join this learning about the quadrilaterals with geogebra.
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