Inverse Functions: Arctangent and Angle Between Two Lines (Part 24)

Here’s a straightforward application of arctangent that, a generation ago, used to be taught in a typical Precalculus class (or, as it was called back then, analytical geometry).

Find the smallest angle between the lines y= 3x and y = -x/2.

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This problem is almost equivalent to finding the angle between the vectors \langle 1,3 \rangle and \langle -2,1 \rangle. I use the caveat almost because the angle between two vectors could be between 0 and \pi, while the smallest angle between two lines must lie between 0 and \pi/2.

This smallest angle can be found using the formula

\theta = \displaystyle \tan^{-1} \left( \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \right),

where m_1 and m_2 are the slopes of the two lines. In the present case,

\theta = \tan^{-1} \left( \left| \displaystyle \frac{ 3 - (-1/2) }{1 + (3)(-1/2)} \right| \right)

\theta = \tan^{-1} \left( \left| \displaystyle \frac{7/2}{-1/2} \right| \right)

\theta = \tan^{-1} 7

\theta \approx 81.87^\circ.

Not surprisingly, we obtain the same answer that we obtained a couple of posts ago using arccosine. The following picture makes clear why \tan^{-1} 7 = \cos^{-1} \displaystyle \frac{1}{\sqrt{50}}.

dotproduct4In tomorrow’s post, I’ll explain why the above formula actually works.

 

Inverse Functions: Arccosine and Dot Products (Part 23)

The Law of Cosines can be applied to find the angle between two vectors {\bf a} and {\bf b}. To begin, we draw the vectors {\bf a} and {\bf b}, as well as the vector {\bf c} (to be determined momentarily) that connects the tips of the vectors {\bf a} and {\bf b}.

dotproduct2Using the usual rules for adding vectors, we see that {\bf a} + {\bf c} = {\bf b}, so that {\bf c} = {\bf b} - {\bf a}

We now apply the Law of Cosines to find \theta:

\parallel \! \! {\bf c} \! \! \parallel^2 = \parallel \! \! {\bf a} \! \! \parallel^2 + \parallel \! \! {\bf b} \! \! \parallel^2 - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta

\parallel \! \! {\bf b} - {\bf a} \! \! \parallel^2 = \parallel \! \! {\bf a} \! \! \parallel^2 + \parallel \! \! {\bf b} \! \! \parallel^2 - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta

We now apply the rule \parallel \! \! {\bf a} \! \! \parallel^2 = {\bf a} \cdot {\bf a}, convert the square of the norms into dot products. We then use the distributive and commutative properties of dot products to simplify.

( {\bf b} - {\bf a} ) \cdot ({\bf b} - {\bf a}) = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta

{\bf b} \cdot ({\bf b} - {\bf a}) - {\bf a} \cdot ({\bf b} - {\bf a}) = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta

{\bf b} \cdot ({\bf b} - {\bf a}) -{\bf a} \cdot ({\bf b} - {\bf a}) = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta

{\bf b} \cdot {\bf b} - {\bf a} \cdot {\bf b} - {\bf a} \cdot {\bf b} + {\bf a} \cdot {\bf a} = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta

{\bf b} \cdot {\bf b} - 2 {\bf a} \cdot {\bf b} + {\bf a} \cdot {\bf a} = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta

We can now cancel from the left and right sides and solve for \cos \theta:

- 2 {\bf a} \cdot {\bf b} = - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta

\displaystyle \frac{ {\bf a} \cdot {\bf b} }{ \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel } = \cos \theta

Finally, we are guaranteed that the angle between two vectors must lie between 0 and \pi (or, in degrees, between 0^\circ and 180^\circ). Since this is the range of arccosine, we are permitted to use this inverse function to solve for \theta:

\cos^{-1} \left( \displaystyle \frac{ {\bf a} \cdot {\bf b} }{ \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel } \right) = \theta

The good news is that there’s nothing special about two dimensions in the above proof, and so this formula may used for vectors in \mathbb{R}^n for any dimension n \ge 2.

In the next post, we’ll consider how this same problem can be solved — but only in two dimensions — using arctangent.

Inverse Functions: Arccosine and Dot Products (Part 22)

Here’s a straightforward application of arccosine, that, as far as I can tell, isn’t taught too often in Precalculus and is not part of the Common Core standards for vectors and matrices.

Find the angle between the vectors \langle 1,3 \rangle and \langle -2,1 \rangle.

dotproductThis problem is equivalent to finding the angle between the lines y = 3x and y = -x/2. The angle \theta is not drawn in standard position, which makes measurement of the angle initial daunting.

Fortunately, there is the straightforward formula for the angle between two vectors {\bf a} and {\bf b}:

\theta = \cos^{-1} \left( \displaystyle \frac{ {\bf a} \cdot {\bf b} }{ \parallel \!\! {\bf a} \!\! \parallel \parallel \!\! {\bf b} \!\! \parallel } \right)

We recall that {\bf a} \cdot {\bf b} is the dot product (or inner product) of the two vectors {\bf a} and {\bf b}, while \parallel \!\! {\bf a} \!\! \parallel = \sqrt{ {\bf a} \cdot {\bf a} } is the norm (or length) of the vector {\bf a}.

