1.4 The Dot Product (continued)

Projections

Work, Force, and Displacement

We know from the previous subsection that there is a third configuration of vectors, which occurs when $\vu$ and $\vv$ are orthogonal. Suppose that $\vu$ and $\vv$ are nonzero orthogonal vectors. What would $\vw_1$ and $\vw_2$ be in this case?

We want to switch the roles of $\vu$ and $\vv$ for the examples in the previous parts. Specifically, for these configuration of vectors, we want to split $\vv$ into parts that are parallel to $\vu$, which we will call $\vz_1$, and orthogonal to $\vu$, which we will call $\vz_2$. On , draw $\vz_1$ and $\vz_2$ for each configuration.

Based on your drawing, is it the case that $\vw_1 = \vz_1$? What about $\vw_2$ and $\vz_2$?

Explain why $\vu=\vw_1 + \vw_2$.

Compute $\vu\cdot \vv$ as $(\vw_1+\vw_2)\cdot \vv$. Simplify your answer as much as possible, using the fact that $\vw_2$ is orthogonal to $\vv$.

Since $\vw_1$ is parallel to $\vv$, there is a scalar $k$ so that $\vw_1 = k\vv$. Substitute $k\vv$ for $\vw_1$ in your answer to the previous part and then solve for $k$.

Give a formula for $\vw_1$ purely in terms of $\vu$ and $\vv$.

Work through this exercise and explain your reasoning step by step.

Let $\vv = \langle 4, -8 \rangle$. Find $\proj_{\vv} \vu$, $\comp_{\vv} \vu$, and $\proj_{\perp \vv} \vu$. Draw a picture to illustrate the vectors involved. Finally, express $\vu$ as the sum of two vectors where one is parallel to $\vv$ and the other is perpendicular to $\vv$.

Now let $\vw = \langle -2,4 \rangle$. Add $\vw$ to the picture you drew in the previous part. Without doing any calculations, find $\proj_{\vw} \vu$. Explain your reasoning.

Find a vector $\vw$ not parallel to $\vz = \langle 3,4 \rangle$ such that $\proj_{\vz} \vw$ has length $10$. Note that there are infinitely many different answers!

Determine the work done by a 25 pound force acting at a $30^{\circ}$ angle to the direction of the object's motion, if the object is pulled 10 feet.

Determine if more work or less work is done if the angle to the direction of the object's motion is $60^\circ$.

When running a sprint, the racers may be aided or slowed by the wind. The wind assistance is a measure of the wind speed that is helping push the runners down the track. It is much easier to run a very fast race if the wind is blowing hard in the direction of the race. So that world records aren't dependent on the weather conditions, times are only recorded as record times if the wind aiding the runners is less than or equal to 2 meters per second. Wind speed for a race is recorded by a wind gauge that is set up close to the track. It is important to note, however, that weather is not always as cooperative as we might like. The wind does not always blow exactly in the direction of the track, so the gauge must account for the angle the wind makes with the track. Suppose a 4 mile per hour wind is blowing to aid runners by making a $38^{\circ}$ angle with the race track. Determine if any times set during such a race would qualify as records.

Molecular geometry is the geometry determined by arrangements of atoms in molecules. Molecular geometry includes measurements like bond angle, bond length, and torsional angles. These attributes influence several properties of molecules, such as reactivity, color, and polarity.

As an example of the molecular geometry of a molecule, consider the methane $\text{CH}_4$ molecule, as illustrated in Figure. According to the Valence Shell Electron Repulsion (VSEPR) model, atoms that surround single different atoms do so in a way that positions them as far apart as possible. This means that the hydrogen atoms in the methane molecule arrange themselves at the vertices of a regular tetrahedron. The bond angle for methane is the angle determined by two consecutive hydrogen atoms and the central carbon atom. To determine the bond angle for methane, we can place the center carbon atom at the point $\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right)$ and the hydrogen atoms at the points $(0,0,0)$, $(1,1,0)$, $(1,0,1)$, and $(0,1,1)$. Find the bond angle for methane to the nearest tenth of a degree.