- #1

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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

- Thread starter zainriaz
- Start date

- #1

- 21

- 0

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

- #2

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Do you know how to calculate the dot product of two vectors?

- #3

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dot product= ABcos(theeta)

cos0=1

cos90=0

now try doing

cos0=1

cos90=0

now try doing

- #4

diazona

Homework Helper

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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.dot product= ABcos(theeta)

cos0=1

cos90=0

now try doing

- #5

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so can i get that easier method to solve this problem

- #6

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i m just confused because of the diferent unit vectors! in both equationsActually, 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.

- #7

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Look up dot product from a textbook or online.

- #8

HallsofIvy

Science Advisor

Homework Helper

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These are twoif A = 2i^+6k^

B=6i^+5j^

find A.B

if A.B=0 then what do u conclude from it?

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!

- #9

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A- B= (6-(-4))i+ (9- 4)j, etc.These are twodifferentquestions!

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

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

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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

diazona

Homework Helper

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They're not really "units," but "unit vectors." But the name doesn't matter as much as knowing what they are.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?

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.

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

[tex]\vec{A}\cdot\vec{B} = \vert A\vert \vert B\vert \cos\theta[/tex]

Think about which direction the various unit vectors point in space and what the angles between them are.

- #12

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thanks a lot

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