You can access this page from the homepage in I-Learn.
Pre-Class Chatter
Feel free to turn on your mic and chat with people or use the zoom chat. Or chat with your group in your Jamboard.
9:00 AM Jamboard Links
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12:45 PM Jamboard Links
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Announcements
Hot-Chocolate Break Room
- Looking for SDL partners?
- Hoping to find a study group (in person or via Zoom)? You can each host your own Zoom meetings without faculty.
- Share anything with the class.
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.
Rapid Recall
- 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 $$\theta = \arccos\left(\frac{10-26-20}{-2\sqrt{26}\sqrt{10}}\right).$$
Group problems
Remember to take turns writing on your group's Jamboard (links at the top of the page). 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 point $(1,2,3)$ 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 $(x,y,z)$.
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