Cross product help (I think)

  • Thread starter Failure007
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  • #1
Hey guys, I normally don't use forums for help but I've been stuck on these questions for some time now and I just want to see if there's anyone out there who can set me in the right direction. These questions are from a grade 12 Calculus and Vectors class and they relate to the cross product.

1. How much work is done by a force, in Newtons, represented by the vector F = (4, 7, 1) if it results in a displacement, in metres, represented by the vector delta(I'm assuming it's delta but the symbol is really small) S = (5, 3, 8).

2.If vectors A and B are perpendicular and A = (4, – 4, 8) and
B = (5, k, 2k), find k.

For the first one I tried applying the cross product to make a new vector until I remembered that finding the Work for F and the new vector would result in a dot product of 0, because of the fact that cross products are perpendicular to both vectors. After realizing it wouldn't work I tried calculating the distance between the vectors but this didn't help either. I'm stumped on both of these and any help would be greatly appreciated, Thanks alot!
 

Answers and Replies

  • #2
rock.freak667
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Actually they are related to the dot product.

Work is F.s

No need to use the cross-product
 
  • #3
So if I multiplied [4, 7, 1]*[5,3,8] and got 49 as my dot product, then that's how much work is needed to result delta S = [5, 3, 8]? Why would they even mention the delta in the first place, because that's the only reason I assumed this question was difficult...
 
  • #4
rock.freak667
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So if I multiplied [4, 7, 1]*[5,3,8] and got 49 as my dot product, then that's how much work is needed to result delta S = [5, 3, 8]? Why would they even mention the delta in the first place, because that's the only reason I assumed this question was difficult...
No idea about the delta, but the dot product gives a scalar, while cross-pdt gives a vector. Work is a scalar so, the only way to get the amt of work done would be to find the dot-product.
 
  • #5
HallsofIvy
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So if I multiplied [4, 7, 1]*[5,3,8] and got 49 as my dot product, then that's how much work is needed to result delta S = [5, 3, 8]? Why would they even mention the delta in the first place, because that's the only reason I assumed this question was difficult...

What do you mean "mention the delta"? I presume what you had was [itex]\delta S= [5, 3, 8][/itex] or [itex]\Delta S= [5, 3, 8][/itex]. They use the notation [itex]\delta S[/itex] or [itex]\Delta S[/itex] because it implies a change in S which is what is necessary to do work: Force applied to move something- to change its position.
 
  • #6
Sorry, I didn't know how to include the triangle in the question so I just called it delta, but what I was having trouble with was finding the work to move from the initial displacement to the delta displacement. So basically I figured there was an initial displacement vector that I had to find, but the only way I knew of creating a vector given two vectors is through the cross product. Does anyone know how to find the initial displacement?
 

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