SLC S23 Week5 || Coordinates&Vectors

fazal-qadir (61)in Teachers & Students • 3 days ago (edited)

Assalamualaikum

Greetings steemians!

Season 23 week 5 , coordinates and vectors. As the topic is very interesting to engage and trigonometry was my favorite subject. I learnd about coordinate and vectors when I was in 9th class back in 2001. Let's recall it and try to solve the given task effectively. I am very late for submitting the required task. Busy schedule as well as the month of Ramdan is going. So very less time I get to solve these questions. Let's not waste more time and start the test.

Task 1

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

Number line:

Number line is used to represent number on a line. Usually it is dividend into two section. In center of the line is called origin which value is zero. On right side of zero the value is written in positive form , while the number on the left side of zero all number are negative.

Graphical representation:

Coordinates plane:

A coordinate plane is two dimensional mesh , which is used for the representation of position of points in a plane.

x-axis (horizontal)
y-axis (vertical) .

Coordinates plane is divided into four quadrants and a origin(0,0). Where value of x and y is zero. At this point both axis intersects .
We count quardents from anti-clockwise direction.

Representation of ordered pairs

We represent axis in ordered pair (x,y)

In I- quadrants the value of x-axis and y-axis become positive. (x,y)
In II-quadrant the value of x is negative while value of y is positive (-x,y)
in III- quadrant the both x and 6y axis values are negative (-x,-y)
In iv- quadrants the value of x is positive while the value of y is negative. (x,-y)

Graphical representation:

As we can see from the graph. The value of ordered pair A, B , C and D. The value and sign of each axis and their corresponding pairs.

. I have learned it when I was in 9th grade. Yeah at that time it was not easy. But now many things are cleared and easy to understanding.

Task 2

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))

Let's take A(2,4) and B(0,0)
Find the distance from A to B , we will apply the formula to find it.

A(x1,y1) ,B(x2,y2) BA= (x1-x2,y1-y2)
Put above values we get BA= (2-0,4-0)
= (2,4)

As we are going coordinates plane :
The vector is called displacement vector . As we have find the displacement as well as the direction. The ordered pair shows both x and y values are positive so it will be in I-quadrant . The direction is up northeast.

Graphical representation:

In this question we are ask to show these vectors A(-3, 11), B(4,7), C(0,4), D(4,0), E(-4,-7), F(11,3) on plane. It is shown below

Graphical representation:

Relationship

Vector AB=(7,-4)
Vector AE= (-1,-18)
Vector CB=(4,3)
Vector CE=(-4,-11)
Vector DF=(7,3)

Task 3

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

First of all plot a, b and c vectors on the plane. When you done with it. a and b like head to tail rule. The vector a head will be the tail of vector b . If we can not add directly in head to tail rule the vector c. We will take the result of a+b with vector c. So we can get the final results as shown in the figure.

Explanation:

Start from the origin (0,0) , plot the point (3,7) which is vector a. The vector a equal to ( 3,7). Now from the head of vector a (3,7 ,add vector b(-1,-3) to reach the (3-1,7-3) is equal to (2,4) .

Now from vector b head at (2,4)
Add vector c which will became (2,4)+(1,5) = (2+1,4+5) =( 3,9)

So resultant vector is a+b+c= (3,9)

task 4

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.

Vector A is (4,4) and Vector B is ( 10,8) .

There coordinates is equal to u(6,4) .

If we want to construct the vector double of the original one. We have to double the coordinate if this one. We need to double the (6,4) . Which has be to ( 12,8 ). To get it we need to find the end points . One is know the other is not know. Know end point is (4,4) to find the other we have to do (4+12,4+8) . We get the other end point which is (16,12).

Graphical representation:

Task 5

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

Initial points a(4,4) b(6,-2)
U=a+b

I took a vector (4,4) and b vector (2,-6) after addition from head to tail rule. I plot a vector and than from the tail of vector a I connect tail of b with head of a and form it. After that I find the resultant vector u. Which is equal to (6,-2)