6.4 – Vectors & Dot Products
Jan 10th, 2011 by corricelli
Hello,
Use this section to post questions/ respond to questions in Chapter 6, Section 4.
NOTE: HW Due on 2/11 has been changed to the following problems: 59, 63, 67, 71-77 odd, 81
Thanks,
Mrs. Corricelli
What is the homework for tonight?
Hello Meg!
7,11,17-27 odd,33,35,41,45,49,53,57
Alright thank you so much!
Alright I think I got most of number 6 from pg 491 but I’m confused on part D. What angle measure are they asking for?….the angle where vector v and u instersect (I think that is 90) or the angle where vector s (v+U) meets the ground.
I think it would be the new angle the ski diver is falling at, so is you drew S from the initial point of u to the terminal point of v it would be the angle of u and s.
I am confused on letter e though, when it says the ski diver is going 30 degrees west, is it the same ski diver or a new situation?
Sam – I think that the question means that the skydiver is falling from the same height, so vector u is the same, and vector v is 30 to the west instead of the 40 to the east which was in the original problem. so basically its the same skydiver but the wind (vector v) has changed
I started working on problem number 6 also, and I have a question about te vector u. Sine vectors have both direction and magnitude, and the skydiver is FALLING, does this mean that its magnitude is – 120 and not positive?
Also, what is the difference between the component form of a vector (as mentioned in part a) and the addition of 2 vectors (mentioned in part b). To sketch s = u + v, would I need to use part a for that? (isnt the sketch just the diagonal of the”rectangle”?)
the* vector u. Since** vectors
How did you guys find velocity for part E?
I did it the way I thought you were supposed to by finding the downward velocity and the velocity of the wind and then adding them together but I got a really weird answer. Am I going about it wrong?
Meg, I think I used a similar approach for part e. I left vector u where it was, and drew v starting from the terminal of u, extending 30 units to west. I was confused if the magnitude of this new vector v was -30, since its on the negative side of the coordinate plane? Can any one tell me if this is right?
However, whether its -30 or +30, when you find the new velocity of the plane, you’ll end up squaring the 120 and -30 (or 30 whichever it is) to get the answer. I got an answer of ALMOST 116 mi/hr. Is that the same “wierd answer” you also got? I was also confused because how could this answer (the hypotenuse) be less than the value of one of the legs (vector U, which is 120?)
WAITT NEVER MIND, the velocity ends up being about 123.69 mi/hr (i made a mistake on the calculator, sorry!!)…. BUT isn’t this still wierd though, in the sense that despite the wind being 30 mi de west, the answer is not much different than the original velocity of the plane?!! (only about 4 mi/hr away!)
Mawra, check your math, I also ended up squaring the 120 and 30 for part e, but got around 123 as the answer, which makes more sense if its the hypotenuse.
Colleen- I took the angle measure in part D to be the angle between S and vector u, so the top angle of the triangle.
Correct me if I’m wrong!
sorry that angle description was probably confusing…I mean the angle i found was the angle from the start of the sky diver’s fall.
Right….?
I think your math is all correct. But you guys both found the speed of the plane, not the velocity.
Speed is scalar because it doesn’t have magnitude and direction, whereas velocity does. In the example problem in the book it had velocity shown as a vector, so my weird answer was which could possibly be reduced to , I’m not entirely sure that is legal though…
sorry it deleted my vectors for some reason that should say:
my weird answer was which could possibly be reduced to
okay it didn’t like it that way either.. One more time haha
my weird answer was (-900, -14400) which could possibly be reduced to (-1, -16)
Haley, thanks! For part D, I agree with you, the angle measure is from the start of the diver’s fall (at the top).
Meg, I understand what you mean by how we’re looking for the velocity, which is a vector. But I don’t understand how it could be positive or ngative, since it’s that diagonal line? How did you get the vector to be (-900, -14400)?
I just followed the same steps that it did in the example problem. I used the component forms, found the velocity of each, and then added them together to get the velocity of the diver
Meg (and those who are still awake),
You’re right, velocity can be expressed as a vector because it has magnitude (speed) and direction.
I ended up getting the square roots of your components, which i found by using the formula: resultant= +
*resultant= vector 1 + vector 2
It doesn’t ask for magnitude, so there’s no need for speed.
For number 63 on tonight’s homework, I don’t understand how there can even be a projection when u and v are parallel and lie on the same parallel. Anyone get it?
Haley, the projection of u onto v would be the same as u because u and v are parallel. So it’s like the examples Mrs. Corricelli was talking about in class where you imagine the sun and the shadow it would make, if the two lines are parallel then the “shadow” of u would be identical to it.