SUNY Geneseo Department of Mathematics
Tuesday, February 14
Math 223
Spring 2023
Prof. Doug Baldwin
The “linearity” question in problem set 3?
The question asks whether the “magnitude” operation on vectors is “linear.” From the definition of “linear” in the problem, this boils down to answering 2 questions. If both answers are “yes,” then magnitude is linear:
There are 2 general ways you can go about answering. If you think the answers are “yes,” then you could try to use the definition of “magnitude” and some algebra to show that the two equations defining linearity hold. Or if you think either answer is “no,” you could just show a “counter-example,” i.e., a pair of vectors or a vector and a scalar for which one of the linearity equations doesn’t hold.
Before you do either of these things, I recommend just trying some examples to get a feel for what the definition and its equations say. Based on these examples, you can decide whether you want to try to prove that the equations always hold, or exhibit a counter-example (which might end up being one of the examples you tried). Experimenting with a few examples is a good way to help yourself understand many things in math.
Based on “Equations for a Line in Space,” “Distance between a Point and a Line,” and maybe a little bit from “Relationships between Lines” in section 1.5 of the textbook.
Lines can be described by a vector equation, parametric equations, or symmetric equations (which are all basically alternative forms of the same thing).
There’s an equation for finding the distance between a point and a line. See below for a derivation and example of it.
Example of how to find the distance between a line and a point?
How far is point (1, 1, 0) from the line r(t) = (1, 2, 1) + t⟨-2, 1, -1⟩?
The distance from a point to a line is the distance in a direction perpendicular to the line. So if you drew a vector from any point on the line (maybe its reference point, for example) to the remote point, and that vector made angle a with the line, the length of the vertical would be the magnitude of the vector times sin a:
And one of the neat rules about cross products is that their magnitudes are equal to the product of the magnitudes of the vectors involved times the sine of the angle between them. So the cross product of the vector to the point and the line’s direction vector is almost the distance we want, except it’s too large by a factor of the magnitude of the direction vector. Fix that problem by dividing by that magnitude:
Now try using this formula to find the actual distance in this problem. There’s a fair bit to calculate (e.g., the vector to the point, a cross product, some magnitudes), but they’re all calculations we’ve seen before:
Recognizing intersecting and skew lines?
Find the intersection, if any, between lines r(t) = (1, 2, -1) + t⟨2, -1, 0⟩ and s(t) = (7, -1, -1) + t⟨0, 1, 2⟩.
The key idea is to find a point, if any, that’s on both lines. At such a point, r and s are equal — but be careful that the parameter values that make them equal might be different, so use different names for the parameter in each to avoid confusion.
To solve this equation, break it down into separate components, which gives you a system of equations:
Now use the third equation to solve for u, and the first to solve for t:
Now it might seem that we’re done, but not quite: we’ve found values for t and u that make the x and z coordinates equal, but we don’t know that the y coordinates are also equal at those values. So plug t and u into the second equation and check that it really holds:
It does, so we’ve discovered that the lines intersect. Plugging t into the equation for r and u into the equation for s finds the point of intersection:
If t and u hadn’t also solved the second equation in this example, we would know that the lines are skew. And if the lines were parallel, we’d be able to find some constant that we can multiply one direction vector by to get the other (but we can’t).
Planes!
Please read “Equations for a Plane” in section 1.5 of the textbook.