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Thread: Vector help

  1. #1
    Senior Member I-Think's Avatar
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    Vector help

    GCE Maths
    With respect to an origin O, the pts. A and B have the following position vectors

    OA=2i+2j+k OB=i+4j+3k

    The line has vector equation
    r=4i-2j+2k+s(i+2j+k)

    I. Prove that the line l does not intersect the line through A and B

    II. Find the equation of the plane containing l and the point A giving your answer in the form ax+by+cz=d
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  2. #2
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    Some explaining please

    Quote Originally Posted by I-Think View Post
    GCE Maths
    With respect to an origin O, the pts. A and B have the following position vectors

    OA=2i+2j+k OB=i+4j+3k

    The line has vector equation
    r=4i-2j+2k+s(i+2j+k) what is s?
    is this "line" a sum of the vectors OA;OB? if so is the point of the exercise to find s?

    I. Prove that the line l does not intersect the line through A and B
    If a and b are vectors it does not matter where on the cordinate system you place them. Not sure what you mean here...?

    II. Find the equation of the plane containing l $\displaystyle \Rightarrow$ what is this? and the point A giving your answer in the form ax+by+cz=d
    $\displaystyle \Rightarrow$ is that the sum of vectors OA;OB?
    Regards H (sorry but really difficult to understand without a picture)..
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  3. #3
    MHF Contributor
    Grandad's Avatar
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    Hello I-Think
    Quote Originally Posted by I-Think View Post
    GCE Maths
    With respect to an origin O, the pts. A and B have the following position vectors

    OA=2i+2j+k OB=i+4j+3k

    The line has vector equation
    r=4i-2j+2k+s(i+2j+k)

    I. Prove that the line l does not intersect the line through A and B

    II. Find the equation of the plane containing l and the point A giving your answer in the form ax+by+cz=d
    First we need to find the vector equation of the line through $\displaystyle A$ and $\displaystyle B$. This can be written as
    $\displaystyle \vec r = \vec{OA}+t\vec{AB}$ ...(1)
    where $\displaystyle t$ is a scalar parameter. Do you understand this bit? It simply says:
    $\displaystyle \underbrace{\vec r=}_{\text{You can move from O to any point on the line ... }}$$\displaystyle \underbrace{\vec{OA}}_{\text{ by a movement to A ...}}\underbrace{+}_{\text{ followed by ...}}\underbrace{t\vec {AB}}_{\text{ a movement in the direction of AB}}$
    Now
    $\displaystyle \vec{OA} = 2\vec i + 2\vec j + \vec k$
    and
    $\displaystyle \vec{AB} = \vec{OB} - \vec{OA}$
    $\displaystyle = -\vec i +2 \vec j +2\vec k$
    So, from (1), the equation of the line through $\displaystyle AB$ is:
    $\displaystyle \vec r = 2\vec i + 2\vec j + \vec k+t(-\vec i +2 \vec j +2\vec k)$
    Next, we need to prove that this line doesn't intersect line $\displaystyle l$. We do this by showing that there are no values of $\displaystyle s$ and $\displaystyle t$ that will bring us to the same point on each of the lines. So, suppose that the lines meet. Then, for some value of $\displaystyle s$ and of $\displaystyle t$:
    $\displaystyle 2\vec i + 2\vec j + \vec k+t(-\vec i +2 \vec j +2\vec k)= 4\vec i-2\vec j+2\vec k+s(\vec i+2\vec j+\vec k)$
    Comparing the coefficients of $\displaystyle \vec i$ and $\displaystyle \vec j$, we get:
    $\displaystyle 2-t = 4+s$
    and
    $\displaystyle 2+2t = -2+2s$
    which gives:
    $\displaystyle s = 0,\;t=-2$
    If we now compare the coefficients of $\displaystyle \vec k$:
    On the LHS: $\displaystyle 1-4 = -3$

    On the RHS: $\displaystyle 2 + 0 = 2$
    Since these are not equal, there are no values of $\displaystyle s$ and $\displaystyle t$ that satisfy the equation, and hence the lines do not intersect.

    Part II

    We have a choice of two methods here. The first uses simple algebra; the second, rather more sophisticated, uses a vector and scalar product.

    For the first method, we find a couple of points on the line $\displaystyle l$ by taking two values of $\displaystyle s$, and then solve a set of simultaneous equations.

    So, with $\displaystyle s = 0$ and $\displaystyle s = 1$, we get the points $\displaystyle (4, -2, 2)$ and $\displaystyle (5, 0, 3)$ lying on $\displaystyle l$, and therefore also in the plane. And the point $\displaystyle A$ also lies in the plane.

    So, if the plane is $\displaystyle ax+by+cz=1 $ (which is simply the equation $\displaystyle ax+by+cz=d$ re-written when divided throughout by $\displaystyle d$, using new values of $\displaystyle a, b, c$), we get:
    $\displaystyle \left \{\begin{array}{l}
    4a-2b+2c=1\\
    5a+0b+3c=1\\
    2a+2b+c=1\\
    \end{array}\right.$
    This gives:
    $\displaystyle \left \{\begin{array}{l}
    a=1\\
    b=\tfrac16\\
    c=-\tfrac43\\
    \end{array}\right.$
    and, when we get rid of fractions, gives the plane:
    $\displaystyle 6x+y-8z=6$
    If you want to see an explanation of the vector method, look at this page.

    Grandad
    Last edited by Grandad; Feb 9th 2010 at 06:59 AM. Reason: I missed out a letter z
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  4. #4
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    Well

    Wow!!
    Grandad is the master
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  5. #5
    Senior Member I-Think's Avatar
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    Grandad, the day I learn to teach as well as you is the day I will be a better person
    Last edited by I-Think; Feb 21st 2010 at 02:29 AM. Reason: Spelling errors, my english teacher will be appalled
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