A simple vector problem with a rich set of approaches

In precalculus, we do a little bit with vectors. And last year and this year, I gave a basic problem to my kids. (I think I found it in some standard textbook.) What I love about this problem is that there are so many ways kids approached it. All essentially the same, but all different enough. Because my kids weren’t all doing it the same way, it has shown me that we are teaching them well. And also, it reminds me that a super basic problem can be a super rich problem.

Approach 1: Since we were in vector land, a few kids solved it like this. They found the vector from P to Q, and then set the magnitude equal to 5 and did the algebra.

Approach 2: Similarly, some students just used the distance formula that they had memorized.

I like that this kid expanded out $(-3-x)^2$ and then later eventually factored. Because that was slightly different than how the student in approach 1 solved it (leaving $(x+3)^2$ as is, and then taking the square root of both sides.

Approach 3: One student found the equation of a circle of radius 5 around the point $(-3,1)$. Then they realized that they were looking for the solution to a system of equations for the circle and the line $y=4$. So they substituted $y=4$ into the equation of the circle and solved!

Approach 4: Most students took a geometric/visual approach. They drew the point $(-3,1)$ and the line $y=4$. Then they drew these two triangles (seeing that the vertical distance from the point to the line was 3 and the diagonal distance from the point to the line was 5, since we want a distance of 5 away from $(-3,1)$. Then they used 3-4-5 right triangles to get the horizontal distance.

All of these were lovely. I enjoyed seeing them all together and drawing some connections among them. Most kids were awed when they saw Approach 3. And since so many students didn’t take a straight up algebraic approach, they were like “ooooh” when they saw Approach 1/2. I supposed what I most like about this is that it really highlights how circles, distance, and vectors are all essentially tied together. I mean kids should know that circles and distance are fundamentally related (but of course they don’t always remember that). But this problem connects those two concepts with something new: vectors. And that the magnitude of vectors simply being an equation involving a circle, secretly.

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Actually, while I’m writing this, I might as well share this other problem that had a couple of approaches. This was the basic question:

Approach 1: Most students took this approach. They drew the vector, and then drew a smaller vector with unit length, and then used similarity to find this new vector (with a scale factor of $\frac{1}{\sqrt{5}}$.

Related to this were students who simply saw the scalar that was multiplied by the vector $<5,12>$ to be a “scale factor” that stretches/shrinks the vector by a particular factor. But why this works is because of this similar triangles argument.

Approach 2: A bunch of students used trig. They first found “the angle” and then realized that angle put on a unit circle would work! The fact that so many students saw the problem this way made me happy. I then asked what if we wanted the vector to have length 2 or 3 (instead of unit length), and they were able to answer it. We also talked about one huge deficit of this approach: you lose exactitude since they approximated the angle with their calculator. Even if they didn’t round, they wouldn’t “see” the square root of 5 pop out, when they would with the similarity argument.

Approach 3: Okay, so strangely this year, none of my students used this approach. But it is related to the similar triangles approach, and in years past, I’ve had students come up with it. So I showed it to them so they could see another approach. It’s an algebraic approach to find the scale factor.

Fin!