# Calculating the magnitude of two vectors

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!

## Answers and Replies

mfb
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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 wanna 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?!

mfb
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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 wanna 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?

mfb
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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 text book 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 plz explain the answer to me?

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

my text book says that the magnitude is never negative!
Good, so you already excluded one of the possible answers.

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 plz

mfb
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ok i did them
Please show your results then.

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

mfb
Mentor
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 plz

arildno
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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, plz spill the beans :(

arildno
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i know the law of cosines, there are no angles or anything that i could use, guyz why are u making this so difficult, plz 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

arildno
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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

arildno
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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.

But, from your expression:
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 -!

arildno
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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)

arildno
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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..