Trouble Understanding Abstract Vectors

[liaso]

Hello,

I am doing self-study to learn about vectors. I'm using a book that has no explanations, unfortunately.

The question I don't understand:

Find the magnitude and direction of (vector)i+(vector)j.

I know how to do this when I have magnitude and direction for each vector. But I have no idea how to do it when I don't have this information.

It tells me the answer is magnitude=(sqrt)2 and the direction is 45(degrees) counterclockwise from the positive x-axis. But I have no idea how they got that answer.

Any help explaining this would be greatly appreciated.

 

[liaso]

Thank you for the link and the term "unit vector" Jeff!! I'll check those out today.

Edited to add - IT MAKES SENSE!! Thank you again so much. I wish the book had told me about unit vectors. It didn't, so I had no idea that i and j were so special. You rock!

 

[HallsofIvy]

Hello,

I am doing self-study to learn about vectors. I'm using a book that has no explanations, unfortunately.

The question I don't understand:

Find the magnitude and direction of (vector)i+(vector)j.

I suspect your book does not actually say that! As you now know, i and j are themselves vectors. But any vector in the xy-plane can be written as a number times i plus a number times j.

I know how to do this when I have magnitude and direction for each vector. But I have no idea how to do it when I don't have this information.

It tells me the answer is magnitude=(sqrt)2 and the direction is 45(degrees) counterclockwise from the positive x-axis. But I have no idea how they got that answer.

Think of this as a right triangle. Draw a line from (0, 0) to (1, 0) and another line from (1, 0) up to (1, 1). The vector itself is i+ j. The horizontal and vertical lines are i and j respectively. But by the Pythagorean theorem, the length of the hypotenuse of the triangle, the length of the vector i+ j is given by [itex]\sqrt{1^2+1^2}= \sqrt{2}[/itex]. Also the tangent of the angle is "opposite side over near side which, here, is 1/1= 1. Measuring the angle in degrees, it is \(\displaystyle \theta= arctan(1)= 45[/itex].

Any help explaining this would be greatly appreciated.

\)

 

[liaso]

You are welcome.

It must be a strange book to give problems involving unit vectors without first explaining them. Maybe the book is designed to supplement a lecture series.

It's a study guide for the CSET exams. So not a textbook. It's been 16 years since I last took a math class. Some of it comes back as I review this book, but other parts seem so foreign. I might need to get a textbook in the end. (I skipped Trig in high school and went straight to Calc, so sometimes I think I might not have actually learned a couple of the things they are talking about in the first place.) The vector chapter was a tough one for me, but between the information you gave me and some very helpful you tube videos, I'm pretty sure I've got it down.

I suspect your book does not actually say that!

Oh, I assure you it does. That's why I was having trouble. Very little information. (I'm a very honest adult, trying to change my career. I teach exam Math for the SAT and ACT, and I'd like to be able to get my credentials to be able to teach math in a school, rather than after school, thus my decision to start studying for the CSET. Even though I don't want to teach trig, I still have to pass that section of the exam.)

As you now know, i and j are themselves vectors. But any vector in the xy-plane can be written as a number times i plus a number times j

I already knew they were vectors. I'll admit I was so frustrated about not knowing their direction angles that I forgot (vector)i is the same as (vector)1i. I appreciate that you wrote in to help me though. Thank you. It's always good to have a rule spelled out; I'm less likely to forget it that way. I have no problem with Pythag and right triangles. I was able to work other vector problems in the same section. What I do really struggle with is finding the angle based on the tangent. I know tan is opp/adj, but I am not allowed a calculator in this exam section, and I don't know how to do tan^-1 without a calculator. So I think this is where skipping Trig is hurting me.

Again, thank you Jeff and HallsofIvy. I really appreciate the help.