# The cross product and dot product of vectors

http://img297.imageshack.us/img297/2527/physicsin9.jpg [Broken]

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

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rock.freak667
Homework Helper
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.

rock.freak667
Homework Helper
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

gabbagabbahey
Homework Helper
Gold Member
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.

rock.freak667
Homework Helper
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 wierd way, idk

thank you for all your help though!

gabbagabbahey
Homework Helper
Gold Member
kAxB is not the same as k(AxB)
Yes it is.:surprised