SLC S23 Week5 || Coordinates&Vectors

_{Photo taken from Pixabay}

Another week, another challenge from @sergeyk, this time we have Coordinates and Vectors. It's not too late to join in if you want, you can also check the full post and how to participate in the following post SLC S23 Week5 || Coordinates&Vectors.

Let's start!

Tell about the number line and the coordinate plane. When did you first learn about coordinates? Was it difficult?

When talking about the number line, we usually mean a horizontal straight line where numbers are placed at a fixed interval. The center of the line is in 0, with negative numbers on the left side of it and positive numbers on the right. I learned about the number line in primary school, 2nd or 3rd grade, at first only with positive numbers. This helps visualize basic operations like additions or subtractions. Negative numbers are typically introduced in 5th grade.

_{Number line}

The coordinate plane is made of 2 perpendicular number lines, both with the center in (0,0).

The vertical number line is called the y axis (OY), with positive numbers above and negative below.
The horizontal number line is the OX axis, which is the same as the number line you learn in primary school.
The plane is split in 4 quadrants

_{Coordinate Plane}

By the time I reached 7th grade, I started using the coordinate plane for more complex geometric problems like slope of a line, equations of shapes, vectors and later, trigonometry and circles.

The learning curve is not too step or hard to learn as long as you pay attention and understand it. I personally hated trigonometry and the coordinate plane, can't say it was difficult to learn, it just wasn't my cup of tea or something I enjoyed doing.

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).
If all vectors are shown, there will be too many of them, so display only a few. For example 5 or 6))

We know that to calculate the vector's coordinates we use the coordinates of the points that make the vector like this:
Example: vector AB = (x_B - x_A, y_B - y_A)

Let's say that vector AB has (x,y) as coordinates. This means that vector BA will have (-x,-y) because it's the reverse of AB. This is useful so I don't have to construct both AB and BA as only the direction changes.

There are a total of 30 vectors that can be constructed using these 6 points. It would be pretty pointless to construct all of them as a vector will always have a reverse so there are only about 15 "unique" vectors. I constructed 9 of them.

_{AB(7, -4), AC(3, -7), EC(4, 11), DB(0, 7), DF(7, 3}

_{AE(-1, -18), DE(-8, -7), FC(-11, 1) and BF(7, -4)}

Place the vectors a(3,7), b(-1,-3), and c(1,5) on the plane.
Construct the vector a + b + c.

We know from the lesson that the coordinates of a vector AB are calculated using this formula: vector AB = (x_B - x_A, y_B - y_A).

(In the table below, I give a short description of how I constructed these vectors.)


	In our case, vector a has (3,7) as the coordinates. If we consider point A with coordinates (1,2) and B(4,9), using the formula we get: a = (4-1,9-2) = (3,7)
	For b(-1,-3), I used points C(7, 6) and D(6, 3) and construct vector CD
	For c(1,5), I used points E(12, 9) and F(13, 14) and vector EF

Let's see all 3 vectors constructed:

Now, we can't do the addition of 3 vectors but we can add 2 of them to obtain vector d and then do c+d.

First, I did the addition of a+b using the triangle method. I used point B as the origin point for vector c resulting in vector BD. This is how is looks before adding vector d:

To find the coordinates for vector d, we simply add the x parts and y parts of the 2 vectors: d = (-1+3,-3+7) = (2,4). And here we can see it:

_{Representation of vector d}

With a+b=d done, it's time to turn our attention d+c. Using the same triangle method, we construct vector c' originating in point D and ending in F.

The resulting vector, e, will have (3, 9) as coordinates and looks like this:

A better look at the vector resulted from the addition of a+b+c:

Place two random points and determine their coordinates.
Create a vector from these points and write its coordinates.
Construct a vector that is twice as large as the created one.

To start, I placed points A and B in random spots on the grid.

Next, I constructed 2 perpendicular lines for each point: one on the OX axis and one on the OY axis. This gave the coordinates for the 2 points: A(4, 6) and B(-4.5,2).

I created vector BA, which will gave (8.5, 4) as coordinates:

_{Vector BA(8.5, 4)}

A vector twice the size of the existed one means that we can simply multiply the coordinates by 2 and we will have the coordinates of the new, bigger vector. This means that the new vector is have (17, 8) as coordinates.

I added points C and D and the perpendicular lines to determine their coordinates. It gave me C(-9, 3) and D(8, 11):

The resulting vector, CD, will have (17, 8) as the coordinates which means it's twice the size of vector BA:

_{Vectors BA and CD}

Construct three arbitrary vectors: a, b, k.
Build the vectors u = a + b and v = b - k.
Then construct u + v and u - v.

I constructed the 3 vectors using the Vector from Point tool and randomly placed them.

Let's start with building vector u = a + b:
Like in Task 3, I used the triangle method by placing vector b at the end of vector a.

Next, I constructed vector u = a + b, with the start in point A and end in D', and by calculating it's coordinates we get u(3, 3):

_{Vector u = a+b}

Now to make v = b - k:
For subtracting the vectors, I used the steps presented by @sergeyk. Added vector k in the starting point of vector b, constructed vector v or GD with (-4,4) as coordinates.

_{Vector v = b - k}

After the 2 operations, this is what we have:

For u + v, I am using the Triangle Method again. The resulting vector will have (-1, 7) as coordinates

_u+v

Similar to b-k, I did u-v.

_u-v

And both constructions in the same photo:

Bonus Task

Bonus task – Find out what a vector projection is, explore it, and construct the projection of one vector onto another.

Vector projection represents the method of finding how much of a vector (a) lies on another one (b). With other words, the projection of a vector is it's shadow on another vector if the light would be perpendicular on it. It's a pretty complex part of vector geometry with complex calculations that need to be done like the scalar product a ⋅ b, the square magnitude b ⋅ b, scalar multiplier and finally the projection.

Example:
Let's say I want to project vector b on vector a where a(6, 2) and b(3, 4), The scalar product is the product of x coordinates of each vector + the product of y coordinates. In this case: b ⋅ a = (6 x 3) + (2 x 4) = 18 + 8 = 26

The magnitude is the coordinates of the vector squared and added: a ⋅ a = 6² + 2² = 36 + 4 = 40

The scalar multiplier is the fraction of (b ⋅ a)/(a ⋅ a) = 26/40 = 13/20.

Let's look at it in GeoGebra:
I started with 3 points: A(0,0); B(6,2) and C(3,4) and constructed vectors AB (a) and AC (b).

Next, I added a perpendicular line from C on vector a and marked the intersection point:

Now let's find what the projection is: proj_ab = (13/20) x (6, 2) = (3.9, 1.3). This means that vector AC' or b' will have (3.9, 1.3) as coordinates. This can also be seen in the image below:

_{Projection of vector b on vector a}

Since I finished this before @ady-was-here I would like to invite him to take part. Let's see what next week's challenge will bring.

See you next time!

_{Posted using SteemPro}