Here are some pictures of quadrics.
The first one shows the quadric x2 + y 2 − z 2 = 1, which is a hyperboloid of
one sheet, together with a line through (5, 5, 7) (which is on the surface because
25 + 25 − 49 = 1) with direction vector v = h3, 4, 5i. The points on this line
have the form (5 + 3t, 5 + 4t, 7 + 5t) and they’re all on the surface. That is, pick
a t, any t, take x = 5 + 3t, y = 5 + 4t, z = 7 + 5t, and compute x2 + y 2 − z 2 .
You’ll get 1. It’s like a magic trick.
But magic is just another word for ‘Hey, wait a minute what’s going on?!’
There’s a reason, and we’re in the business of getting and giving reasons.
Algebraically, if you square out (5 + 3t)2 + (5 + 4t)2 − (7 + 5t)2 , you get
25 + 30t + 9t2 + 25 + 40t + 16t2 − 49 − 70t − 25t2 . Collecting terms, that all
simplifies to...25 + 25 − 49 + 30t + 40t + 9t2 + 16t2 − 25t2 = 1 + 0t + 0t2 = 1.
There’s a deeper question here. How would you find the a, b, and c given the
point (5, 5, 7) and the requirement that the line lie in the surface? The short
answer is that you’d write down the algebraic form of what you were asking
for, set out another requirement such as a + b = 1 to avoid infinitely many
equivalent direction vectors for your line and focus on just one, and then grind
out the algebra. The long answer involves executing the grinding. One of my
hopes for the course is that you’ll get to where you can turn the short answer,
which is always ”figure out what you want, translate it into some more suitable
form, and work out the implications”, into the long answer.
Moving on to the next surface, let’s look at the paraboloid x2 − y 2 = z. This
is one of those saddles we talked about (or will have talked about) in class. The
point (2, 1, 3) is on the surface, because if you put x = 2, y = 1, and z = 3 and
compute x2 − y 2 , you get 4 − 1 = 3 = z. So, you want a line through (2, 1, 3)
so that any point on that line will be on the surface? Not just any surface
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will allow for such a thing. Spheres, for instance, do not have lines that run
along them. But you can always write down the algebra for what would have
to happen, and see if there’s a way to choose the details so that it does happen.
Here, that means looking for (a, b, c) so that 2 + at = x, 1 + bt = y, and
3+ct = z works. Taking a = 1, b = 1, and c = 2 emerges from the calculation as
one option. So the line r = h2, 1, 3i + th1, 1, 2i lies entirely on this saddle-shaped
quadric surface. Easter Egg!
And here’s the picture.
Note: In an earlier edition of this page, x2 + y 2 − z 2 had a typo.
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