There's an example on "Elementary Geometry of Differentiable Curves:An Undergraduate Introduction" by Gibson on page 9 which is:
"Example 1.10 Let be a non-zero vector, and let be a line with equation . By a direction for the line we mean any
vector orthogonal to : it is an example of a 'tangent' vector.(Chapter 2) Thus we could take . Alternatively, we could choose
any two distinct points on the line, and take : the relation then follows from ,
on subtraction. On this basis we say that two lines , are orthogonal when the corresponding 'normal' vectors
, are orthogonal, or equivalently the corresponding 'tangent' vectors are orthogonal:
either way, the condition is that ."
My problem is: How did the author picked and to be normal vectors from the two equations?
Can anyone kindly tell me how and why are these normal vectors? And also how did the author(I know he's is 100% right) find out that and
are orthogonal? I can't seem to see the big picture.
It's in analogy to normals to planes. If you look at a plane
a normal vector to this plane is . It's 'normal' to every line through the plane. Now if you set you obtain the intersection of the plane with the -plane. A normal vector is , which can be seen as lying in the -plane. All you do now is disregard the third dimension and you get your situation.
i think it's important to note that the equation:
only is a line in a space of two dimensions. in more dimensions, say n, what you get is an affine hyperplane (with n-1 dimensions).
let's look at this from a "high-school perspective".
say we have a line given by:
ax + by = c. because of the peculiarities of x as a "preferred independent variable", we have two cases to consider:
case 1) b = 0.
here, our line is: ax + c = 0. again, we have two cases to consider:
case 1a) a = 0.
in this case, c must be 0, as well, and any (x,y) will suffice (0x + 0y + 0 = 0 is true for any point in the plane). so our "line" is actually the entire plane! however, this case is actually ruled out by taking N to be non-zero (if N = 0 were allowed, then every vector in the plane is indeed orthogonal to the origin. strange, but true).
case 1b) a ≠ 0 (which must be true if b = 0).
here, x = c/a, which is a vertical line. so a tangent vector is (x,0) (for any x) and a normal vector is (0,y) for any y. it is perfectly acceptable to choose x = y = a.
case 2a) a = 0, b ≠ 0. this is similar to case 1b) and you can work out the details yourself. (0,b) is a tangent vector, and (-b,0) is a normal vector (you could use (b,0) as well...at every point on a line, there are "two" normals. usually we pick an orientation of the plane, to choose a "positive" one (the one at 90 degrees, instead of -90 degrees)).
case 2b) a ≠ 0, b ≠ 0. this is the "interesting one".
typically, we define two lines:
y = m1x + b1
y = m2x + b2
to be perpendicular if m1m2 = -1.
so suppose we have our lines as:
a1x + b1y = c1
a2x + b2y = c2, what does perpendicularity mean in terms of the a's,b's and c's?
putting the two lines in slope-intercept form we have:
y = (-a1/b1)x + c1/b1 and
y = (-a2/b2)x + c2/b2
hmm...note that the slopes don't involve "c's" at all, just a's and b's.
thus m1m2 = -1 becomes:
(-a1/b1)(-a2/b2) = -1 that is:
a1a2 = -b1b2 or:
a1a2 + b1b2 = 0.
one often identifies the direction of a line with its slope: so a "parallel" vector to the first line going through the origin lies on the line:
y = m1x = (-a1/b1)x. we can choose any x we please, so let's choose x = -b1.
this gives us a parallel vector (x,y) = (-b1,a1).
now all we need to do is verify that N = (a1,b1) lies on "the perpendicular line":
y = m2x. so we plug in the value x = a1 into this formula:
y = m2a1 = (-a2/b2)a1 = (-a1/b1)(-a2/b2)(-b1)
= (-1)(-b1) = b1.
one caveat: these vectors, T and N actually lie in the "tangent space at the point P on the line L" (which is where the "c" comes in). so technically, we should write:
N = (a,b)P, T = (-b,a)P to indicate the "tails" of the vectors (arrows) start at the point P, rather than at the origin. geometrically, this means translating the origin to the point P on the line.