The cross product and dot product of vectors

[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

 

[rock.freak667]

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|<br /> \begin{array}{ccc}<br /> i &amp; j &amp; k\\<br /> a_1 &amp; a_2 &amp; a_3\\<br /> b_1 &amp; b_2 &amp; b_3<br /> \end{array}<br /> \right|

 

[Xiience]

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]

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

 

[Xiience]

(-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

 

[Xiience]

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]

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 j≠ j 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]

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)

 

[Xiience]

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!

 

[gabbagabbahey]

kAxB is not the same as k(AxB)

Yes it is.