Announcements
- Assessment Friday, starts on the second week. The quiz will be in class on Friday. You are welcome to use a blank Mathematica notebook or any calculator you wish to perform computations with on the exam. You are not allowed to consult AI. You may use one handwritten page of notes that you've created yourself.
Self-Directed Learning Projects
- Have a specific plan that helps you develop deeper understanding. Think about Bloom's Taxonomy (see 1 and 2), and focus your efforts towards the highest levels.
- Carry out the plan, making modifications as needed (follow new leads, keep in the time constraints, etc.).
- Create something based on what you learned (Bloom's taxonomy).
- Share your work publicly.
- Complete a short steward report to reflect on your learning process.
Group Discussion
Problem set
Did you upload your prep work to Canvas? If not, please do so now.
Briefly discuss which problems from the problem set you would like to talk about as a group.
- Let each person have a turn presenting something from their preparation.
- Take turns leading a discussion about what you have prepared.
- I'll assign groups to write up a nice solution to share with the class.
- If you finish early, it's OK. Work on the practice problems below.
Extra Practice
The goal is to have a discussion about how to do things. This is most easily accomplished when you stick together as a group. Avoid "divide and conquer", which can be effective for getting a bunch of things done, but won't facilitate discussion.
- Give a vector equation of the line that passes through the points $(1,2,3)$ and $(-2,4,9)$ (all distances are in meters, and times in minutes).
- Modify your vector equation from the previous part so that the speed of an object that is tracked with this equation is 3 meters per unit time.
- An object starts at $P=(1,2,3)$ and each unit of time its displacement is 2 units in the direction of $\vec v=(-4,5,1)$. Give an equation for the position $(x,y,z)$ at any time $t$.
- Use the law of cosines ($c^2=a^2+b^2-2ab\cos\theta$ or $\vec u\cdot \vec v = |\vec u||\vec v|\cos\theta$) to find the angle between each pair of vectors below.
- $(-2,1)$ and $(1,3)$.
- $(2,3)$ and $(-1,4)$
- $(\pi,e)$ and $(\sqrt{17},c)$
- $(1,2,3)$ and $(-7,2,1)$
- $(1,2,3)$ and $(x,y,z)$.
Brain Gains
- Give a vector of length 4 that is parallel to the vector $\vec v = (-1,2,-2)$.
Solution
A quick answer is $$\frac{4}{3}(-1,2,-2)=\left(-\frac{4}{3}, \frac{8}{3}, -\frac{8}{3} \right).$$ The length of $\vec v$ is $|\vec v| = \sqrt{(-1)^2+(2)^2+(-2)^2} = \sqrt{9} =3$. A unit vector is then $$\hat v = \frac{\vec v}{|\vec v|} = \frac{(-1,2,-2)}{3}=\left(-\frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \right).$$ The requested vector is then $$\vec w = 4\hat u = \frac{4}{3}(-1,2,-2)=\left(-\frac{4}{3}, \frac{8}{3}, -\frac{8}{3} \right).$$
- Give a vector equation of a line that passes through $(1,2)$ and is parallel to $(3,4)$.
Solution
$\vec r(t)=\left<3,4\right>\,t + \left<1,2\right>$
- Use the law of cosines ($c^2=a^2+b^2-2ab\cos\theta$) to find the angle between the vectors $\vec u = (-1,5)$ and $\vec v = (2,4)$.
Solution
The vectors have lengths $a=|\vec u| = \sqrt{(-1)^2+(5)^2} = \sqrt{26}$ and $b=|\vec v| = \sqrt{(2)^2+(4)^2} = \sqrt{20}$. The difference is $\vec v-\vec u = (3, -1)$, and has length $c = |\vec v - \vec u| = \sqrt{10}$. From the law of cosines, we get $10 = 26+20-2\sqrt{26}\sqrt{10}\cos \theta$. Solving for $\cos\theta$ gives $\cos \theta = \frac{10-26-20}{-2\sqrt{26}\sqrt{10}}$, which means $$\theta = \arccos\left(\frac{10-26-20}{-2\sqrt{26}\sqrt{10}}\right).$$
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