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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Rapid Recall
- Give a vector of length 10 that is parallel to the vector $\vec v = (4,-3)$.
Solution
The length of the original vector is 5. A vector of length 10 parallel to the original is $$\frac{10}{5}(4,-3)=(8,-6).$$
- Let $\vec u = (2,3,-1)$ and $\vec v = (-4,1,7)$. Compute the dot product $\vec u\cdot \vec v$.
Solution
The dot product is $$ \vec u\cdot\vec v = (2,3,-1)\cdot(-4,1,7) = -8+3-7 = -12. $$
- Using the same vectors as above, compute $\vec u\cdot \vec u$ and the magnitude $\left|\vec u\right|$.
Solution
The dot product is $$ \vec u\times\vec u = (2,3,-1)\cdot(2,3,-1) = 4+9+1 = 14. $$ The magnitude is $$ |\vec u|=\sqrt{4+9+1} = \sqrt{14}. $$
- What is the angle between $\vec u = (1,2,3)$ and $\vec v = (1,1,-1)$.
Solution
The dot product gives $\vec u\cdot\vec v = 1+2-3=0$. Since the dot product is zero, and the law of cosines in dot product form is $\vec u\cdot\vec v=|\vec u||\vec v|\cos\theta$, we know that $0 = |\vec u||\vec v|\cos\theta$. The only way the product on the right can be zero is if either one of the vectors has zero length (not true), or we have $\cos\theta = 0$. Since the latter is true, this means $\theta =\pi/2$ (or alternately $\theta = 90^\circ$).
Group problems (Jamboards)
The law of cosines, in dot product form, is $$\vec u\cdot \vec v = |\vec u||\vec v|\cos\theta.$$
- Find the angle between each pair of vectors below, using the dot product law above.
- $(-2,1)$ and $(1,3)$
- $(2,3)$ and $(-1,4)$
- $(-5,1)$ and $(2,10)$
- $(1,2,3)$ and $(-7,2,1)$
- $(1,2,3)$ and $(x,y,z)$.
- Give a set of values for $x,y,z$ above so that the angle is 90 degrees. Then give another. Then give another.
- The (vector) projection of $\vec P$ onto $\vec Q$ is given by the formula $$\ds \text{proj}_\vec Q\vec P = \frac{\vec P\cdot \vec Q}{\vec Q\cdot \vec Q}\vec Q.$$ Compute the projection of $\vec P$ onto $\vec Q$.
- $\vec P = (-2,1)$ and $\vec Q = (1,3)$
- $\vec P = (2,3)$ and $\vec Q = (-1,4)$
- $\vec P = (-5,1)$ and $\vec Q = (2,10)$
- If you ever finish early as a group, you can use the link below to open up the problem set and discuss some of the problems due tomorrow.
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