Section 13.5 -- Equations of Lines and Planes

When we have a line in the old XY plane, we know it by knowing one of its points and then its direction, or slope. The same holds true here. Suppose we have two vectors, r and r₀. These will be the position vectors of points P and P₀, that is, r is the vector from the origin to P, and r₀ is the vector from the origin to P₀. Now, suppose that a new vector a is the vector from P₀ to P, or from the end of r₀ to r. The triangle law gives us that r = r₀ + a.

Now. Let's suppose that we're looking for some line L, and we'll let some vector v be parallel to this line L, and now let's also suppose that a is inside of L, that is, at some point, r and r₀ end at random points on L, and the part of L that lies inside this "boundary" is a. Because v || a and v || L (|| means "is parallel to"), then we can say that a = tv, that is, some scalar t times the vector v gives us the same magnitude as a.

We replace a with tv, then, and we have r = r₀ + tv. This is our equation for the vector of L. We call this a vector equation of L. Our parameter is t. Now, let's give some values to these vectors.

We'll say v = <a,b,c>, then tv = <ta,tb,tc>, and we'll say r <x,y,z>= and r₀ = <x₀, y₀, z₀>. Our vector equation then becomes

<x,y,z> = <x₀ + ta, y₀ + tb, z₀ + tc>. Because L and a are parallel, then their corresponding components are equal.

This means that we have three scalar equations:

x = x₀ + at
y = y₀ + bt
z = z₀ + ct

And these are our parametric equations (see Chapter 11 for more info).

Now, wtf? Let's give a quick example here.

Suppose we have some P₀ = (1,1,1) and we start at the origin, and we want to find the vector parallel to i + j + k. We have r₀ = <1,1,1> and v = <1,1,1>, so we multiply v by t to get , and then we simply add up tthe components: x = x₀ + at = 1 + t, y = y₀ + bt = 1 + t, z = z₀ + ct = 1 + t. Thus, our new r is now <1 + t, 1 + t, 1 + t>. Bam. And by plugging in a new t, say, t = 3, then we can find all the new points on this vector r, like (4,4,4) in our case of t = 3. It's simple. You are given P₀ and v, and you find r.

Now, what if we don't like the parameter t? Let's get rid of it! To do so, we must assume that a,b,c =/= 0, and then we solve each of those scalar equations for t. This gives us

t = (x - x₀)/a = (y - y₀)/b = (z - z₀)/c. These are called symmetric equations. We employ these when we are GIVEN both P and P₀, allowing us to find r and r₀ and thus v (which is r - r₀). This is very useful later when we are trying to find where L intersects various planes and, ultimately, when we need to find some boundary conditions for when we'll apply calculus later on.

Now, if only ONE of a,b,c = 0, then we can still eliminate t in the same way, and we simply remove the entire part of the equation where the letter = 0. Suppose b = 0. Then we just say y = y₀, and that leaves us with the x and z parts equal each other still. This also implies our line lies in vertical or horizontal planes, which can simplify things just a wee bit.

Anyway, usually, we don't care about entire lines. We only wanna know what's going on in a piece of the line. We can do this very simply.

We know that r = r₀ + tv, and we know v = r₁ - r₀. We just substitute back in--r = r₀ + t(r₁ - r₀), and we have

r = (1 - t)r₀ + tr₁, where 0 < t < 1. r₁, by the way, is the same as r, except we've narrowed it a bit to just be from 0-1 for t. This lets us describe a tiny piece of the line and ignore the dumb crap beyond our parameter.

And so, we have all our equations for solving for lines. Now we just have one more little geometric concept. In the XY axes, you knew two lines existed: intersecting lines and parallel lines. All lines either were parallel or they intersected at some point way off in space. In 3-D, there is a third kind of line called a skew line. Skew lines do not intersect and they are not parallel. The easiest way to look at this is by having two people side by side in the middle of a platform on a staircase, where the stairs run up and down from this platform. One runes up, the other runs down. They will never hit each other 'cuz they're to each others' right and left, and they aren't going in the same direction at ALL. Skew lines work like this.

Now for some PLANES! This is not a separate topic. It builds straight from the lines we've just talked about.

So, we have this vector a which is formed by r - r₀. What we want to do now is try and describe a plane it will be located it in. More specifically, we're interested in the direction of this plane. Is it flat, is it slanted, if so what's its slope, etc. To do that, we need our arbitrary point P₀ on the plane and then we need something called the normal vector. We define as n, and it is simply the orthogonal (perpendicular) vector to the plane itself. That is, if you have a plane (like a sheet of paper) lying flat horizontally, n points straight up. If it's vertical, n points to the right or something.

Because n is orthogonal to our vector a, we say that n · a = 0. a = r - r₀, so n · (r - r₀) = 0. But we should just rewrite that as this--

n · r = n · r₀. This is called the vector equation of a plane.

But vectors are often a bit confusing to deal with, so we oftentimes want a scalar representation. To do that, we simply replace n with , r with , and r₀ with ₀, y₀, z₀>. This gives us

· ₀, y - y₀, z - z₀> which can be rewritten as

a(x - x₀) + b(y - y₀) + c(z - z₀) = 0. This is the scalar equation of the plane.

