[Core 4] 3D Coordinate Geometry

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    I found the first two parts easy, I got s = -1, t = 1, a = 4, and b = -3 which I assume is correct as I managed to show the position vector in part 2.

    However, I really don't understand part 3. I don't even know where to start. I assume it involves Pythagoras somewhere but I'm not sure where to go with that.

    Any help would be appreciated. Thanks in advance.
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    (Original post by DarkEnergy)
    Any help would be appreciated. Thanks in advance.
    You know that every single point on \ell_1 is of the form (9 + 3s, 2 - 4s, -1) for some s. So once you've found the required s, you've found the point.

    You know the distance between two points in terms of Pythagoras. Here, the distance will be:

    \displaystyle \sqrt{((9+3s) - 6)^2 + ((2 - 4s) - 6)^2 + \cdots} = 10

    Then square both sides, expand out the brackets, and solve for s (it'll be a quadratic!) and then plug it back into the point.
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    (Original post by Zacken)
    You know that every single point on \ell_1 is of the form (9 + 3s, 2 - 4s, -1) for some s. So once you've found the required s, you've found the point.

    You know the distance between two points in terms of Pythagoras. Here, the distance will be:

    \displaystyle \sqrt{((9+3s) - 6)^2 + ((2 - 4s) - 6)^2 + \cdots} = 10

    Then square both sides, expand out the brackets, and solve for s (it'll be a quadratic!) and then plug it back into the point.
    I got 25s^2 + 40s - 71 = 0 as my quadratic which doesn't seem quite right as it would give weird numbers for my vector. I solved s on my calculator to be 1.0654 or -2.665. Is this right before I continue further? Thanks!

    edit: just realised i put 40s instead of 50s.. gonna recalculate
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    (Original post by DarkEnergy)
    I got 25s^2 + 40s - 71 = 0 as my quadratic which doesn't seem quite right as it would give weird numbers for my vector. I solved s on my calculator to be 1.0654 or -2.665. Is this right before I continue further? Thanks!
    Yup, you've definitely mixed up somewhere. You should have (3 + 3s)^2 + (-4-4s)^2 + (-1 - -1)^2 = 100 \iff 25s^2 + 50s - 75 = 0
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    (Original post by DarkEnergy)
    I got 25s^2 + 40s - 71 = 0 as my quadratic which doesn't seem quite right as it would give weird numbers for my vector. I solved s on my calculator to be 1.0654 or -2.665. Is this right before I continue further? Thanks!
    Check your expansion again.
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    (Original post by Zacken)
    Yup, you've definitely mixed up somewhere. You should have (3 + 3s)^2 + (-4-4s)^2 + (-1 - -1)^2 = 100 \iff 25s^2 + 50s - 75 = 0
    (Original post by RDKGames)
    Check your expansion again.
    Thanks.. can't believe I'm that rusty. I got s = 1 and s = -3, so I used s = 1 to get x = 12, y = -2, z = -1 as my final position vector. Is this correct?
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    (Original post by DarkEnergy)
    Thanks.. can't believe I'm that rusty. I got s = 1 and s = -3, so I used s = 1 to get x = 12, y = -2, z = -1 as my final position vector. Is this correct?
    Yep, either one would be okay.
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    (Original post by Zacken)
    Yep, either one would be okay.
    Thanks for your help!
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    (Original post by DarkEnergy)
    Thanks for your help!
    You're welcome!
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    how much harder is c4 geometry to c1 geometry?
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    (Original post by gomc)
    how much harder is c4 geometry to c1 geometry?
    Depends on the person. Subjectivity and all that, etc...
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    (Original post by gomc)
    how much harder is c4 geometry to c1 geometry?
    Different geometry approach. C4 deals with vectors and C1 deals with your normal every day x-y plane.
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    I think a simpler way to do these types of questions is to realise that a distance of 10 is 2 of the  \begin{pmatrix} 3\\ -4 \\ 0 \end{pmatrix} direction vectors. So 10 units either side of A on the line  \ell _1 is the position vector  \begin{pmatrix} 6 \\ 6 \\ -1 \end{pmatrix} \pm 2 \begin{pmatrix} 3 \\ -4\\ 0 \end{pmatrix} .
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    (Original post by gomc)
    how much harder is c4 geometry to c1 geometry?
    I didn't do very good in C1 last year (78 UMS) but I'm finding C4 to not be too bad so far, although my work ethic is a lot better this year so it might come down to that too.
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    (Original post by B_9710)
    I think a simpler way to do these types of questions is to realise that a distance of 10 is 2 of the  \begin{pmatrix} 3\\ -4 \\ 0 \end{pmatrix} direction vectors. So 10 units either side of A on the line  \ell _1 is the position vector  \begin{pmatrix} 6 \\ 6 \\ -1 \end{pmatrix} \pm 2 \begin{pmatrix} 3 \\ -4\\ 0 \end{pmatrix} .
    I don't understand what you mean by the first part
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    (Original post by DarkEnergy)
    I don't understand what you mean by the first part
    If you start at any part on the line (A lies on the line), to get to any other point of the line, you need to move in the direction of (3, -4, 0). (which is why it's called the direction vector). If you move off in any other direction, you're no longer on the line.

