Brain Gains
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 $\vec Q$. Then begin finding 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 $\vec Q$.
5. How much work is done by $\vec P$ through a displacement $\vec Q$?
Solution
The work done by $\vec P$ through a displacement $\vec Q$ is just the dot product $$\vec P\cdot \vec Q=(3,4)\cdot(-2,5) = \underbrace{(3)(-2)+(4)(5)}_{\text{sufficient}} = -6+20 = 14.$$
Group problems
Let $P=(3,4)$ and $Q=(2,0)$.
- Using the formula $$\ds \text{proj}_\vec Q\vec P = \frac{\vec P\cdot \vec Q}{\vec Q\cdot \vec Q}\vec Q,$$ compute $\text{proj}_\vec Q\vec P$, the project of $\vec P$ onto $\vec Q$. We may also write this as $\vec P _{\parallel \vec Q}$, which we read as, "The vector component of $\vec P$ that is parallel to $\vec Q$."
- Draw $\vec P$, $\vec Q$ and $\text{proj}_\vec Q\vec P $ on the same grid with their base at the origin.
- Add to your picture the vector difference $\vec P_{\perp \vec Q} = \vec P - \text{proj}_{\vec Q}\vec P$, which we call "The vector component of $\vec P$ that is orthogonal to $\vec Q$." How is $\vec P_{\perp \vec Q}$ related to the other vectors?
- Now compute the projection of $\vec Q$ onto $\vec P$ (so swap which vector is projected onto the other).
- Draw $\vec P$, $\vec Q$ and $\text{proj}_\vec P\vec Q $ on the same grid with their base at the origin, and add to your picture the vector difference $\vec Q_{\perp \vec P} = \vec Q - \text{proj}_{\vec P}\vec Q $.
- How much work is done by $\vec P$ through a displacement $\vec Q$?
If you finish the above, then repeat it with $P=(3,4)$ and $Q=(-2,5)$.
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