Hello gentlemen - Marquette University High School



Hello gentlemen. I hope that you are enjoying the weather. The storm sure is beautiful, looking at it from inside a warm house through an insulated window.

I’d like to review the material that we covered in class yesterday:

Vectors in the Plane (6.3).

I started the homework in class. Here I’d like to provide additional examples for you to look at. I’ll begin on page 417 of the text.

In this set of exercises, we are asked to write the component form of the vector. This means, more or less, that we are supposed to find the coordinates that the vector stops at when it begins at the origin.

Three and 4 are simple examples. Four’s harder, let’s start there. We can see from the graph that the initial point is at the origin, and that the terminal point is (4, –2). We can also see that the vector terminates (that is ends) in Quad IV. This tells us the signs for the component form are < +, – >.

So now we need to find < (x, (y >. Recall that (x means the change in x’s or the difference of the x’s. Notation for component form of a vector uses angles brackets instead of parenthesis.

So the component form of this vector is < 4 – 0, -2 – 0> or .

Now let’s find the magnitude. Magnitude is length, more or less. The text does provide some “formal” formulas, but I’d rather that you consider what you already know, that is the Pythagorean Theorem. (In any right triangle, square each leg and add the squares. This sum is the hypotenuse squared.) “But wait! There’s no right triangle!!” said Stu Dent. “Well, it’s on the Cartesian Plane… Draw one!” I reply.

If we draw a line from the terminal point to either axis, then we form a right triangle. Since we know the component form of the vector, , we know the length of the horizontal side, that is 4, and the length of the vertical side, that is 2. So, [pic][pic][pic].

Let’s try a harder one, like 6.

We need to find the component form first, which means that we need to move the vector so it starts at the origin (see red vector on the exercise at right).

Now you can just look at this image and see that the vector, in component form, is shifted one up and one to the left. Which you can possibly visualize -- now that the initial point is at the origin -- the terminal point is now (4, 6), which means the component form is . Again, the drawing confirms that the terminal point is in the Quadrant I, so the signs are +, +. Using the formula: or .

To find the magnitude, use your prior knowledge of the Pythagorean Theorem. Hence, construct a right triangle, as shown at left.

The change in the x’s (or horizontal distance) is 4. The change in y’s (or vertical distance) is 6. We can confirm this by comparing these values to the component form. Now, [pic] [pic]. So ||v|| = [pic]

If you don’t have a graph, you could always draw one. It’s legit to draw one. Or you could work through the process with a little more thought without a picture. Let’s look at 9.

So, component form is , or , or rewritten with common denominators < (-4-5)/2, (-3-2)/2>. Simplify and then the component form is < -9/2, -5/2 >.

Now for the magnitude: [pic][pic][pic][pic][pic].

[pic]

[pic]

13. –v (shown in red above). 14 is 3u, and is shown below. The u vector was measured, and then lined up 3 times, so that the resulting vector is 3u (the thin red line).

[pic]`

15. u + v

Use the 2 given vectors to draw a parallelogram. The diagonal that has an endpoint at the origin is the sum of the vectors, that is the red vector shown is the sum.

[pic]

16. u – v. The –v vector is shown in blue. Then draw the parallelogram and add u and –v. The solution is shown in red.

[pic]

My biggest hint to you with working with these exercises is “don’t over-think it, just be careful”.

Vector Operations

Operations in math are (for example) addition, subtraction, multiplication, and division.

19. Let u = and v =

a. Find u + v.

To find the new vector, add the x’s and then add the y’s. Order doesn’t matter because addition is commutative. Hence u+v = = .

b. Find u – v.

Well if you find –v first, then order won’t matter. Subtraction is adding the opposite. To find –v, use the scalar of -1: -1 = . Then = .

c. Find 2u – 3v.

Find the scaled vectors first. Hence 2u = 2 = and -3v = -3, . Now add 2u and -3v: = .

d. Find v + 4u.

Again, find the scaled vector first: 4u = 4 = . Then add it to v: = .

