# The cross product and dot product of vectors

• Xiience
It's a matter of convention as to whether you group the scalar with the vector or with the cross product.
Xiience
http://img297.imageshack.us/img297/2527/physicsin9.jpg

i've been working with the AxB in the first one, and found that |A||B|sin(theta) = A x B, and i thought i had found my theta to be 1 degree, but i don't believe that's right. also, when i attempted to do the dot product with the C vector, i got completely lost..these are the last 2 problems on my homework, and any advice towards completely them would be so, so greatly appreciated

Thank you so much

Last edited by a moderator:
For $\vec{A}=a_1 i+a_2j+a_3k$ and $\vec{B}=b_1i+b_2j+b_3k$

$$\vec{A}\times \vec{B} = \left| \begin{array}{ccc} i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{array} \right|$$

okay, so using that cross product, I've got (-7.00)[C(7, -8, 0) . (-36, 72, -48)]
(7.00)(-828)
=-5796

does that look right? i believe i did the dot product right, a1b1+a2b2+a3b3

hm, that definitely does not look right.

What was your vector for $6 \vec{A} \times \vec{B}$ ?

(-48, 168, -54)

i think i did it wrong :( i distributed through (2i, 3j, -4k) x (-3i, 4j, 2k)

what i think my textbook said was i x j = k , ect.

i've managed to pull something up on matrices, like the way you showed it. going to try to work it out that way now

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i got -9156 as my answer, and it is wrong =\

using matrices
( i j k) ( 3 4) ( 2 -4) (2 3 )
(2 3 4) ( 4 2) i - ( -3 2) j + (-3 4) k
(-3 4 2)
that led my 6(A x B) to be (132, 48, 8), then i did a1a2 + b1b2 + c1c2 with my C vector and my A x B vector, then i multiplied what i got from that by -7, and it got me -9156, and it is wrong :(

edit- those matrices look horribly sloppy, but you get the idea

Xiience said:
i think i did it wrong :( i distributed through (2i, 3j, -4k) x (-3i, 4j, 2k)

what i think my textbook said was i x j = k , ect.

This method is fine; but the way you wrote your vectors is incorrect: A=2i+3j-4k or (2,3,-4) but not (2i, 3j, -4k).

And you did end up calculating the product incorrectly.

When you distribute the cross product you need to be mindful of the order in which you write the cross products; i x jj x i...You should get:

A x B=(2i+3j-4k) x (-3i+4j+2k)= -6(i x i)+8(i x j)+4(i x k)-9(j x i)+12(j x j)+6(j x k)+12(k x i)-16(k x j)-8(k x k)=22i+8j+17k

And so 6A x B=132i+48j+102k.

that led my 6(A x B) to be (132, 48, 8), then i did a1a2 + b1b2 + c1c2 with my C vector and my A x B vector, then i multiplied what i got from that by -7, and it got me -9156, and it is wrong :(

Well, the last component of your 6A x B was incorrect, but that shouldn't have impacted your final answer since C has no z-component.

Still, you have somehow incorrectly calculated the dot product; 7*132-8*48=924-384=540 which does not give you -9156 when you multiply it by -7.

Xiience said:
i got -9156 as my answer, and it is wrong =\

using matrices
( i j k) ( 3 4) ( 2 -4) (2 3 )
(2 3 4) ( 4 2) i - ( -3 2) j + (-3 4) k
(-3 4 2)
that led my 6(A x B) to be (132, 48, 8), then i did a1a2 + b1b2 + c1c2 with my C vector and my A x B vector, then i multiplied what i got from that by -7, and it got me -9156, and it is wrong :(

edit- those matrices look horribly sloppy, but you get the idea

Find 6A and then cross that vector with B

kAxB is not the same as k(AxB)

i figured out the first one, and the second one my roommate got and explained it to me, he did it a really weird way, idk

thank you for all your help though!

rock.freak667 said:
kAxB is not the same as k(AxB)

Yes it is.

## What is the difference between the cross product and dot product of vectors?

The cross product of two vectors results in a vector that is perpendicular to both of the original vectors, while the dot product results in a scalar value that represents the projection of one vector onto the other.

## How do you calculate the cross product of two vectors?

The cross product can be calculated by taking the determinant of a 3x3 matrix composed of the unit vectors and the components of the two vectors. This can also be expressed in terms of the magnitude of the vectors and the angle between them.

## What is the physical significance of the cross product?

The cross product is often used in physics to calculate torque, angular momentum, and magnetic fields. It also has applications in 3D graphics and computer vision.

## What is the relationship between the cross product and the dot product?

The cross product and dot product are related through the distributive property, but they have different geometric and mathematical properties. They can also be used together to calculate the angle between two vectors.

## Can the cross product be applied to more than two dimensions?

No, the cross product is only defined for three-dimensional vectors. However, there are generalizations of the cross product for higher dimensions, such as the vector triple product, which is defined for four-dimensional vectors.

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