Two vectors are at right angles (perpendicular) to each other when they satisfy the equation
Results/ Analysis
By plotting the vector equation:,
After plotting this vector equation, the first step is to find various points in time. That is to substitute values for (t) to find the coordinates of the object.
t = 0, ,
,
t = 1, ,
,
t = 2, ,
,
t = 3, ,
,
t = 4, ,
From these points for t 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.
A graph can be created by plotting those vectors with the equation. For .
figure 1.
It can be seen from the graph that the 5 points that were plotted are linear and move in the same direction. The initial point being (5, -3) and each successive point moving in the direction of 1 across and 3 up according to the direction vector .
To determine the parametric equation for this vector, we split the vector equation into the x and y components.
and
By knowing the value of t, this can be substituted into the 2 parametric equation and the coordinate for (x, y) can be found.
Now, we are going to plot another vector equation: for comparing with the first equation.
After plotting this vector equation, the first step is to find various points in time. That is to substitute values for (t) to find the coordinates of the object.
t = 0, ,
,
t = 1, ,
,
t = 2, ,
,
t = 3, ,
,
t = 4, ,
Again from these points for t 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 with the equation. For .
By comparing those two equations, we are able to see there is a pattern that although the value of (t) changes, the vectors 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 and
In this time, we chose the point and from the vector equation:. Where r = 1, r = 3. After that, we draw the line L that those two point can pass through it.
figure 3
The slope of these two point is same as the slope of the equation . (Refer to figure 2)
We assume that the vector is so
= td {distance = time × speed}
Now by drawing the diagram
t ≥ 0
We are trying to find out that all the equation of L can explain by the general formula above. Therefore, we can use another two point and from the vector equation:. Where r = 1, r = 3. Another graph can be drawn below.
Figure 5
Also, the slope of these two point is same as the slope of the equation . (Refer to figure 1)
Again, by using the diagram
Figure 6
We can find out that the formula t ≥ 0 is also fix on this two point and from the vector equation:.
Therefore, is the general formula for determining the vector with AB two points.
The line L is every all points which extend along a direction vector, in other words all scalar multiples of a direction vector, which extend from a particular point in space defined by its own direction vector.
t is a scalar multiple of arbitrary value, any value in the set of all real numbers. is a direction vector describing the lines direction. It and its multiples are being vector added to the constant, which is a vector describing the point which the vector passes through.
If a line L passes through a point U (h, k) in a direction d. Then we can create a vector equation which is the general position vector for point R on line L.
By looking to the diagram,
Figure 7
Now, changing the velocity equation in the form
. Line L passed through a point U (h, k) in a direction
Suppose a plane in space has normal vector
And that is includes the fixed point . U( u, k) is any other point in the plane.
Now is perpendicular to n
where the RHS is a constant.
could also be written as , which implies .
If a plane has normal vector and pass through
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 sub into , and then provide the thoery above.
where t = 1
