Vector geometry: Proof of a trapezium and cross and dot product help

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SUMMARY

This discussion focuses on vector geometry, specifically the calculations involving cross products and dot products of vectors. The participants clarify the definition of the cross product, emphasizing that the square of a vector refers to its dot product with itself, resulting in a scalar value. The conversation also addresses the proof of a trapezium formed by vectors AB, BC, and CD, confirming that the conditions for parallel sides are met through vector relationships. The final calculations yield a result of 108 for the squared magnitudes of the cross products.

PREREQUISITES
  • Understanding of vector operations, specifically cross products and dot products.
  • Familiarity with vector notation and properties, including magnitude and direction.
  • Knowledge of geometric properties of trapeziums and conditions for parallel sides.
  • Basic algebraic manipulation skills for vector equations.
NEXT STEPS
  • Study the properties of the cross product in vector algebra.
  • Learn how to compute dot products and their geometric interpretations.
  • Explore the geometric properties of trapeziums and other quadrilaterals in vector form.
  • Practice solving vector equations involving multiple operations and their implications.
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Students studying vector geometry, mathematics educators, and anyone seeking to deepen their understanding of vector operations and geometric proofs.

FilipaE
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Homework Statement


#1 Given that the angle between the vectors a and b is 2Pi/3 and |a|=3 and |b|=4 calculate:
(axb)^2 [(2a+b)x(a+2b)]^2

#2 Given three unit vectors, a, b, c such that a+b+c=0 find (a dot b) + (b dot c) + (c dot a)

#3 Given AB=a+2b BC=-4a-b CD= -5a-3b where a and b are any two vectors, prove that ABCD is a trapezium


Homework Equations



Given above

The Attempt at a Solution


#1 I know that axb = |a||b|sinx
and axa = 0 and axb = -b x a but i do not understand how the powers or constants affect this?

#2 No idea

#3 A trapezium has one set of parallel sides only and i know if vectors are parallel they are a multiple of each other, however none of these seem to be multiples of each other?
 
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FilipaE said:
#1 I know that axb = |a||b|sinx

This is false. The length of a\times b equals |a||b|\sin\theta.

Anyway, what do you mean with (a\times b)^2?? What does it mean to take the square of a vector??
 
I am using x as the cross product, meaning the cross product of a and b is |a||b|sinx, is this not correct?
 
and (a x b)^2 means the cross product of a and b all sqaured
 
FilipaE said:
I am using x as the cross product, meaning the cross product of a and b is |a||b|sinx, is this not correct?

No, that is not correct. Review your definition of cross product.
 
FilipaE said:
and (a x b)^2 means the cross product of a and b all sqaured

Yes, I get that: it means that you take the square of the cross product. But the cross product is a vector. What do you mean with the square of a vector?
 
This is the definition of cross product online |a x b|=|a||b|sinx
I have no idea what you mean vectors is a very confusing topic for me
 
FilipaE said:
This is the definition of cross product online |a x b|=|a||b|sinx
I have no idea what you mean vectors is a very confusing topic for me

Yes, the definition |a\times b|=|a||b|\sin x is correct. But writing a\times b=|a||b|sin x is not correct. Do you see the difference?
 
Oh i see, so how does the formula differ without the modulus sign? and what happens where there is addition involved as in Q1 part 2
 
  • #10
Do you understand that a\times b is a vector and that |a\times b| is a scalar??
 
  • #11
yes i do now, but given this information how would i write the vector a x b?
 
  • #12
Before you do that, you got to understand what (a\times b)^2 is. What does it mean to square a vector?
 
  • #13
Is it the dot product of the two vectors, creating a scalar?
 
  • #14
FilipaE said:
Is it the dot product of the two vectors, creating a scalar?

Yes.

So you need to find (a\times b)\cdot (a\times b).

Given a vector v, what is the definition of v.v?
 
  • #15
|v|^2?
 
  • #16
So (a\times b)\cdot (a\times b)=|a\times b|^2. Now you should be able to calculate this.
 
  • #17
Thankyou! I have got the answer 108?

So for [(2a+b) x(a+2b)]^2 I do [(2a+b)x(a+2b)] dot [(2a+b)x(a+2b)] = |(2a+b)x(a+2b)|^2

I know to do this i need to use the distributive law. Is it

2a x a + 2a x2b + b xa + b x2b?
Where 2a x a = 0 and bx2b=0?
 
  • #18
So the answer is 108? Thankyou!
For [(2a+b) x(a+2b)]^2 is this the same as
(2a+b) x(a+2b) dot (2a+b) x(a+2b) = |(2a+b) x(a+2b)|^2 ??

I know i have to use some sort of distributive law here but i am not sure on how to do it would it be 2a x a + 2a x 2b + b x a + b x 2b ? where 2a x a = 0 and b x 2b =0??
 
  • #19
Sorry didnt think the first message had sent!
 

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