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    Hi everyone!

    I'm doing a question and i'm struggling on it, can someone help?

    Referred to a fixed origin O, the points A, B and C have position vectors (9i - 2j + k), (6i + 2j + 6k) and (3i + pj + qk) respectively, where p and q are constants.

    a) Find, in vector form, an equation of the line l which passes through A and B. (2 marks)


    My method:

    using r = a + t(b-a) :

    r = (9i - 2j + k) + t(-3i + 4j + 5k)

    this answer is correct.

    Given that C lies on l,
    b)Find the value of p and q. (2 marks)


    My method:

    i thought i could compare the (-3i + 4j + 5k) part of the equation of the line to the position vector of C which is (3i + pj + qk).

    so the i coefficient is multiplied by -1 to get from -3 => 3
    so i'd guess that the j and k's are therefore -4 and -5 respectively......

    obviously i'm wrong. the markscheme is handwritten by my teacher and i think she's written p= 6 and q= 11.

    how do i do it?
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    Use the coord you do know (the i coord - which is 3) and equate it with your line l

    So:

    9(i) + -3t(i) = 3

    (i)'s can be ignored but i've put them in to help your understanding.

    This will give you t.
    Then use that to work out the other two coords.
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    sub t = 2 into l.
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    You need to rearrange the equation of the line you've found into
    l = (9-3t)i + (4t-2)j + (1+5t)k

    you can then compare the coordinates of C with this line; i.e.

    9-3t = 3
    4t-2 = P
    1+5t = Q

    From the first one you get t=2, so you sub that in and get p and q
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    You dont need to rearrange anything.

    r = (9i - 2j + k) + t(-3i + 4j + 5k)

    9i+t(-3i) = 3i, so t = 2
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    bugzy, gyrase and tgg208 - thanks u three for the help.
    i cannot believe i spent 10 minutes staring at the question !!!!
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    i've been trying to the next parts of the question in the first post, but i'm struggling again

    here's what i managed:

    c) calculate, in degrees, the acute angle between OC and AB.

    Vector OC = (3i + 6j + 11k)

    Vector AB = Vector OB - Vector OA
    = (-3i + 4j +5k)

    then i found a.b = 70

    then i found |a| = sqroot of 166 and |b| = sqroot of 50

    then i plugged the values into the scalar dot product formula to get the correct answer of 40 degrees (nearest degree).

    My problem is the next part....



    The point D ies on AB and is such that OD is perpendicular to AB.
    d) Find the position vector of D.


    my attempt:
    make Vector OD = c

    c = (xi + yj +zk)

    then i made the scalar product of a.c and b.c = 0 (because the vector OD is perp.)

    a.c = 9x - 2y - z = 0
    b.c = 6x + 2y + 6z =0

    then i did simultaneous equations to solve firstly for x and i got x = 1/3.

    i checked the MS and the coefficient of x isn't 1/3 its meant to be 36/5, so im already wrong

    how do i do this guys ???? thanks so much

    by the way, the answer is meant to be 36/5i + 2/5j -2k

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    (Original post by sheena18)
    i've been trying to the next parts of the question in the first post, but i'm struggling again

    here's what i managed:

    c) calculate, in degrees, the acute angle between OC and AB.

    Vector OC = (3i + 6j + 11k)

    Vector AB = Vector OB - Vector OA
    = (-3i + 4j +5k)

    then i found a.b = 70

    then i found |a| = sqroot of 166 and |b| = sqroot of 50

    then i plugged the values into the scalar dot product formula to get the correct answer of 40 degrees (nearest degree).

    My problem is the next part....



    The point D ies on AB and is such that OD is perpendicular to AB.
    d) Find the position vector of D.


    my attempt:
    make Vector OD = c

    c = (xi + yj +zk)

    then i made the scalar product of a.c and b.c = 0 (because the vector OD is perp.)

    a.c = 9x - 2y - z = 0
    b.c = 6x + 2y + 6z =0

    then i did simultaneous equations to solve firstly for x and i got x = 1/3.

    i checked the MS and the coefficient of x isn't 1/3 its meant to be 36/5, so im already wrong

    how do i do this guys ???? thanks so much

    by the way, the answer is meant to be 36/5i + 2/5j -2k

    You didn't need to find a.c or b.c, there's a much simpler way.

    \vec{AB} = \begin{bmatrix} -3\\4\\5\end{bmatrix} and \vec{OD} = \begin{bmatrix} x\\y\\z\end{bmatrix} = \begin{bmatrix} 9-3t\\-2+4t\\1+5t\end{bmatrix}

    Since OD is perpendicular to AB, the dot product of both is zero, i.e. \vec{AB}.\vec{OD} = 0

    \Rightarrow \begin{bmatrix} -3\\4\\5\end{bmatrix}.\begin{bmat  rix} 9-3t\\-2+4t\\1+5t\end{bmatrix} = 0

    Therefore, [-3(9-3t)]+[4(-2+4t)]+[5(1+5t)]=0
    Solve the equation, find the value of t and put it in the equation for OD and that will be your position vector for D.
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    thank you so so much for replying!
    i can see what you did, but one part that i don't understand is how the vector equation for OD is (9-3t)i + (-2 + 4t)j + (1 + 5t)k ?

    i think you used the r = a + tb formula, with a = (9i - 2j + k) and b = (-3i + 4j + 5k).

    would you please be able to explain the thought processes you went through to get the vector equation of OD, as i wouldn't know how to go about it
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    (Original post by sheena18)
    thank you so so much for replying!
    i can see what you did, but one part that i don't understand is how the vector equation for OD is (9-3t)i + (-2 + 4t)j + (1 + 5t)k ?

    i think you used the r = a + tb formula, with a = (9i - 2j + k) and b = (-3i + 4j + 5k).

    would you please be able to explain the thought processes you went through to get the vector equation of OD, as i wouldn't know how to go about it
    Well basically the vector equation of vector OD is the same as the vector equation for point D. And if a point lies on a line, then it's vector equation is the same as the equation of the line. Hence I used the equation of line AB (the whole equation within one bracket) to find vector equation for point D which is the same as vector equation of vector OD.

