Math Lab and Academic Services
Did you know that you can get a tutor for any class on campus, for free? The university pays the tutors, and wants to support every student with any class. The following student contacted me, and he has hours this semester in the 2-5pm range. I would highly suggest getting a group together that meets either before or after class, and then inviting him to come a couple times a week to join your group. It's one of the best study methods I know of.
- Cameron Rose - ros14011@byui.edu - 702-613-1297
Rapid Recall
1. Give a vector equation of the line that passes through $P=(3,4)$ and $Q=(-2,5)$.
Solution
A vector from $P$ to $Q$ is $\vec {PQ}= (-2-3,5-4) = (-5,1)$, which is a direction vector for the line. Two options for a vector equation of the line are $$(x,y) = (-5,1)t+(3,4) \quad \text{or}\quad (x,y) = (-5,1)t+(-2,5).$$ Any nonzero multiple of the direction vector would work as well, as well as any other form of writing equations of lines. As an example, we could have written this as $$ x\hat i +y\hat j = (-10t+3)\hat i + (2t+4)\hat j \quad\text{or}\quad \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}5\\-1\end{pmatrix}t+\begin{pmatrix}3\\4\end{pmatrix}. $$
2. Compute the dot product of $\vec P$ and $\vec Q$. $\quad$ Bonus: find the angle between $\vec P$ and $\vec Q$
Solution
We compute $$\vec P\cdot \vec Q=(3,4)\cdot(-2,5) = \underbrace{(3)(-2)+(4)(5)}_{\text{sufficient}} = -6+20 = 14.$$ The angle is $$\theta = \cos^{-1}\left(\frac{\vec P\cdot \vec Q}{|\vec P||\vec Q|}\right) = \cos^{-1}\left(\frac{(3)(-2)+(4)(5)}{\sqrt{3^2+4^2}\sqrt{(-2)^2+5^2}}\right) = \cos^{-1}\left(\frac{14}{5\sqrt{29}}\right) .$$
3. Give two vectors that are orthogonal to $\vec P$.
Solution
We need a vector $S=(x,y)$ so $\vec P\cdot \vec S=0$, or in other words $3x+4y=0$. Two options are $(0,0)$ and $(4,-3)$. All other options (since we are in 2D) are a multiple of $(4,-3)$, so you might have $(-4,3)$ or $(-8,6)$ as well as many other options.
4. 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 $\vec Q$.
Solution
We compute $$\ds \text{proj}_\vec Q\vec P = \frac{\vec P\cdot \vec Q}{\vec Q\cdot \vec Q}\vec Q = \frac{(3)(-2)+(4)(5)}{(-2)^2+5^2}(-2,5) = \frac{14}{29}(-2,5) .$$ Note that this is a little less than half $\vec Q$.
Group problems
Let $P=(3,4)$ and $Q=(-2,5)$.
- Draw $\vec P$, $\vec Q$ and $\text{proj}_\vec Q\vec P $ on the same grid, all with their base at the origin. Try your best to give the $x$ and $y$ directions the same scale, otherwise you won't be able to see the connections among vectors.
- Add to your picture the vector difference $\vec P - \text{proj}_{\vec Q}\vec P$. What is the angle between this difference and $\vec Q$.
- Now compute the projection of $\vec Q$ onto $\vec P$.
- Then draw $\vec P$, $\vec Q$ and $\text{proj}_\vec P\vec Q $ on the same grid, all with their base at the origin, and add to your picture the vector difference $\vec P - \text{proj}_{\vec P}\vec Q $.
- How much work is done by $\vec P$ through a displacement $\vec Q$?
- For the last problem, start by drawing the following figure:
- place the lower left corner of a regular hexagon at the origin, with the bottom side along the $x$-axis.
- label the bottom side $\vec a=\langle 1,0\rangle$, the lower left side $\vec b$, and the lower right side $\vec c$, (with the tail of $\vec c$ at the tip of $\vec a$.
- label the interior angle between the bottom side and the lower right side as $\theta$.
- Find $|\vec a|$, $|\vec b|$, $|\vec c|$, and $\theta$. (hint: use geometry)
- Compute $\vec a \cdot \vec c$
- Compute $\vec a \cdot \vec b$
- Draw $\text{proj}_{\vec a} \; \vec b$ and $\text{proj}_{\vec a}\; \vec c$.
- Draw $\text{proj}_{\vec b}\; \vec c$.
- Find the $x$ component of $\vec a + \vec b + \vec c$
- Draw two random vectors on your chalk board, with their base at the same point. Label them $\vec u$ and $\vec v$. Then draw both $\text{proj}_\vec v\vec u $ and $\text{proj}_\vec u\vec v $.
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