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C4 WJEC Vectors help?

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    Hi there, I've been told vectors are easy but I really can't get my head around them Here's a question I'm trying to do from the 2007 paper:

    9. (a) The position vectors of the points A and B, relative to a fixed origin O, are i + 3j – 2k and 3i + 6j + k, respectively.
    (i) Find AB.
    (ii) Find the vector equation of the line AB.
    (iii) The vector equation of the line L is r = 2i + 3j + 7k + μ (i + j + 4k).
    Given that L and AB intersect, find the position vector of the point of intersection.
    (b) Find the angle between the vectors i + 2j – k and 3i – j + 2k.


    for 9) i) am I right in thinking you have to use (i - 3i)^2 + (3j - 6j)^2 + (-2k - k)^2 and then take the square root of that? So 3j + 3k - 2?

    For 9) ii) I think you have to use "r = a + tb". a and b here would be the lines i + 3j – 2k and 3i + 6j + k? What is t?

    And as for the rest I have no idea. Thanks in advance
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    (Original post by Draco56)
    Hi there, I've been told vectors are easy but I really can't get my head around them Here's a question I'm trying to do from the 2007 paper:

    9. (a) The position vectors of the points A and B, relative to a fixed origin O, are i + 3j – 2k and 3i + 6j + k, respectively.
    (i) Find AB.
    (ii) Find the vector equation of the line AB.
    (iii) The vector equation of the line L is r = 2i + 3j + 7k + μ (i + j + 4k).
    Given that L and AB intersect, find the position vector of the point of intersection.
    (b) Find the angle between the vectors i + 2j – k and 3i – j + 2k.


    for 9) i) am I right in thinking you have to use (i - 3i)^2 + (3j - 6j)^2 + (-2k - k)^2 and then take the square root of that? So 3j + 3k - 2?
    No, to find the direction vector, you just have to subtract OA from OB or the other way round.
    For 9) ii) I think you have to use "r = a + tb". a and b here would be the lines i + 3j – 2k and 3i + 6j + k? What is t?
    a could be either position vector of A or position vector of B. b is the direction vector of AB(which you're being asked in part (i)), and t is just a parameter.

    And as for the rest I have no idea. Thanks in advance
    Put vector equation of AB equal to vector equation of L, and find values of parameters μ and t by solving the simultaneous equations. Do you know the formula to find angle between two lines? The one which uses dot/scalar product?
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    Cheers! Gimme a minute and I'll see what answer I get.

    Okay so:

    a) i) 2i + 3j = 3k
    ii) r = i + 3j - 2k + t(2i + 3j + 3k)
    iii) -i - 5k
    b) 96.3

    Correct?
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    (Original post by Draco56)
    Cheers! Gimme a minute and I'll see what answer I get.

    Okay so:

    a) i) 2i + 3j = 3k
    ii) r = i + 3j - 2k + t(2i + 3j + 3k)
    iii) -i - 5k
    b) 96.3

    Correct?
    All are correct, except your answer to (b). Did you use \cos \theta = \dfrac{\vec{AB}\cdot \vec{L}}{|\vec{AB}||\vec{L}|}? That gives me an angle of 31.3 degrees.
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    (Original post by Zishi)
    All are correct, except your answer to (b). Did you use \cos \theta = \dfrac{\vec{AB}\cdot \vec{L}}{|\vec{AB}||\vec{L}|}? That gives me an angle of 31.3 degrees.
    I used that formula, but why AB and L? The question tells us to use i + 2j - k and 3i - j + 2k? Unless those are AB and L in this case and I did the calculations wrong. I did:

    (1 x 3) + (2 x -1) + (2 x -1)
    ----------------------------------------------
    sqrt(1^2 + 2^2 - 1^2)*sqrt(3^2 - 1^2 + 2^2)

    Then took the inverse cos.
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    (Original post by Draco56)
    I used that formula, but why AB and L? The question tells us to use i + 2j - k and 3i - j + 2k? Unless those are AB and L in this case and I did the calculations wrong. I did:

    (1 x 3) + (2 x -1) + (2 x -1)
    ----------------------------------------------
    sqrt(1^2 + 2^2 - 1^2)*sqrt(3^2 - 1^2 + 2^2)

    Then took the inverse cos.
    Ah, I didn't read the question again, so I thought that it was asking for the angle between AB and L. Anyway, you answer of 96.3 degrees is correct then.
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    Ah I see Thanks a lot for the help!

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