Vector dot product homework help?

In summary: That's the way i learnt it!In summary, the dot product of two vectors is calculated by taking the product of the components of each vector and adding them together. This can be done using the distributive property. The result is a scalar value, which represents the magnitude of the vectors multiplied by the cosine of the angle between them.
  • #1
zainriaz
21
0
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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  • #2
Do you know how to calculate the dot product of two vectors?
 
  • #3
dot product= ABcos(theeta)
cos0=1
cos90=0

now try doing
 
  • #4
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.
 
  • #5
so can i get that easier method to solve this problem
 
  • #6
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
 
  • #7
Look up dot product from a textbook or online.
 
  • #8
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!
 
  • #9
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.

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
thanks a lot
 

1. What is a vector dot product?

A vector dot product is a mathematical operation that takes two vectors and produces a scalar quantity. It is also known as the scalar product or inner product.

2. How do I calculate the dot product of two vectors?

To calculate the dot product, you need to multiply the corresponding components of the two vectors and then add the products together. For example, if vector A is (a1, a2, a3) and vector B is (b1, b2, b3), the dot product would be a1b1 + a2b2 + a3b3.

3. What is the geometric interpretation of the dot product?

The dot product can be interpreted as the product of the lengths of the two vectors and the cosine of the angle between them. This means that the dot product is positive if the vectors are pointing in the same direction, negative if they are pointing in opposite directions, and zero if they are perpendicular.

4. How is the dot product related to the angle between two vectors?

The dot product is related to the angle between two vectors through the trigonometric identity cos(θ) = A · B / (|A| * |B|). This means that the dot product can be used to find the angle between two vectors.

5. What are the applications of the dot product?

The dot product has various applications in mathematics, physics, and engineering. It is commonly used in vector calculus, mechanics, and computer graphics. It can also be used to find projections, angles, and distances in vector spaces.

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