20200923

The image in the cave looks all black to the human eye. However, lots of materials look black when used. These cave markings, detailing a map of the surrounding area, could have been made all at once, or by different groups of people as more information was learned. Is there a way to determine if the materials used to make the map are all the same?

We use cameras designed to capture wave lengths that we can't see, and take 6 different snap shots (capturing different types of information). For each pixel in the picture, the cameras record six numbers. We can think of these numbers as a 6D vectors, such as the 3 simplified vectors below.

• $\vec v_1 = (2, 1, 5, 3, 4, 2)$
• $\vec v_2 = (4, 2, 10, 6, 8, 4)$
• $\vec v_3 = (4, 2, 1, 3, 6, 5)$

How do we recognize if two vectors represent the same material? First, the numbers can increase proportionally with brightness. Double the brightness, and the numbers coming back can double. So the same material may register as different vectors.

• Can you tell that $v_2$ is twice $\vec v_1$?

We can use the law of cosines $$\vec u\cdot\vec v = |\vec u||\vec v|\cos\theta$$ to compute the angle between each six dimensional vector. The engineering team which worked on this project used a method called $k$-means clustering to group vectors together based off these angles, and determined that several different materials were used.

We'll do this one together, and look at a visual solution using Mathematica and a file from I-Learn 215-Tech-Introduction.nb.

The (vector) component of $\vec F$ that is parallel to $\vec d$. Note that this vector is just the projection of $\vec F$ onto $\vec d$, so $\text{proj}_{\vec d}\vec F$.

The length of this vector is sometimes called the scalar component of $\vec F$ that is parallel to $\vec d$.

The vector component of $\vec F$ that is orthogonal to $\vec d$.

Since $\vec F_{\parallel \vec d}$ and $\vec F_{\perp \vec d}$ must sum to $\vec F$, then $\vec F_{\perp \vec d} = \vec F - \vec F_{\parallel \vec d} = (6,7)-(2,-1) = (4,8)$.

Note that $\vec F_{\parallel \vec d} \cdot \vec F_{\perp \vec d}=0$, which should be the case (these two vectors are always orthogonal).

The work done by $\vec F$ through $\vec d$ is the dot product $$\vec F\cdot \vec d = |\vec F||\vec d|\cos\theta = (10 N)(5 m)\cos 0 = 50 Nm.$$ This is one of the key application of the dot product.

Note that when the direction of the force and displacement are the same, then $\cos \theta = 1$, which gives the work as simply $W = |\vec F||\vec d|$. This is the most common introductory explanation in a physics course. In general, we have $$W = |\vec F||\vec d|\cos \theta = \vec F\cdot \vec d.$$ The dot product was created around 1900, with one benefit being that it simplifies work computations, precisely when the force and displacement are NOT in the same direction.

Work is given as the dot product of $\vec F$ and $\vec d$. Thus, $W=(3,5)\cdot(-2,1)=-6+5=1$ Nm .