Exercise 23 is written in what is called a “linear combination”. Use the coefficients and your knowledge of the component form. The linear combination of u = i + j is equal to the component form of u = . The linear combination of v = 2i – 3j is the same as the component form v = .

Unit Vectors

A unit vector has a magnitude of 1 unit. So to find a unit vector in the direction of the given vector, just use a scalar that is the reciprocal of the magnitude. For example, exercise 28: . If we remember that this is the smallest integer Pyth. Triple, then we know the magnitude is 5. If not, then we can use the Pythagorean Theorem, or a sketch, or both. The opposite reciprocal of this magnitude is 1/5. So the unit vector is 1/5, or .

To find a vector in the same direction of a given vector, but with a new magnitude:

1. Find the unit vector of the directional vector.

2. Use the given magnitude as a scalar.

Let’s look at 38. Finding the unit vector of u = 2i – 3j or u = . I need to find the reciprocal of the magnitude: [pic]. So the unit vector of u is [pic]. I could clean this up now (by rationalizing the denominator and distributing the scalar value), or later. Since I know the next step is just distributing the new scalar, I’ll wait.

The desired magnitude is 10. So [pic][pic][pic].

On a personal note, it really takes a long time to scan and type/write all this stuff up. Please let me know (1) if you read it, and (2) if it was any help.

Linear Combination

1. Find component form

2. Rewrite as linear combination

Exercise 42.

Step 1: = .

Step 2: u = 3i +8j.

Hmmm... I think I’ll use Geometer’s SketchPad for the sketch (only because I can).

Which one? Hmmm... 49 looks wicked, let’s do that one.

Analytic solution: Let’s see u = and w = . If v = ½(3u +w), then I should probably find 3u first. So 3u = 3 = . Now v = ½ = ½ = [pic]

Geometric solution: Let’s see. I need [pic] and [pic]. So, [pic] = [pic],

and [pic].

[pic]

Let’s look at 52. What this exercise is asking me is what are the coordinates for a segment that makes a 135º angle and is 8 units long. But wait, then the magnitude is 8, and the direction angle is 135º. Wasn’t too much to do there. (

Let’s try another, like 54. v=< -4, -7>, so a graph looks like:

So [pic][pic][pic].

The angle direction, using our right-triangle trig knowledge can be found using the reference angle and the 3rd Quad. I used GSP to find the measure of the reference angle (at right)…

Let’s find it analytically.

[pic]. I’m using Graphcalc for the calculation:

arctan (7/4)

1.05165021254838

ans(radians->degrees)

60.2551187030581

ans+180

240.255118703059

So the direction angle is 240.255º.

Let’s look at 64. Geometric Solution: Analytic Solution:

Referring to the unit circle, we know that the intersection between a 30º angle and the unit circle are [pic], and the intersection of a 90º angle and the unit circle are [pic]. These coordinates are actually the values for the unit vector made by the given angle. Now we just need to multiply by the scalar, that is the magnitude, such that [pic] =[pic] and [pic]. Now we add them to find the sum: [pic][pic].

The analytic and geometric solutions are the same.

Let’s look at 67. Quick sketch:

[pic]Ok, maybe this wasn’t the best exercise. It’s easy to see (on the sketch) that since the x and y values are the same on each of the vectors, we are using two 45-45-90 triangles. So 45º + 45º = 90º. So much for the Law of Cosines.

Try 68: I’m tired of GSP. I’ll use just right triangles this time.

V=. It’s in Quad I.

So tan V = 1/3.

1 arctan (1/3) = 0.32175055439664

3 ans(radians->degrees) = 18.4349488229219º

U=. It’s in Quad IV.

2 So tan U = -1/2

-1 arctan (-1/2) = -0.4636476090008

ans(radians->degrees) = -26.5650511770777

So, the angle between the two vectors is 18.4349488229219º + 26.5650511770777º = 44.9999999999996º.

Hmmmm…well…the directions told us to use the Law of Cosines, but I didn’t because I didn’t need it.

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download