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    two equations of lines:

    L1 (Point A) r = (9,5,2) + \lambda (-9,8,-2)

    L2 (Point B) r = (31,-3,20) + \mu (2,a,7)

    the first part was to find a which i did as the dot product of the direction vectors is zero ,because the lines are perpendicular so:

    (-9,8,-2) * (2,a,7) = a - 32

    therefore a = 32

    so we can re write eq 2 as

    r = (31,-3,20) + \mu (2,32,7)

    So this is the part i don't understand:

    It says find the coordinates of the point C where L1 and L2 intersect?

    How do I do this?
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    (Original post by a10)
    two equations of lines:

    L1 (Point A) r = (9,5,2) + \lambda (-9,8,-2)

    L2 (Point B) r = (31,-3,20) + \mu (2,a,7)

    the first part was to find a which i did as the dot product of the direction vectors is zero ,because the lines are perpendicular so:

    (-9,8,-2) * (2,a,7) = a - 32

    therefore a = 32

    so we can re write eq 2 as

    r = (31,-3,20) + \mu (2,32,7)

    So this is the part i don't understand:

    It says find the coordinates of the point C where L1 and L2 intersect?

    How do I do this?
    Your value of a is incorrect, check the dot product again.

    To find the intersection (after you've found a correctly), you seek \lambda and \mu such that the following equation holds:

    \begin{pmatrix} 9 \\ 5 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -9 \\ 8 \\ -2 \end{pmatrix} = \begin{pmatrix} 31 \\ -3 \\ 20 \end{pmatrix}  + \mu \begin{pmatrix} 2 \\ a \\ 7 \end{pmatrix}

    Can you see how this gives 3 equations in terms of \lambda and \mu? Use two of them to find values that work and then verify them in the 3rd equation.
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    (Original post by Farhan.Hanif93)
    Your value of a is incorrect, check the dot product again.

    To find the intersection (after you've found a correctly), you seek \lambda and \mu such that the following equation holds:

    \begin{pmatrix} 9 \\ 5 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -9 \\ 8 \\ -2 \end{pmatrix} = \begin{pmatrix} 31 \\ -3 \\ 20 \end{pmatrix}  + \mu \begin{pmatrix} 2 \\ a \\ 7 \end{pmatrix}

    Can you see how this gives 3 equations in terms of \lambda and \mu? Use two of them to find values that work and then verify them in the 3rd equation.
    Okay thanks, i did this with the previous question and worked out lambda and mew. But when i have the values what do i do with them to get the C co-ordinates.

    EDIT: I worked out a to be 4?

    \begin{pmatrix} 9 \\ 5 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -9 \\ 8 \\ -2 \end{pmatrix} = \begin{pmatrix} 31 \\ -3 \\ 20 \end{pmatrix}  + \mu \begin{pmatrix} 2 \\ 4 \\ 7 \end{pmatrix}
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    (Original post by a10)
    Okay thanks, i did this with the previous question and worked out lambda and mew. But when i have the values what do i do with them to get the C co-ordinates.

    EDIT: I worked out a to be 4?
    Yep.

    \begin{pmatrix} 9 \\ 5 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -9 \\ 8 \\ -2 \end{pmatrix} = \begin{pmatrix} 31 \\ -3 \\ 20 \end{pmatrix}  + \mu \begin{pmatrix} 2 \\ 4 \\ 7 \end{pmatrix}
    Just substitute the value of lambda into L1, or mu into L2, they should give the same result - the coordinates of C.

    In fact it's a useful check to do both.
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    (Original post by ghostwalker)
    Yep.



    Just substitute the value of lambda into L1, or mu into L2, they should give the same result - the coordinates of C.

    In fact it's a useful check to do both.
    i think i did something wrong on the last part the answers seem a bit obsurd,

    lambda = -54/10

    mew = -179/5

    EDIT: i just checked it in the third equation it seems iv done something wrong :/

    I'll try again:

    9-9\lambda = 31 + 2\mu (multiply this eq by 2)

    18-18\lambda = 62 + 4\mu

    5 + 8\lambda = -3+4\mu

    subtracting the two:

    13 - 26\lambda = 65



-26\lambda = 52



\lambda =-2
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    (Original post by a10)
    I had:

    9-9\lambda = 32 + 2\mu (multiply this eq by 2)
    Where did the 32 come from? It's not correct.
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    (Original post by ghostwalker)
    Where did the 32 come from? It's not correct.
    OMG :mad2: hahaha thanks!! it's still early in the morning :cry2:
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    (Original post by a10)
    i think i did something wrong on the last part the answers seem a bit obsurd,

    lambda = -54/10

    mew = -179/5

    EDIT: i just checked it in the third equation it seems iv done something wrong :/

    I'll try again:

    9-9\lambda = 32 + 2\mu (multiply this eq by 2)

    18-18\lambda = 64 + 4\mu

    5 + 8\lambda = -3+4\mu

    subtracting the two:

    13 - 26\mu = 67



-26\mu = 54



\mu =-27/13
    this is full of arithmetic mistakes.

    firstly, its 31, not 32, and -18 - 8 isnt 10. [EDIT: you've already corrected that while I was typing!]
    your answer is an integer. and of course you switched from lambda to mu halfway through so make sure you get that right or it will mess you up!
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    (Original post by c471)
    this is full of arithmetic mistakes.