If we collect all the terms together, we can rewrite this one step further as

ax + by + cz + d = 0, where d = -(ax₀ + by₀ + cz₀), and this is called the linear equation in xyz.

Now for a stupid little property. Two planes are parallel if their normal vectors are parallel. It makes sense if you draw it. If two normal vectors are pointing straight up, they are parallel, and this means that their planes lie flat on the ground (like a sheet of paper), which makes them parallel to each other. However, we don't really care too much about planes being parallel, and we're instead curious about planes that aren't parallel. We want to find the angle between them. To do so, we consider only their normal vectors, and it's simply the dot product equation. That is,

cos = (n₁ · n₂) / (||n₁|| ||n₂||).

Another thing that is useful is figuring out the line of intersection. When two lines intersect, they intersect at a point. Well, when two planes intersect, they intersect at a line. This will be line L, and to find line L, we simply go back up to the beginning of this section, and we find some vector v parallel to this line L. To find v, we simply take the cross product of the two normal vectors. Thus,

v = n₁ x n₂.

We can then apply symmetry equations and other things to figure out the overall equation of the line or points on it.

Lastly, suppose we have a random point P₁ in space and we want to know the distance between it and our plane. To do that, we need our P₀ and our normal vector. If we draw it, we can force P₁ to lie on n, and then P₀ lies anywhere on the plane, and all three are connected to make a triangle. By doing so, we see that the distance D is simply the scalar projection of b onto n, where b is the vector from P₀ to P₁. Thus,

D = |comp_nb| = (|n · b|) / ||n||.

We can rewrite this...

D = (|a(x₁ - x₀) + b(y₁ - y₀) + c(z₁ - z₀)|) / Sqrt(a² + b² + c²) , we distribute the a,b,c and regroup according to 0s and 1s:

D = (|(ax₁ + by₁ + cz₁) - (ax₀ + by₀ + cz₀)|) / Sqrt(a² + b² + c²) , we recognize that all the 0th terms are actual points, and so they're given, and so we can rewrite the entire 0th part as some constant d:

D = (|ax₁ + by₁ + cz₁ + d|) / Sqrt(a² + b² + c²)

And that's it! Now let's work some examples!!

------

1) Find a vector equation and parametric equations for the line through point (1,0,-3) and parallel to the vector 2i - 4j + 5k.

This is easy. r = r₀ + tv. v is 2i - 4j + 5k = <2,-4,5> and r₀ = <1,0,-3>. tv = <2t, -4t, 5t>, so now we just add to get r = <2t + 1, -4t, 5t - 3>. The vector equation then is r = (2t + 1)i + (-4t)j + (5t - 3)k.

The parametric equations are x = 2t + 1, y = -4t, and z = 5t - 3.

2)  Find the distance from the point (2,8,5) to the given plane x - 2y - 2z = 1.

It's the last formula.  D = |ax₁ + by₁ + cz₁ + d| / sqrt(a² + b² + c²).  Here we have a = 1, b = -2, c = -2, d = -1 (from the plane), x1 = 2, y1 = 8, z1 = 5 (from the given point).

So, we just plug in.  D = |1*2 + -2*8 + -2*5 + -1| / sqrt(1² + (-2)² + (-2)²) = |2 - 16 - 10 - 1| / sqrt(1 + 4 + 4) = |-25| / sqrt(9) = 5/3.

3)  Find the angle between the planes x + y + z = 0 and x + 2y + 3z = 1.

We'll call the normal vectors of these n1 = <1,1,1> and n2 = <1,2,3>.  Then the cos(θ) = (n1 · n2) / (|n1|*|n2|)

That we just plug in things.  n1 . n2 = 1*1 + 1*2 + 1*3 = 1 + 2 + 3 = 6.  |n1| = sqrt(1² + 1² + 1²) = sqrt(1 + 1 + 1) = sqrt(3).  |n2| = sqrt(1² + 2² + 3²) = sqrt(1 + 4 + 9) = sqrt(14).  sqrt(3)*sqrt(14) = sqrt(42).

So, cos(θ) = 6/sqrt(42), then θ = cos^-1(6/sqrt(42)) = 0.3876 radians.

-------------

Why do you care?

Trajectory from impact is very important from this as an example I can immediately think of.  Magnetic fields are what I'm getting at.  You may induce a magnetic field such that it creates a "magnetic plane" over an area, and you'll have a particle between two of these planes cocked at an angle and such that the particle is also charged and is SUPPOSED to move around, but you don't want it touching the planes or any sort of wall.  But you need to keep that magnetic field to induce electrical current.  Drafting equations gives us an application of magnetic force, then.

For those of us building large skyscrapers in cities that are already polluted with very high buildings and have to deal with winds (looking at you, Chicago), this is also fairly important because it's a good idea to know the distance between X point on building A and Y point on building B (where X and Y are defined because these will deal with maximal wind shear, vorteces, whatever the problem is), and sometimes this location will fluctuate a little depending on the time of year, so you can't just simply take one measurement and be done, but taking the measurements of the "vectors" of the buildings (basically, the height) gives you a sort of "blueprint" any time you need to measure forces of "wind interaction" between the two buildings at ANY point on either building.

And wind can change its nature after dealing with an obstruction, like Building B.