Vector dot product homework help?

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Homework Help Overview

The discussion revolves around calculating the dot product of vectors in a three-dimensional space, specifically focusing on the vectors A and B defined with different unit vectors. Participants are exploring the implications of the dot product being zero and the operations involving vector addition and subtraction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the dot product using the formula and question the relevance of angles in the absence of provided values. There is confusion regarding the different unit vectors in the vectors A and B, prompting inquiries about how to approach the problem.

Discussion Status

Some participants have provided guidance on the dot product calculation and the properties of unit vectors, while others express confusion about the different unit vectors involved. Multiple interpretations of the problem are being explored, particularly regarding the implications of the dot product being zero.

Contextual Notes

There is mention of textbook definitions and the need for clarity on vector operations, as well as the confusion stemming from the use of different unit vectors in the problem setup.

zainriaz
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if A = 2i^+6k^
B=6i^+5j^
find A.B
if A.B=0 then what do u conclude from it?


one more question
if A = 6i^+9k^
and B=-4i^+4j
then find A+B,A-B,A.B
 
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Do you know how to calculate the dot product of two vectors?
 
dot product= ABcos(theeta)
cos0=1
cos90=0

now try doing
 
jyothsna pb said:
dot product= ABcos(theeta)
cos0=1
cos90=0

now try doing
Actually, I wouldn't use that way here because the angles are not provided in the problem. The other way of calculating dot products will be much easier.
 
so can i get that easier method to solve this problem
 
diazona said:
Actually, I wouldn't use that way here because the angles are not provided in the problem. The other way of calculating dot products will be much easier.

i m just confused because of the diferent unit vectors! in both equations
 
Look up dot product from a textbook or online.
 
zainriaz said:
if A = 2i^+6k^
B=6i^+5j^
find A.B
if A.B=0 then what do u conclude from it?
These are two different questions!
A.B= 2(6)+ 6(5)

If A.B= 0 then, using the formula jyothsna pb gave: [itex]A. B= |A||B| cos(\theta)[/itex] so at least one of these is true: |A|= 0, |B|= 0, [itex]\theta= \pi/2[/itex].


one more question
if A = 6i^+9k^
and B=-4i^+4j
then find A+B,A-B,A.B[/QUOTE]
A- B= (6-(-4))i+ (9- 4)j, etc.

If you are expected to do these as homework, certainly the definitions of A+ B, A- B, and A.B must be in your textbook!
 
HallsofIvy said:
These are two different questions!
A.B= 2(6)+ 6(5)

If A.B= 0 then, using the formula jyothsna pb gave: [itex]A. B= |A||B| cos(\theta)[/itex] so at least one of these is true: |A|= 0, |B|= 0, [itex]\theta= \pi/2[/itex].


one more question
if A = 6i^+9k^
and B=-4i^+4j
then find A+B,A-B,A.B
A- B= (6-(-4))i+ (9- 4)j, etc.

If you are expected to do these as homework, certainly the definitions of A+ B, A- B, and A.B must be in your textbook![/QUOTE]

i no it is the dot product question
but the confuzing thing is that in A=the units r i^and k^ but in B=the units r i^and j^
how can we solve it when the units r different?
 
  • #10
i^ , j^ n k^ are perpendicular vectors so as per the formula i gave their dot product ( whatever b the product) will b 0.

that is i.j , i.k , j.k are all 0
now try doing
remember its scalar product so it does not have production
 
  • #11
zainriaz said:
but the confuzing thing is that in A=the units r i^and k^ but in B=the units r i^and j^
how can we solve it when the units r different?
They're not really "units," but "unit vectors." But the name doesn't matter as much as knowing what they are.

So here's your problem: you have the following two vectors:
[tex]\vec A=2\hat i+6\hat k[/tex]
[tex]\vec B=6\hat i+5\hat j[/tex]
and you need to take the dot product. To do this, you need to use the distributive property. Think about this: if you were multiplying
[tex](m + n)(p + q)[/tex]
you would know how to expand that out, right? You'd get
[tex]mp + mq + np + nq[/tex]
Stop me if you're not 100% comfortable with that.

To take the dot product of two vectors, you do the same thing, except that you replace regular multiplication with the dot product. So in the example above, if m,n,p,q were vectors instead of regular variables, what you get would be
[tex](\vec m + \vec n)\cdot(\vec p + \vec q) = \vec m\cdot\vec p + \vec m\cdot\vec q + \vec n\cdot\vec p + \vec n\cdot\vec q[/tex]
Does that make sense? Please ask if it doesn't.

If that does make sense, try doing the same thing with your two vectors [tex]\vec A[/tex] and [tex]\vec B[/tex]. You should get four terms, and each one should contain a product of two unit vectors, like [tex]\hat i\cdot\hat i[/itex] or [tex]\hat j\cdot\hat k[/tex], etc.<br /> <br /> Now you need to evaluate those products of unit vectors, i.e. replace them with their numeric values. The relevant identities have been given already in this thread, and you should also be able to look them up in your textbook or other reference. But you should also think about why they're true. They come from the definition<br /> [tex]\vec{A}\cdot\vec{B} = \vert A\vert \vert B\vert \cos\theta[/tex]<br /> Think about which direction the various unit vectors point in space and what the angles between them are.[/tex]
 
  • #12
thanks a lot
 

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