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In this task, we are going to show how any two vectors are at right angles to each other by using patterns with vectors.

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Introduction

Introduction

In this task, we are going to show how any two vectors are at right angles to each other by using patterns with vectors.

This will be achieved by firstly plotting two vectors and discussing any similarities and difference between them. Then two randomly selected points will be chosen and a line drawn through them. Different ways of representing this vector will be explained. Finally, the general form of a vector equation will be used to determine and prove how two vectors are perpendicular to each other.

The use of the Geogebra computer software will be used to graph each vector although the methods used will be explained.

Vectors are used to display the magnitude and direction for a path of an object. Examples include velocity, acceleration, force, displayment, momentum and weight. A vector can be written in 3 different forms:

Velocity vector: image02.pngimage02.png

Parametric equation:  x= a+ct           y=b+dt

Cartesian equation: image46.pngimage46.png

Vector can be drawn with an initial point (a, b) and then a direction vector image58.pngimage58.png. The line that goes through the set of points has an arrow at the end to show its direction.  

...read more.

Middle

image66.pngimage66.png ,

image67.pngimage67.png,

t = 3,     image68.pngimage68.png ,

image70.pngimage70.png,

t = 4,     image71.pngimage71.png ,

image72.png

Again from these points for timage52.pngimage52.png it is now possible to plot them on a set of Cartesian axes. This was done on the Geogbra computer program. It follows the idea of simply points on the x, y axes when based at the origin.

Another graph can be created by plotting those vectors image53.pngimage53.png with the equationimage73.pngimage73.png. For  image55.pngimage55.png. image00.png


By comparing those two equations, we are able to see there is a pattern that although the value of (t) changes, the vectors image74.pngimage74.png are collinear. This means they lie on a same straight line with the same slope.

In the first equation, the slope of it is lying on the positive side, but the second slope is lying on the negative side. It is considering to their vectors direction image75.pngimage75.png and image76.pngimage76.png

In this time, we chose the point image77.pngimage77.pngand  image78.pngimage78.pngfrom the vector equation:image61.pngimage61.png. Where r = 1, r = 3. After that, we draw the line L that those two point can pass through it.

image79.pngfigure 3

The slope of these two point is same as the slope of the equation image80.pngimage80.png.  (Refer to figure 2)

We assume that the vector image74.pngimage74.png is image81.pngimage81.png so

image81.pngimage81.png = td {distance = time × speed}

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Conclusion


image13.png
t is a scalar multiple of arbitrary value, any value in the set of all real numbers. image15.pngimage15.png is a direction vector describing the lines direction. It and its multiples are being vector added to the constantimage16.pngimage16.png, which is a vector describing the point image16.pngimage16.png which the vector passes through.

If a line L passes through a point U (h, k) in a direction dimage17.png. Then we can create a vector equation image18.pngimage18.pngwhich is thegeneral position vector for point R on line L.

By looking to the diagram,

image19.pngFigure 7

Now, changing the velocity equationimage20.pngimage20.pngin the form image21.pngimage21.png

image22.png

image23.png

image24.png

image25.pngimage25.png. Line L passed through a point U (h, k) in a direction image26.pngimage26.png

Suppose a plane in space has normal vector image28.pngimage28.png

And that is includes the fixed point image29.pngimage29.png. U( u, k) is any other point in the plane.

Now image30.pngimage30.png is perpendicular to n

image31.pngimage31.png

image32.png

image33.png

image34.pngimage34.png where the RHS is a constant.

image35.pngimage35.png could also be written as image36.pngimage36.png, which impliesimage37.pngimage37.png.

If a plane has normal vector image38.pngimage38.png and pass through image40.pngimage40.png

then is has equation -bu + ak = -bu1 + ak1 = d,     where d is a constant. This is called the Cartesian equation of the plane.

Now, we are trying to subimage41.pngimage41.pngintoimage42.pngimage42.png, image43.pngimage43.pngand then provide the thoery above.

image35.png

image44.pngimage44.png        where  t = 1image01.png

...read more.

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