# Calculating the magnitude of two vectors

• UnD3R0aTh
In summary, the conversation discusses displacement vectors and their magnitudes. It is determined that the magnitude of the difference vector, S-T, can vary depending on the magnitudes of S and T. The concept of orthogonal vectors is also mentioned, where the magnitude of the difference vector remains the same regardless of addition or subtraction. The conversation also explores the relationship between the magnitudes of S, T, and S-T, with the hint of using the law of cosines to determine the length of S-T in the general case of a triangle.

#### UnD3R0aTh

1. Two displacement vectors, S and T, have magnitudes S = 3 m and T= 4 m. Which of the following could be the magnitude of the difference vector S - T ? (There may be more than one correct answer.) (i) 9 m; (ii) 7 m; (iii) 5 m; (iv) 1 m; (v) 0 m; (vi) - 1 m

2. Vector principles

3. shouldn't this problem tell me the direction of the vectors? the magnitude of the resultant vector is not always the sum or difference of the magnitude of the vectors unless they are parallel!

shouldn't this problem tell me the direction of the vectors?
It does not have to. The problem statement does not ask about the actual magnitude, it just asks what is possible.
the magnitude of the resultant vector is not always the sum or difference of the magnitude of the vectors unless they are parallel!
Right, but you can still say something about the possible values for the magnitude of the difference.
Can you have two vectors with a magnitude of 1 each, where the difference has a magnitude of 1000?

no i don't think that's possible, but i have a side question to make sure i understand what i read in my textbook, then i will try to solve this problem, if i have a displacement vector of magnitude 1km north, then another 2km east, adding the two will result in a vector 63 degrees east of north with magnitude = square root of 5, but what if i want to subtract the two vectors, i will have to flip vector 1 km then slide it so that it's tail lies on the head of vector 2km, then again the resultant vector's magnitude is going to be square root of 5, is that correct? how come the addition and the subtraction of two right angle vectors have the same magnitude?!

UnD3R0aTh said:
no i don't think that's possible
Correct.

but i have a side question to make sure i understand what i read in my textbook, then i will try to solve this problem, if i have a displacement vector of magnitude 1km north, then another 2km east, adding the two will result in a vector 63 degrees east of north with magnitude = square root of 5, but what if i want to subtract the two vectors, i will have to flip vector 1 km then slide it so that it's tail lies on the head of vector 2km, then again the resultant vector's magnitude is going to be square root of 5, is that correct? how come the addition and the subtraction of two right angle vectors have the same magnitude?!
This is special for orthogonal vectors - flipping the sign of one vector is like mirroring the whole setup, and mirroring does not change lengths (magnitudes) of vectors.

well then it should be 7 m because as far as i can understand the magnitude doesn't depend on whether it's addition or subtraction, the direction is what changes correct? and the magnitude should be close to the sum of the magnitude of the two vectors correct?

UnD3R0aTh said:
well then it should be 7 m because as far as i can understand the magnitude doesn't depend on whether it's addition or subtraction
In general, it does.
and the magnitude should be close to the sum of the magnitude of the two vectors correct?
Why should it be close to the sum?

Did you consider some examples, like
S=(0,1), T=(1,0)
S=(5,2), T=(4,2)
S=(5,0), T=(-4,0)
...? This will help to get a feeling for the problem.

T=(-4.0) my textbook says that the magnitude is never negative! and sorry but i gave it everything that i have, i don't see any difference between what u said above and the problem, the problem remains i can't tell unless the vectors are drawn! can u please explain the answer to me?

My examples are vectors (in 2D), not magnitudes of vectors.
You can draw them on a plane, if you like.

my textbook says that the magnitude is never negative!

oh those are coordinates, ok i will draw them and add and subtract and see how it goes!

ok i did them, I'm not really sure what to make of it! more hints please

UnD3R0aTh said:
ok i did them

What did you get for S-T? Which magnitudes did you get for S, T and S-T?

S=(5,2), T=(4,2) magnitude = 1
S=(0,1), T=(1,0) magnitude = square root of 2
S=(5,0), T=(-4,0) magnitude = 9

i can't find any relationship between the magnitude of the vectors and the magnitude of the difference vector

Where are the magnitudes of S and T?
There is no fixed relationship, that is the important result. There are lower and upper bounds, however. Take two straight objects with a very different length, put them on your desk (such that their ends touch) and see which length between the non-touching ends you can get. What is the maximum, what is the minimum?

i'm sorry i don't have a ruler with me atm, and i have a lot to go through still, i should finish this vectors chapter, plus another chemistry chapter :( more hints please

Do you remember the..law of cosines?

Think a bit about how your question might be related to that..

i know the law of cosines, there are no angles or anything that i could use, guyz why are u making this so difficult, please spill the beans :(

UnD3R0aTh said:
i know the law of cosines, there are no angles or anything that i could use, guyz why are u making this so difficult, please spill the beans :(
Now, think geometrically about vectors S, T, S-T

What sort of geometrical SHAPE will these typically look like when you draw them?

Hint: Draw S and T both from the origin. Where will you find S-T?

they could be a straight line, they could be a triangle, they could be anything

Anything? When you add them vectorially?