    firstly, its 31, not 32, and -18 - 8 isnt 10. [EDIT: you've already corrected that while I was typing!]
    your answer is an integer. and of course you switched from lambda to mu halfway through so make sure you get that right or it will mess you up!
    haha yeahh thanks I've fixed it now xD
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    (Original post by a10)
    OMG :mad2: hahaha thanks!! it's still early in the morning :cry2:
    np
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    (Original post by ghostwalker)
    np
    okay so i got \mu = -2 and \lambda = -2

    i subbed them into the L1 and L2 equations and i have

    On the LHS i have (9,5,2)(18,-16,4) do i multiply these now to get C?
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    (Original post by a10)
    okay so i got \mu = -2 and \lambda = -2

    i subbed them into the L1 and L2 equations and i have

    On the LHS i have (9,5,2)(18,-16,4) do i multiply these now to get C?
    No,

    The definition of L1 is r=(9,5,2)+lambda(-9,8,-2)

    You just sub the value of lambda in, and then r will be the point of intersection.

    I.e. =(9,5,2)-2(-9,8,-2)
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    (Original post by ghostwalker)
    No,

    The definition of L1 is r=(9,5,2)+lambda(-9,8,-2)

    You just sub the value of lambda in, and then r will be the point of intersection.

    I.e. =(9,5,2)-2(-9,8,-2)
    i did that but then i don't understand how we then get the co-ordinates of C because we subbed both values of lambda and mew into either side?
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    (Original post by a10)
    i did that but then i don't understand how we then get the co-ordinates of C because we subbed both values of lambda and mew into either side?
    Not sure what you're subbing the values of the parameters into.

    You only need to sub lambda into L1, and this give you r (the position of C)
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    (Original post by Farhan.Hanif93)
    Your value of a is incorrect, check the dot product again.

    Can you see how this gives 3 equations in terms of \lambda and \mu? Use two of them to find values that work and then verify them in the 3rd equation.
    hi,

    EDIT: Could you help on the question below? Please
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    Example:

    \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}  + \mu \begin{pmatrix} 0 \\ 1 \end{pmatrix}

    In this case \lambda = 2 and \mu = 1 .

    Given that those parameters take those values, both equations refer to the point \begin{pmatrix} 1 \\ 0 \end{pmatrix} + 2 \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}  + 1 \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}

    So it doesn't really matter which one you use, as they will both reach the same value.
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    (Original post by aznkid66)
    Example:

    \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}  + \mu \begin{pmatrix} 0 \\ 1 \end{pmatrix}

    In this case \lambda = 2 and \mu = 1 .

    Given that those parameters take those values, both equations refer to the point \begin{pmatrix} 1 \\ 0 \end{pmatrix} + 2 \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}  + 1 \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}

    So it doesn't really matter which one you use, as they will both reach the same value.
    thanks

    how do you solve this question?

    Point X is the point on L1 such that the vector CX(with the arrow at the top) is perpendicular to the vector AB (again with the arrow at the top, don't know how to do this in latex haha)

    a) Find the co-ordinates of X
    b) Find the shortest distance of C from L1


    A= \begin{pmatrix} 4 \\ 2 \\ -5 \end{pmatrix}
    B= \begin{pmatrix} 10 \\ 2 \\ -7 \end{pmatrix}
    C= \begin{pmatrix} 12 \\ 5 \\ 9 \end{pmatrix}

    I worked out AB and subbed that in L1 to get the equation of the line it came out to be:

    r= \begin{pmatrix} 4 \\ 2 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 6 \\ 0 \\ -2 \end{pmatrix}

    From here i tried to work out vector CX which will be x-c so:

    X= \begin{pmatrix} x \\ y \\ z \end{pmatrix}
    C= \begin{pmatrix} 12 \\ 5 \\ 9 \end{pmatrix}

    Vector CX:

    \begin{pmatrix} x \\ y \\ z \end{pmatrix} - \begin{pmatrix} 12 \\ 5 \\ 9 \end{pmatrix} = \begin{pmatrix} x-12 \\ y-5 \\ z-9 \end{pmatrix}

    Is this correct?? What do i do next im stuck :/

    After trying the dot product i get:

    6x -2z= 54?
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    Wait, so are A and B also on the line?

    Well, to find the coordinates of X...

    Let the coordinates of X be (x,y,z). What is vector CX? What is vector AB? Knowing what CX dot AB is, can you make an equation of x, y, and z?

    I don't know how you found the equation of the line, but knowing that...

    (x,y,z) is a point on the line, so you can write the x, y, and z components in terms of lambda. Now, can you evaluate your previous equation (hint: substitution)?

    There is also a method pertaining to the cross product, but I'm not to fond of that one. You can post for more information if you're more familiar with that method, or just curious.
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    (Original post by aznkid66)
    Wait, so are A and B also on the line?

    Well, to find the coordinates of X...

    Let the coordinates of X be (x,y,z). What is vector CX? What is vector AB? Knowing what CX dot AB is, can you make an equation of x, y, and z?

    I don't know how you found the equation of the line, but knowing that...

    (x,y,z) is a point on the line, so you can write the x, y, and z components in terms of lambda. Now, can you evaluate your previous equation (hint: substitution)?

    There is also a method pertaining to the cross product, but I'm not to fond of that one. You can post for more information if you're more familiar with that method, or just curious.
    Could you please check...I've edited it but don't know how to continue? I tried the dot product being zero since it said perpendicular but then u get two unknowns x and z???
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    (x,y,z) is a point on the line, which means that there exists a lambda where r=(x,y,z)

    For that lambda, in terms of lambda according to the equation of the line x=...?
    Similarly, in terms of lambda z=...?
 
 
 
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