We need to have at least 2 direction vectors to show the direction of the plane, and then a point to know where does the plane lie exactly. We multiply the 2 direction vectors with different constants, to show that any direction vector proportion to that ratio is also a direction vector. Similarly, this form of the plane equation is not unique. Again, this form can be written in the ijk form, in which looks ugly and long.
There is another vector equation of the plane. Though not named properly, I call it the ‘normal’ form. We first find the normal vector of a plane, i.e., a vector which is normal to both the direction vectors. You obtain the normal vector by getting the cross product of b and c. Suppose that the normal vector is (a, b, c), the normal form of the equation will be
Where d is constant which determines the position of the plane. d has a significant meaning. If the normal vector (a, b, c) is a unit vector (magnitude = 1), then d is the perpendicular distance from the plane to the origin. For 2 planes, if their values of d have opposite signs, it means that they are at the opposite sides of the origin. Finding the value d is simple: Just plug in a point lying in the plane into x, y, z, then you get it.
This cartesian form is unique, unlike the other forms. This is the most common form of the equation of planes used. You can see that this equation is linear, and that the equation
y = mx + c, or x = a are all equations of planes in 3 dimensional space.
So to sum up, to construct a plane equation, you need one of these information:
1. 3 points lying on the plane.
2. 2 points lying on the plane, and 1 directional vector.
3. 2 lines lying on the plane.
4. a point lying on the plane, and the normal vector of a plane.
There is a fast way to get the equation of the plane when 3 points are given. I haven’t tried this before, but you could make use of the determinants below to find your equation:
LINE LIES IN / PARALLEL / INTERSECT A PLANE
We shall now discuss how to determine whether a line lies in / is parallel to / intersects a plane. Given the equations of the line and plane to be
We first find whether the direction vector of the line is parallel to the plane. In other words, we want to know whether the direction vector of the line is perpendicular to the normal vector of the plane. By taking b • n, if the answer is zero, then the line is parallel to the plane. We might want to know whether the parallel line actually lies in the plane. We can do this by substituting the position vector of the line into r2, and if LHS = RHS, then indeed the line lies in the plane, and is otherwise if the equality doesn’t hold.
You should be able to solve for t, which satisfies all the 3 parametric equations. Then finally, to find the point of intersection, we substitute t back into the line equation to find (x, y, z).
PLANE PARALLEL TO / INTERSECTS ANOTHER PLANE
Since the cartesian equation is unique, 2 planes can only coincide one another if they have the same plane equation. 2 planes are parallel only if they have the same normal vector, which is also easy to find. Planes that are not parallel have to intersect somewhere, and we can determine the line of intersection. Consider 2 plane equations below:
We need to solve this system of linear equations to find x, y and z. Recall the Chapter on Matrices, this system of equations have infinitely many solutions. As usual, let one of them be t, solve for x, y and z in terms of t, and then just substitute a value for t to get a random position vector. The line equation is thus found.
DISTANCE FROM A POINT / PARALLEL LINE TO A PLANE
Notice that there is something different in my equation. It is ‘-d’ instead of ‘+d’ because I made use of the cartesian equation ax + by + cz = d instead of ax + by + cz + d = 0. Please DO NOT CONFUSE THEM.
If you want to find the distance between a parallel line to the plane (note that the line has to be ‘parallel’ to have a ‘distance’…), you substitute the position vector of the line (x, y, z) into the above equation, and you get it.
DISTANCE BETWEEN 2 PLANES
I will explain why this makes sense. Firstly, you should recall that the values d/|n| and e/|n| are the perpendicular distances from the planes to the origin. Also remembering that the distance really depends on whether both the planes lie on the same side of the origin, or the other (same sign or different sign). You subtract them, then take the modulus because distance is never negative.
ANGLE BETWEEN LINE AND PLANE
Note that if you used cos θ, you would have gotten the angle between the line and the normal vector instead.
ANGLE BETWEEN PLANES
The angle between 2 planes is actually the same angle between the 2 normal vectors. So given 2 planes with normal vectors m and n respectively, we can find the angle between 2 planes by using the dot product,
Recall that this is the same formula to find the angle between 2 lines.
Now that you know how to construct planes, you might be curious as in how 3D shapes are constructed. Again, you could make use of the applet I shared with you in the previous post, from the drop down menu of new graph, choose z = f(x, y) surfaces. Fiddle around it and have fun creating awkward shapes. This is obviously out of your syllabus, but let me just give you some equations for some very common shapes in 3D:
End of all those hard core Mathematics, at last! The next few chapters will be on Statistics, and so life will be much easier thereafter. ☺