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Calc Vector (dot product)

  • Thread starter iclutcha
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  • #1
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Homework Statement


The cosine of the angle between vector a and vector b is 4/21. Find p.

vector a = 6i + 3j - 2k
vector b = -2i + pj - 4k

Homework Equations



(Vector a) . (Vector b) = abcos(theta)
(Vector a) . (Vector b) = axbx + ayby + azbz
mag(a) = root(ax^2 + ay^2 + az^2)

The Attempt at a Solution



mag(a) = root(36 + 9 + 4) = 7
mag(b) = root(4 + p^2 + 16) = root(p^2 + 20)

(Vector a) . (Vector b) = -12 + 3p + 8 = 3p - 4

(Vector a) . (Vector b) = abcos(theta)
3p - 4 = 7[root(p^2 + 20)](4/21)
3p - 4 = [4/3][root(p^2 + 20)]

Square both sides

9p^2 - 24p + 16 = 16/9p^2 + 320/p

Collect like terms, and set equation to 0

(65/9)p^2 - 24p - 176/9

Sub into quadratic equation

p = 4
p = -44/65

[CORRECT ANSWER: P = 4]

Thanks for the help ;)
 
Last edited:

Answers and Replies

  • #2
jbunniii
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Homework Statement


The cosine of the angle between vector a and vector b is 4/21. Find p.

vector a = 6i + 3j - 2k
vector b = -2i + pj - 4k

Homework Equations



(Vector a) . (Vector b) = abcos(theta)
(Vector a) . (Vector b) = axbx + ayby + azbz
mag(a) = root(ax^2 + ay^2 + az^2)

The Attempt at a Solution



mag(a) = root(36 + 9 + 4) = 7
mag(b) = root(4 + p^2 + 16) = root(p^2 + 20)

(Vector a) . (Vector b) = -12 + 3p + 8 = 3p - 4

(Vector a) . (Vector b) = abcos(theta)
3p - 4 = 7[root(p^2 + 20)](4/21)

kinda confused here... i did some attempts and i always end up with a quadratic, if someone could just give me some insight, it would be greatly appreciated. Thanks :)
What you have done so far looks OK. Why are you surprised to obtain a quadratic? There are two angles such that cos(theta) = 4/17, so you should expect that there will be two valid solutions.
 
  • #3
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What you have done so far looks OK. Why are you surprised to obtain a quadratic? There are two angles such that cos(theta) = 4/17, so you should expect that there will be two valid solutions.
Well its not so much the quadratic that troubles me, as the solution. When i solve it i get a number with decimals, and it is not near the final answer which the book gives. (4)

If someone could finish off my solution that would be great, as when i did it, it was incorrect.
 
  • #4
Cyosis
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Show your work so we can see where you went wrong.
 
  • #5
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Show your work so we can see where you went wrong.
I just tried it again, edited the first post, no quadratic but i could have done something wrong.
 
  • #6
Cyosis
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There will be a quadratic. You have a p on one side and a sqrt(p) on the other side. To solve it you raise both sides to the power two. As a result you will have a p^2 on one side and a p on the other. If you don't show us your work we can't tell you where you go wrong.
 
  • #7
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There will be a quadratic. You have a p on one side and a sqrt(p) on the other side. To solve it you raise both sides to the power two. As a result you will have a p^2 on one side and a p on the other. If you don't show us your work we can't tell you where you go wrong.
I showed my work in the first post
 
  • #8
Cyosis
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Seems I am going blind, sorry. I'll check it out.

Edit: You edited didn't you!
 
  • #9
Cyosis
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You're squaring incorrectly, [itex](3p - 4)^2=9p^2-24p+16[/itex].
 
  • #10
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You're squaring incorrectly, [itex](3p - 4)^2=9p^2-24p+16[/itex].
heh, yeah i did square it wrong that time... but on my paper i did it jsut like that, but after i use the quadratic formula on the final outcome, its still wrong... maybe you have better luck?
 
  • #11
Cyosis
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Well I have calculated it and i get p=-44/65 and p=4. It is not me who needs to practice this though so regardless whether I do it correctly or not we still need to see where you went wrong! It is a bit of a messy calculation but still.
 
  • #12
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Well I have calculated it and i get p=-44/65 and p=4. It is not me who needs to practice this though so regardless whether I do it correctly or not we still need to see where you went wrong! It is a bit of a messy calculation but still.
EDIT: nvm i got it, dumb mistake.
 
Last edited:

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