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In this short paper I will try to review two methods of writing the equations of a line in space. They are called the parametric form and the vector (or point-vector) form. We will develop both models by looking at the familiar two space equations on a plane.
When you first learned to graph lines you probably learned to graph a single point, and then use the slope to find additional points on the line. Lets try that with the line with a slope of 2/3 that passes through the point (4, 1). We begin with the knowledge that the point contains (4,1), and we know to find another point I could go right three and up two, so another point on the graph would be (4+3, 1+2) = (7,3). Of course we could have done our slope steps twice; that is, go over 3 and up two and over three and up two again. Then we would get (10,5) . In terms of our new vector language, we can find any new point on the line by starting at any point on the line and moving in the direction of the vector [3, 2]. Of course we would still be on the line if we moved of [3,2] or 4 times the vector [3,2], or in-fact, any distance t times [3,2]. So any point (x,y) that is on the line must be some number t of whole or partial steps in the direction of [3,2] from the point (4,1). We write this in vector form as (x,y)=(4,1)+ t(3,2) . Note that when you plug in values of 1 or 2 for t, you get out the points we found above. If we wanted to find points on the other side of (4,1) we would just let t be negative. This format is called the vector (or point-vector) equation of a line.
In 3-space we could do the same thing with a point in space going in the direction of a particular vector. For example if we wanted to write the equation of a Line Perpendicular to a plane through a given point we could follow the method of the following example:
Find the equation of the line perpendicular to the plane 3x-y+2z=12 that passes through the point (2,1,4). To write the line we need to know a vector in the direction of the line, and a point on the line. The point is given (2,1,4) and from the equation of the plane we know that a vector perpendicular to it has direction [3, -1, 2]. So the equation of the line would be (x,y,z) = (2,1,4) + t[3, -1, 2] (Note that it is not common to use parenthesis and brackets in the same equation, but we use them to separate out the beginning point and the direction vector. Now we can find more points on our line by evaluating the expression for different values of t. When t=2 we get the point (x,y,z) = (2,1,4) + 2 [3, -1, 2] = (10, -1, 8).
It is easy to see from the above that we actually were evaluating three different expressions when we found the point (x,y,z) at t=2. If you did it like I did, you first calculated x, then y, and then found the value of z. If we break the vector form apart into these three separate equations, they are called the parametric equations of the line. In the example above the parametric equations of the line are given by :
X=2+3t;
Y=1-t;
Z= 4+2t.
The variable t is called a parameter (beside the measure) and is the source of the term parametric equation.
You can graph 2-D parametric equations on the Ti-83+ calculator. Under the mode button, select {Par} on the fourth line down (beside Func). We previously wrote the equation in vector form for the line with a slope of 2/3 that passes through the point (4, 1) and saw that it was (x,y)=(4,1)+ t(3,2). We want to break this up into its component parts and write the parametric form as x=4+3t and y=1+2t.
Once you are in parametric mode, the y= screen looks different. It now has two lines for each equation, one for x and one for y. Enter the two parametric equations as shown on the image below (left). [the {x,t,q,n} button that used to make an x variable now makes a t] . The {windodw} screen has also changed. It now permits setting for not only x and y, but for t as well. For the moment, set the t-values as shown in the 2nd image. Now graph the image.
Notice that you only get points beginning at (4,1) and moving up and to the right. Where is the rest of the line? To plot the points in the opposite direction we need to use values of t that are negative. Adjust the Tmin in the Window to 5 and graph again. Use the trace and check several points on the line with T= different integer values.
Finding the intersection of a line and a plane:
For a given point and plane in space there are three possibilities, the line may lie completely in the plane, intersect it at a single point, or be parallel to it. We examine each case.
Line parallel to Plane: If a line is parallel to a plane it is then perpendicular to the vector that is perpendicular to the plane. This means the dot product of the direction vector of the line, and the vector perpendicular to the plane must be zero. Lets show an example. Let the plane be given by 3x-y+2z=25. The perpendicular to the plane has direction [3, -1, 2]. Suppose the line is given by (x,y,z)= (4,1,5) + t[1,5,1] . The line has direction [1,5,1]. Since the dot product of [1,5,1] and [3,-1,2] =0 we conclude that the vectors are perpendicular, and by extension that the line is parallel to the plane. Now
Finding if the line lies in the plane. Of course any line in the plane would also be perpendicular to the vector perpendicular to the plane, so it may be that the line (4,1,5)+t[1,5,1] is not just parallel to the plane , but lies completely on it. If this is true, then every point on the line, including the given point (4,1,5) will be on the plane. We test this by testing if the point (4,1,5) is a solution of the equation of the plane, 3x-y+2z=25. In this case 3(4)-(1)+2(5)= 21, not 25, and so the point is not on the plane, but lies a distance of EMBED Equation.3 , or about 1.069 units away from it. Since the line is parallel to the plane, every other point should also be the same distance from the plane. Why dont you check the point when t=1 to see if it is also the same distance from the plane.
Lines intersecting the plane: If the line is not parallel to the plane, then it must intersect it in some point. Here is an example of how to find the point of intersection. We will again use the plane 3x-y+2z=25 and this time let the line be given by (x,y,z) = (4,1,3) + t[2, -1, -2]. We check the dot product of [2, -1, 2] and [3, -1, 2] and find it is 11, not zero, so the line and the plane are not parallel and must intersect.
Let us stop for a minute and imagine that we knew the point of intersection. If it is on the line given, then there must be some t so that the x value is 4+2t, and the y value is 1-t for the same value, and in fact the z value should also be given by 3-2t. if we just knew what that t-value was, we could find x, y, and z, and this x,y,z would be on the plane 3x-y+2z=25. But wait! (sense the excitement.) .. if we combine these two ideas, we can get an equation for t. If we put 4+2t in place of the x in 3x-y+2z=25, and do the same with 1-t for y, and 3-2t for z, we get 3 (4+2t)- (1-t) + 2 (3-2t)=25.. which can be simplified and solved for t and if we know t, we can find x, y, and z. In this case the simplified equation becomes 3t+17=25. And solving for t we get t=8/3. From (x,y,z) = (4,1,3) + t[2, -1, -2] and letting t=8/3 we get (x, y, z) = (28/3, -5/3,-7/3) for the point of intersection.
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