Group Discussion
Problem set
Did you upload your prep work to Canvas? If not, please do so now.
- We will continue doing this. It enables many ways for you to collaborate that aren't possible without the upload.
- I will start putting which problems to work on in the Title of the corresponding discussion board.
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.
Practice
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 $\vec Q$, and then draw $\vec P$, $\vec Q$ and $\ds \text{proj}_\vec Q\vec P$ on the same axes.
- $\vec P = (-2,1)$ and $\vec Q = (1,3)$
- $\vec P = (2,3)$ and $\vec Q = (-1,4)$
- $\vec P = (-5,1)$ and $\vec Q = (2,10)$
Brain Gains
- 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 $\theta = 90^\circ$).
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