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**radhakrishnamurty padyala****Member**- Registered: 2014-01-22
- Posts: 8

If **a(i)** and **b(i)** represent like parallel vectors, and a triplet of **a(i)** forms a triangle, then does the corresponding triplet of **b(i)** also form a similar triangle? If it does not, does it lead to a paradox?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,882

hi radhakrishnamurty padyala

Welcome to the forum.

I'm not entirely sure what you mean but I'll have a go.

Let's say

and, as you want b(i) to be parallel,

then will a triangle for 'a' produce a triangle for 'b' ? .... yes it will.

Let's have

and

then we'll have

and

and as you can see from the diagram below, the 'bs' make another triangle.

Will it always happen?

Yes, because b(x) will be parallel to a(x) for all x. The 'bs' will make an enlargement of the 'as'. So they are similar.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**radhakrishnamurty padyala****Member**- Registered: 2014-01-22
- Posts: 8

Thank you for the reply.

The solution implicitely demands that the ratios of magnitudes of the corresponding vectors a(i) and b(i), be a constant (When such is the case, it is evident that we can always form two similar triangles with the vector triplets). But no such demand is imposed in the statement of the problem.

Therefore, the issue is: Is it possible to form similar triangles with the two vector triplets a(i) and b(i) such that a(i), b(i) are like parallel vectors, a(i) triplet forms a triangle and where the ratios of the magnitudes of the corresponding vectors is not a constant?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,882

hi

Then I'm confused about what you mean by a triplet. Please give some examples.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**radhakrishnamurty padyala****Member**- Registered: 2014-01-22
- Posts: 8

Thanks again.

**F** = ∑**F(i)** = m(i)**a(i)** =0, is an example where a triplet of **F(i)**s forms a triangle, but the corresponding triplet of **a(i)**s doesn't, unless m(i) = constant. The parallelogram law of addition of force vectors doesn't demand m(i) to be constant for its validity. **F** is force, m is mass on which the force acts producing an acceleration **a**. **F** and a are like parallel vectors.

Triplet of vectors is just three vectors.

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**bob bundy****Moderator**- Registered: 2010-06-20
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Yes I understood that three vectors are involved. But I wanted examples, let's say 3 for 'a' and 3 for 'b', with numbers. Thanks.

Otherwise I cannot understand what constraints are involved.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**radhakrishnamurty padyala****Member**- Registered: 2014-01-22
- Posts: 8

Thanks again

We can consider the example given by you with a modification of one of the vectors, say, the vector **b(2)**. So, the three values for '**a**' and 3 for 'b' are:

**a(1)** = (2,1)

**a(2)** = (4,4)

**b(1)** = (4,2)

**b(2)** = (6,6)

*Last edited by radhakrishnamurty padyala (2014-02-06 01:04:03)*

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**bob bundy****Moderator**- Registered: 2010-06-20
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hi

I had assumed that 'parallel' meant b = ka where k has a single scalar value for all i.

In your example (4,2) = 2x(2,1) but (6,6) = 1.5x(4,4)

That's ok as it clears up my misunderstanding about 'parallel'.

And you have answered your own question: the triangles formed by the two sets of vectors are not similar.

What is the paradox that now concerns you?

Bob

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**radhakrishnamurty padyala****Member**- Registered: 2014-01-22
- Posts: 8

To me, there appears a paradox and I sought help. If you don't see it, it is ok. But if you see it any time and if you have a solution, please let me know.

Thanks.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,882

hi radhakrishnamurty padyala

Let me explain how it seems to me. Perhaps that will solve this puzzle.

If you make up three vectors at random, it is unlikely that they will make a triangle.

This is because the third side of a triangle depends on the other two.

If, say two sides are represented by **p** and **q**, then the third is represented by -(**p**+**q**)

So now let's consider parallel vectors.

If **b** is a vector parallel to **a** then that means that **b** = k**a** where k is some scalar.

So if a triangle is given by vectors **a1**, **a2** and -(**a1**+**a2**) then a similar triangle would have vectors k**a1**, k**a2** and -k(**a1**+**a2**).

But it is also possible to make a line parallel to **a1**, say, **b1** = k**a1** and to make a second line parallel to **b2**, say, **b2** = j**a2**, where k is not equal to j.

In that case the two triangles, **a1**, **a2**, -(**a1**+**a2**) and k**a1**, j**a2** and -(k**a1**+j**a2**) would not be similar because -(k**a1**+j**a2**) is not parallel to -(**a1**+**a2**).

How does that sound ?

Bob

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**radhakrishnamurty padyala****Member**- Registered: 2014-01-22
- Posts: 8

Thanks.

I understand what you say. But I also see that you don't see what I see.

Since the question is related to the basics of vectors, I might try posing the problem in a different fashion.

When do we say **two** given vectors linearly independent? OR,

Do ** two arbitrarily chosen intersecting vectors** form a pair of linearly independent vectros?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,882

hi

Have a look at

http://en.wikipedia.org/wiki/Linear_independence

Thus, two vectors **a** and **b**, are linearly dependent if and only if

k**a** + j**b** = **0 ** <=> **b** = scalar . **a**

If they are not dependent, then they are linearly independent.

If the two vectors intersect then they are not parallel. => **b** ≠ scalar . **a** => they are independent.

Bob

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