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Math Help - Vectors

  1. #1
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    Vectors

    I am doing part c) of the following question and was wondering if anyone can tell me what they get for position vectors of A and B as I'm not sure mine are correct.

    Vectors-screen-shot-2010-06-28-14.27.08.png

    Thanks!
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    Quote Originally Posted by maggiec View Post
    I am doing part c) of the following question and was wondering if anyone can tell me what they get for position vectors of A and B as I'm not sure mine are correct.

    Click image for larger version. 

Name:	Screen shot 2010-06-28 at 14.27.08.png 
Views:	19 
Size:	51.4 KB 
ID:	18033

    Thanks!
    Dear maggiec,

    Take, L_1:\underline{r_{1}}~and~L_2:\underline{r_{2}}

    First write the vector equation of AB. ( \overline{AB}=\underline{r_{2}}-\underline{r_{1}})

    Since AB is perpendicular to L1 and L2, \overline{AB}.\underline{r_{2}}=0~and~\overline{AB  }.\underline{r_{1}}=0

    By the resulting equations you will be able to find the corresponding \lambda~and~\mu values.

    Hope you would be able to continue.
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    Quote Originally Posted by Sudharaka View Post
    Dear maggiec,

    Take, L_1:\underline{r_{1}}~and~L_2:\underline{r_{2}}

    First write the vector equation of AB. ( \overline{AB}=\underline{r_{2}}-\underline{r_{1}})

    Since AB is perpendicular to L1 and L2, \overline{AB}.\underline{r_{2}}=0~and~\overline{AB  }.\underline{r_{1}}=0

    By the resulting equations you will be able to find the corresponding \lambda~and~\mu values.

    Hope you would be able to continue.
    Proceeding with the method you have described is taking a very long time and leaving me \lambda~and~\mu to the power of 4 that are very hard to work out! Especially as this question is form a non-calculator exam.

    I think there must be and easier way to work this out.

    The way I had done it, is by working out \overline{AB}=\underline{r_{2}}-\underline{r_{1}}

    and then establishing a relationship between the line \overline{AB} and the position vector (2,1,2)
    This is possible since before i had proved that this position vector was perpendicular to both L_1:\underline{r_{1}}~and~L_2:\underline{r_{2}} and hence it must be parallel to the line \overline{AB}

    with this I equated the i and k components of \overline{AB}

    and doubled the component j of \overline{AB}

    However the numbers I'm getting seem really odd.
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    Quote Originally Posted by maggiec View Post
    Proceeding with the method you have described is taking a very long time and leaving me \lambda~and~\mu to the power of 4 that are very hard to work out! Especially as this question is form a non-calculator exam.

    I think there must be and easier way to work this out.

    The way I had done it, is by working out \overline{AB}=\underline{r_{2}}-\underline{r_{1}}

    and then establishing a relationship between the line \overline{AB} and the position vector (2,1,2)
    This is possible since before i had proved that this position vector was perpendicular to both L_1:\underline{r_{1}}~and~L_2:\underline{r_{2}} and hence it must be parallel to the line \overline{AB}

    with this I equated the i and k components of \overline{AB}

    and doubled the component j of \overline{AB}

    However the numbers I'm getting seem really odd.
    Dear maggiec,

    For the moment I can't think of an easy method. But if you follow my method you will get \lambda^2~and~\mu^2 terms not \lambda^4~and~\mu^4 terms.

    The method you have used is incorrect. It is not reasonable to equate i and j components of AB and the vector (2,1,2)
    Last edited by Sudharaka; June 28th 2010 at 08:12 AM.
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    Quote Originally Posted by Sudharaka View Post
    Dear maggiec,
    Unfortunately that isn't working. I have looked at the product and \lambda~and~\mu both become 0.

    Do you want to check that my \overline{AB} vector is correct?


    i get: (13.5+7mu-3lambda)i + (-8.5+8mu+4lambda)j + (8.5-11mu+lambda)k
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  6. #6
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    Hello maggiec
    Quote Originally Posted by maggiec View Post
    Unfortunately that isn't working. I have looked at the product and \lambda~and~\mu both become 0.

    Do you want to check that my \overline{AB} vector is correct?


    i get: (13.5+7mu-3lambda)i + (-8.5+8mu+4lambda)j + (8.5-11mu+lambda)k
    I agree with your working, and also your method - which is to say that this vector is parallel to 2\vec i + \vec j + 2\vec k.