    So since the magnitude of (3, -4, 0) is 5. That means whenever you move in that direction by one "step" you move 5 units of distance away from A. So to get to the point on the line that's 10 units away from A, you start at A and move two "steps" in the direction (3, -4, 0) to get to the required point on the line. This can either be two units forward or two units backwards.

    NB: I personally prefer this method myself, but it's not a very general one and I thought it'd confuse you a bit at first - which is why I went with my previous explanation, since you were already thinking along the lines of Pythagoras.
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    (Original post by Zacken)
    If you start at any part on the line (A lies on the line), to get to any other point of the line, you need to move in the direction of (3, -4, 0). (which is why it's called the direction vector). If you move off in any other direction, you're no longer on the line.

    So since the magnitude of (3, -4, 0) is 5. That means whenever you move in that direction by one "step" you move 5 units of distance away from A. So to get to the point on the line that's 10 units away from A, you start at A and move two "steps" in the direction (3, -4, 0) to get to the required point on the line. This can either be two units forward or two units backwards.

    NB: I personally prefer this method myself, but it's not a very general one and I thought it'd confuse you a bit at first - which is why I went with my previous explanation, since you were already thinking along the lines of Pythagoras.
    Ah right cheers, I get it now. On a side note, how many marks do you think this whole question would be worth in an exam?
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    (Original post by DarkEnergy)
    Ah right cheers, I get it now. On a side note, how many marks do you think this whole question would be worth in an exam?
    Probably around 3 marks
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    (Original post by B_9710)
    I think a simpler way to do these types of questions is to realise that a distance of 10 is 2 of the  \begin{pmatrix} 3\\ -4 \\ 0 \end{pmatrix} direction vectors. So 10 units either side of A on the line  \ell _1 is the position vector  \begin{pmatrix} 6 \\ 6 \\ -1 \end{pmatrix} \pm 2 \begin{pmatrix} 3 \\ -4\\ 0 \end{pmatrix} .
    There is a more general version of this (very useful) approach, based on the realisation that l_1 can be restated as:

    \begin{pmatrix} 6\\6\\-1\end{pmatrix}+u\begin{pmatrix} 3\\-4\\0\end{pmatrix}

    (Remember that any point on the line will do for the first part of the vector line equation, and we might as well use the point A). A general point on the line is then

    \begin{pmatrix} 6+3u\\6-4u\\-1\end{pmatrix}

    If we use that to represent some other point on the line, which we'll call B, the vector extending from A to B is

    \vec{AB}=\begin{pmatrix} 6+3u\\6-4u\\-1\end{pmatrix}-\begin{pmatrix} 6\\6\\-1\end{pmatrix}=\begin{pmatrix} 3u\\-4u\\0\end{pmatrix}

    It's then quite easy to find the value(s) of u for which |AB|=10 and then to substitute those value(s) back into the (revised) line equation.
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    (Original post by old_engineer)
    (..)
    Thank you very much!
 
 
 
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