The basic case is that of a TRIANGLE, with two extreme cases as straight lines.

Agreed?

Now, how can you, in the general case of the triangle compute the length of S-T? What formula would you use?

ok drawing some random vectors, making a triangle, applying the law of cosines, i reached that the difference vector is equal to the square root of (25-24cos theta) so the magnitude of the vector depends on the theta obviously and there's no way to tell the magnitude without knowing more information

UnD3R0aTh said:
ok drawing some random vectors, making a triangle, applying the law of cosines, i reached that the difference vector is equal to the square root of (25-24cos theta) so the magnitude of the vector depends on the theta obviously and there's no way to tell the magnitude without knowing more information
Good so far!
Yes, you cannot say which SPECIFIC number the length might be, because theta is not given.

What is the MAXIMAL value it can have? (What theta choice does that correspond to?)
What is the MINIMAL value it can have?(What theta choice does that correspond to?)
Is it possible to find a theta value that will yield any value WITHIN the interval given by the answers to the two previous questions?

well since cosine fluctuates between 1 and -1, the minimal and maximum angles that will correspond to a minimal and maximum value of cosine are 0, and 180, when theta=0 the vector difference is 1, when theta equals 180, the vector difference is 49 (is that solution really possible?) that makes the possible solutions 1, 5 , 7 ,9 all except -!

Haven't you forgotten a square root here? • 1 person
oops sorry yeah, it's all the values between 1 and 7 then, correct? (u are an excellent helper btw, do reply to all of my threads :D)

UnD3R0aTh said:
oops sorry yeah, it's all the values between 1 and 7 then, correct?
Yep!
So now you know what values the magnitude CAN have, and therefore which alternatives in the exercises can be correct answers, and which of the alternatives can't ever be correct..

thank you so much :)

UnD3R0aTh said:
(u are an excellent helper btw, do reply to all of my threads :D)
Thank you! So is mfb. He would have nudged you right along the same path if I hadn't barged in.

Before you go to the next exercise, read for yourself now that you know how to solve it HIS answers once again, and see how he prepares the way.. • 1 person
ok i have a side question that i solved but i want to check the answer with you, is it possible for two vectors to have the same magnitude but different components? yes because changing in theta will result in change in components, but it's not possible to have the same components and different magnitudes because same components implies same magnitudes, correct?

1. " yes because changing in theta will result in change in components"
Yes. It is easiest to visualize vectors of the same magnitude as being the radius vectors in a circle about the origin. Changing theta from 0 to 360 degrees gives all the vectors that exist which have the same magnitude (i.e, the length of your circle's radius!)

2. " but it's not possible to have the same components and different magnitudes because same components implies same magnitudes,"
Again correct. Since the magnitude can be calculated using just those two components, the same calculation with the same two component must yield the..same magnitude.
2.

• 1 person
ok i tried the same method to verify the last statement in the problem, i calculated the magnitude of both vector D and E, i multiplied vector D by 2, i drew a triangle and i used the law of cosines, the lower bound is 16.9, the upper bound is 37.5, but the problem says that the value can never be higher than 16.9 but it might be smaller, what have i done wrong? (problem in the attachments)

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UnD3R0aTh said:
ok i tried the same method to verify the last statement in the problem, i calculated the magnitude of both vector D and E, i multiplied vector D by 2, i drew a triangle and i used the law of cosines, the lower bound is 16.9, the upper bound is 37.5, but the problem says that the value can never be higher than 16.9 but it might be smaller, what have i done wrong? (problem in the attachments)

Now, there are some important differences in this problem relative to the last one:
1. The components are GIVEN. That means the angle between the two vectors are necessarily fixed, so you do not have a range of thetas to look at.
2. But: The problem, precisely because you have the components (you didn't have that in the previous problem!) is easily solved:
Just COMPUTE vector 2D-E, and then determine its magnitude with the Pythagorean theorem.

UnD3R0aTh said:
they all have the same magnitude correct?

Yes.

arildno said:
Now, there are some important differences in this problem relative to the last one:
1. The components are GIVEN. That means the angle between the two vectors are necessarily fixed, so you do not have a range of thetas to look at.
2. But: The problem, precisely because you have the components (you didn't have that in the previous problem!) is easily solved:
Just COMPUTE vector 2D-E, and then determine its magnitude with the Pythagorean theorem.

did u read the last statement about the evaluation? what does it mean?

Yes.
It is a very nebulous statement, but it is critical to highlight something said previously:
"recall from Section 1.7 that subtracting a vector is the same as ADDING the negative of that vector"

Now, keeping THAT statement in mind, look at what "Evaluate" actually says:
"Our answer is in the same order of magnitude as the larger components that appear in that SUM"

That means, the components in SUM 2D+(-E) (prior to summing them, we have components 12, -4, 6, 5, -2, -8)

The biggest component here is the number "12", agreed?
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So, the book is simply saying that when we subtract two vectors A and B from each other, we can look at the biggest components in A and B and say that the magnitude of the difference vector might be roughly as large as the biggest components in A or B (but never too much beyond that), but that magnitude CAN be much smaller (If A is a HUGE vector, but B=A, then A-B has zero magnitude!)