 For this particular example,

\theta = \cos^{-1} \left( \displaystyle \frac{\langle 1,3 \rangle \cdot \langle -2,1 \rangle }{ \parallel \!\!\langle 1,3 \rangle \!\! \parallel \parallel \!\!\langle -2,1 \rangle \!\! \parallel } \right)

\theta = \cos^{-1} \left( \displaystyle \frac{ (1)(-2) + (3)(1) }{ \sqrt{ (1)^2 + (3)^2} \sqrt{ (-2)^2 + 1^2 }} \right)

\theta = \cos^{-1} \displaystyle \frac{1}{\sqrt{50}}

\theta \approx 81.87^\circ

In the next post, we’ll discuss why this actually works. And then we’ll consider how the same problem can be solved more directly using arctangent.

Engaging students: Dot product

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Candace Clary. Her topic, from Precalculus: computing a dot product.

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  1. How can this topic be used in your students’ future courses in mathematics or science?

 

The dot product in algebra is defined as the magnitude and direction of two different vectors, multiplied together. After algebra, the students will start working with vectors. In calculus they will start seeing vectors and finding cross products and dot products of those vectors. Once they get to a linear algebra class, they will begin to work with matrices. Matrices can be seen as vectors, and the dot product of these can then be computed. The dot product can also be used in geometry. The dot product is in geometry can be used to find the angle between two vector, and it can be used to find the length of a vector, with the angle in between known. Computing the dot product of vectors requires the students to remember things like order of operations, and how to multiply several numbers. Knowing how to compute a dot product can help students in physics classes, chemistry classes, and other types of science classes.

 

 

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How could you as a teacher create an activity or project that involves your topic?

One activity that I could do as a teacher is by using big sheets of graphing paper. I can ask the students to work in pairs, and have them draw vectors on a piece of poster board graph paper. They would need to draw three or more vectors, and label them to let other students know what their vectors are. After they have drawn three or more, they will pass it to another group. These groups will then determine the dot product of the vectors that were drawn. They will be required to show their work on the side, neatly, and be able to explain how they got their answers. After the work has been completed, they will need to graph the dot products of the vectors in a different color. Once all the groups are done, the posters will be hung around the room and the class will take a gallery walk to looks at the posters and take notes on the solutions so they are able to see it many times. These posters will then stay up in the classroom for most of the unit for reference.

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1. How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic? Note: It’s not enough to say “such-and-such is a great website”; you need to explain in some detail why it’s a great website.

This video on YouTube will be great to engage the kids. This teacher intrigues me, he is so hyper when it comes to math and really explains it in a simple way to understand. In this video, he breaks the topic down and shows many different ways to compute the dot product of a vector. I like the fact that he states the properties of the vectors before he starts to talk about computing them. I also like that he keeps them up on the board and on the screen while he uses a numerical example. He also shows how we can use the dot product to find the angle between two vectors. He does this in the second part of the video, which means I can cut the video where I need, depending on what topic I am teaching. I think that this teacher does a great job of explaining, and even though this is an educational video, where material is taught, I think kids will learn from it.

 

Engaging students: Vectors in two dimensions

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Derek Skipworth. His topic, from Precalculus: vectors in two dimensions.

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A. How could you as a teacher create an activity or project that involves your topic?

While it may be a cop-out to use this example since I am developing it for an actual lesson plan, I will go ahead and use it because I feel it is a strong activity.  I am developing a series of 21 problems that will be the base for forming the students’ treasure maps.  There will be three jobs: Cartographer, the map maker; Lie Detector, who checks for orthogonality; and Calculator, who will solve the vector problems.  The 21 problems will be broken down into 7 per page, and the students will switch jobs after each page.  The rule is that any vectors that are orthogonal with each other cannot be included in your map.  There are three of these on each page, so each group should end up with a total of 12 vectors on their map.  Once orthogonality is checked by the Lie Detector, the Calculator will do the expressed operations on the vector pairs to come up with the vector to be drawn.  The map maker will then draw the vector, as well as the object the vector leads to.  Each group will have their directions in different orders so that every group has their own unique map.  The idea is for the students to realize (if they checked orthogonality correctly) that, even though every map is different, the sum of all vectors still leads you to the same place, regardless of order.

 

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B. How does this topic extend what your students should have learned in previous courses?

Vectors build upon many topics from previous courses.  For one, it teaches the student to use the Cartesian plane in a new way than they have done previously.  Vectors can be expressed in terms of force in the x and y directions, which result in a representation very similar to an ordered pair.  It gets expanded to teach the students that unlike an ordered pair, which represents a distinct point in space, a vector pair represents a specific force that can originate from any point on the Cartesian Plane.

Vectors also build on previous knowledge of triangles.  When written as \langle x,y \rangle, we can find the magnitude of the vector by using the Pythagorean Theorem.  It gives them a working example of when this theorem can be applied on objects other than triangles.  It also reinforces the students trigonometry skills since the direction of a vector can also be expressed using magnitude and angles.

 

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E. How can technology be used to effectively engage students with this topic?

The PhET website has one of the best tools I’ve seen for basic knowledge of two dimensional vector addition, located at http://phet.colorado.edu/en/simulation/vector-addition.  This is a java-based program that lets you add multiple vectors (shown in red) in any direction or magnitude you want to get the sum of the vectors (shown in green).  Also shown at the top of the program is the magnitude and angle of the vector, as well as its corresponding x and y values.

What’s great about this program is it puts the power in the student’s hands.  They are not forced to draw multiple sets of vectors themselves.  Instead, they can quickly throw them in the program and manipulate them without any hassle.  This effectively allows the teacher to cover the topic quicker and more effectively due to the decreased amount of time needed to combine all vectors on a graph.