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Math Help - Finding scalars from vectors

  1. #1
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    Finding scalars from vectors

    (Note: the bold letters are vectors.)

    "Find the scalars, α and β and such that
    (i) 3i + j is parallel to αi - 4j
    (ii) 3i + β(j - k) is parallel to i - 4j+ 4k
    (iii) 3i + γ(j + 3k) has the same length as 12i - 5k"

    I first assumed that since the vectors are parallel, they would be equal to each other. So, as for (i), and assuming that v = 3i + j, I found |v| as assumed that would be the length and thought that finding the length of the other vector through solving it that way would be the right approach. However, I realised that it isn't.

    I did have a peep at the answers although, that's all that was there so that was hardly any help. (If it is needed - (i)-12 (ii)-12 (iii) 4). I tried looking over my course notes and even the appropriate section from my textbook but, I can't find anything that is similar to this question.
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    Quote Originally Posted by cottontails View Post
    (Note: the bold letters are vectors.)

    "Find the scalars, α and β and such that
    (i) 3i + j is parallel to αi - 4j
    3i + j = k(\alpha i - 4 j) where k is a scalar.

    Therefore:

    3 = k \alpha .... (1)

    1 = -4k .... (2)

    Solve equations (1) and (2) simultaneously to get \alpha.
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    I now no longer have any difficulties with (i) and (ii) however, with (iii), I cannot get the correct answer.
    My working (using t instead of k, since there is a k vector in the equation):
    -> t(3i + γj + 3γk) = 12i - 5k
    --> 3t=12, γt=0, 3γt=-5
    Whenever I solve the equations simultaneously so, t=4 and substituting that into 3γt=-5, I get γ=-5/12.
    Although, am I doing something wrong there since shouldn't that also equal the same (-5/12) with substituting t=4 in γt=0?
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  4. #4
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    Quote Originally Posted by cottontails View Post
    [snip]

    (iii) 3i + γ(j + 3k) has the same length as 12i - 5k

    [snip]
    Quote Originally Posted by cottontails View Post
    I now no longer have any difficulties with (i) and (ii) however, with (iii), I cannot get the correct answer.

    My working (using t instead of k, since there is a k vector in the equation):

    -> t(3i + γj + 3γk) = 12i - 5k

    --> 3t=12, γt=0, 3γt=-5

    Whenever I solve the equations simultaneously so, t=4 and substituting that into 3γt=-5, I get γ=-5/12.

    Although, am I doing something wrong there since shouldn't that also equal the same (-5/12) with substituting t=4 in γt=0?
    Get the length, that is, magnitude, of 3i + γ(j + 3k) and equate it to the length, that is, magnitude, of 12i - 5k. Then solve for γ. I don't see what the trouble can be here, apart from not reading the question properly.
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