    I have done this, and I also get rather nasty numbers. I get the following equations for \lambda and \mu:
    18\mu - 4\lamda + 5 = 0 and 18\mu + 11 \lambda -61 = 0
    which give (if my working is correct):
    \lambda = \dfrac{33}{13} and \mu = \dfrac{67}{234}
    Is that what you make it?

    Grandad
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  7. #7
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    Quote Originally Posted by Grandad View Post
    Hello maggiecI agree with your working, and also your method - which is to say that this vector is parallel to 2\vec i + \vec j + 2\vec k.

    I have done this, and I also get rather nasty numbers. I get the following equations for \lambda and \mu:
    18\mu - 4\lamda + 5 = 0 and 18\mu + 11 \lambda -61 = 0
    which give (if my working is correct):
    \lambda = \dfrac{33}{13} and \mu = \dfrac{67}{234}
    Is that what you make it?

    Grandad
    Yeah!! that's exactly what i get!

    But from the answers I've been giving for this question, putting this back into the equation for each line doesn't give me the correct \overline{OA} and \overline{OB} vectors.

    These are meant to be: \overline{OA}=(-5i - 7.5j + k) and \overline{OB}=(8i - j + 14k)

    Sudharaka:

    I get to a point where 202 - (56/3) \lambda^2 + (3304/39) \lambda + (1981/39) + 234.((26 \lambda^2-118 \lambda - 70.75)/78)^2 - 14 \lambda = 0

    that squared after the bracket makes me think I will get \lambda^4 terms although I haven't expanded on it yet.
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  8. #8
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    Hello maggiec
    Quote Originally Posted by maggiec View Post
    Yeah!! that's exactly what i get!

    But from the answers I've been giving for this question, putting this back into the equation for each line doesn't give me the correct \overline{OA} and \overline{OB} vectors.

    These are meant to be: \overline{OA}=(-5i - 7.5j + k) and \overline{OB}=(8i - j + 14k)

    Sudharaka:

    I get to a point where 202 - (56/3) \lambda^2 + (3304/39) \lambda + (1981/39) + 234.((26 \lambda^2-118 \lambda - 70.75)/78)^2 - 14 \lambda = 0

    that squared after the bracket makes me think I will get \lambda^4 terms although I haven't expanded on it yet.
    The answer you've been given is definitely incorrect. The point A doesn't lie on the line L_1.

    With \lambda = -1, the point -5\vec i -7.5 \vec j + \vec k lies on the line
    \vec r = -2\vec i +11.5 \vec j +\lambda(3\vec i + 4 \vec j - \vec k).
    I suspect a typo somewhere.

    Grandad
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    Quote Originally Posted by Grandad View Post
    Hello maggiecThe answer you've been given is definitely incorrect. The point A doesn't lie on the line L_1.

    With \lambda = -1, the point -5\vec i -7.5 \vec j + \vec k lies on the line
    \vec r = -2\vec i +11.5 \vec j +\lambda(3\vec i + 4 \vec j - \vec k).
    I suspect a typo somewhere.

    Grandad




    Ok, I found the typo. I am attaching the answer I've been given and I have circled the mistake in red:


    Vectors-screen-shot-2010-06-28-18.59.39.png


    So this means the correct answer is as follows? ;

    \overline{OA}=((73/13)i + (35/26)j - (33/13)k)

    and

    \overline{OB}=((1580/117)i + (619/117)j + (626/117)k)

    ?

    Thanks
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  10. #10
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    Hello maggiec
    Quote Originally Posted by maggiec View Post
    Ok, I found the typo. I am attaching the answer I've been given and I have circled the mistake in red:


    Click image for larger version. 

Name:	Screen shot 2010-06-28 at 18.59.39.png 
Views:	7 
Size:	111.9 KB 
ID:	18034


    So this means the correct answer is as follows? ;

    \overline{OA}=((73/13)i + (35/26)j - (33/13)k)

    and

    \overline{OB}=((1580/117)i + (619/117)j + (626/117)k)

    ?

    Thanks
    Yes - I agree with your answers!

    I don't know - making me do all this arithmetic at my time of life!

    Grandad
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