    Edit: i don't know whether that's going to make much sense, but I couldn't explain any better and Yes I used r = a+tb formula
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    (Original post by pisces_abhi)
    Well basically the vector equation of vector OD is the same as the vector equation for point D. And if a point lies on a line, then it's vector equation is the same as the equation of the line. Hence I used the equation of line AB (the whole equation within one bracket) to find vector equation for point D which is the same as vector equation of vector OD.

    Edit: i don't know whether that's going to make much sense, but I couldn't explain any better and Yes I used r = a+tb formula
    firstly, i really appreciate the efforts u've gone to, to help!
    i've got some questions with some stuff (im sorry )

    blue bit: point D lies on both OD and AB, so if the vector equation of the line is the same as the vector equation for a point, wouldn't OD and AB have the same vector equation? (oh, im not undermining u at all, i know i'm wrong, im asking why im worng lol)

    red bit: i suppose i dont get this because i dont get the part before

    im really sorry for this (i'll rep u tomorrow if it makes u feel any less irritated with my dumbness )
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    Are you sure the answer isnt 36/5, 2/5 and 4 because i got 0.6 as t
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    hearty90: my markscheme is handwritten by my teacher, so you could well be right.
    and pleaaaaaaase how did you find the vector equation for OD?
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    (Original post by sheena18)
    firstly, i really appreciate the efforts u've gone to, to help!
    i've got some questions with some stuff (im sorry )

    blue bit: point D lies on both OD and AB, so if the vector equation of the line is the same as the vector equation for a point, wouldn't OD and AB have the same vector equation? (oh, im not undermining u at all, i know i'm wrong, im asking why im worng lol)

    red bit: i suppose i dont get this because i dont get the part before

    im really sorry for this (i'll rep u tomorrow if it makes u feel any less irritated with my dumbness )
    I apologise. I'm really awful with the vector jargon. I can never seem to use the appropriate words.

    Well basically the vector equation of vector OD is the same as the vector equation for point D. And if a point lies on a line, then it's vector equation is the same as the equation of the line.
    (Original post by sheena18)
    blue bit: point D lies on both OD and AB, so if the vector equation of the line is the same as the vector equation for a point, wouldn't OD and AB have the same vector equation? (oh, im not undermining u at all, i know i'm wrong, im asking why im worng lol)
    The position vector of a point (as you must know) is the distance from the origin (O). Therefore it's okay to say that position vector of pt. D is the same as vector for line OD (can't put this in better words i'm sorry )

    And actually I was wrong when I said vector equation of a line is the same as that of a point on it. That's wrong, because on a point the parameter (lambda/t) is a constant, while for a line that lambda can be anything since the line goes on forever and the point, well, is just a point (it has fixed coordinates).

    Edit: Since the parameter is a constant for pt. D, we just used the equation of the line to find the coordinates of D in terms of the parameter. And as I said before, vector OD is the same as position vector of D (which we just found from the equation of line). Tell me if it makes any sense this time and I'll try once again afterwards.

    Don't worry abt the rep btw. Don't really care much abt it
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    i just did what pisces abhi said lol..use that:

    [-3(9-3t)]+[4(-2+4t)]+[5(1+5t)]=0

    i got 50t=30
    t=0.6
    then sub it in2 vector OD.
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    (Original post by hearty90)
    i just did what pisces abhi said lol..use that:

    [-3(9-3t)]+[4(-2+4t)]+[5(1+5t)]=0

    i got 50t=30
    t=0.6
    then sub it in2 vector OD.
    You're right. I didn't actually carry out the working. I think your teacher has made a tiny mistake in the mark scheme. Be quick in pointing out the mistake to your teacher sheena, before someone else steals your moemnt of glory :p:
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    abhi, no need to aplogise! its my fault i dont understand, ur explanations are very good, its me who's not getting it lol

    i get the position vector of D is = vector OD, but im afraid that still doesn't help.

    with all of the help of your explanations i tried to understand finding vector equation OD

    could you find it by doing the triangle law and doing:
    OA + tAB = OD (i dumped the t in there because we don't know where D is on AB)
    where OA = -9i - 2j + k
    and AB = -3i + 4j + 5k


    actually, the reason why im not getting how to get vector equation OD, might be because i'm not understanding the vector equation formula, so i don't really understand what to sub in as a and b

    is a basically position vector and is b the direction it moves in?
    im really really really sorry for taking up your time like this, i feel really guilty!
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    I understand how to create the vector equation and how to find P and Q and so fourth. But in vectors I really can't get my head around a lot of the stuff. Like there's a question saying 'Point D is the point on l closest to the origin, O. FInd the co-ordinates of D'. No idea. The questions are so ridiculous, yet are going to lose me marks.

    Is there any where/way I can easily learn vectors and cram it in before the 12th?? I'm quite a good crammer, but can't get my head around them.
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    finaaaaallllllllllyyy, i get what you mean

    i feel like swearing and jumping up and down with happiness at the same time lol

    abhi, im seriously indebted to you - u spent an hour trying to explain this to me, if there's ever anything i can help with bio/chem/maths (haha)/french/medicine/kings
    then feel free to PM, i do owe u!
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    jarve99, as u can see by this thread, i am in no position to offer advice on C4 vectors lol,
    i think just keep at it, and go through the chapter in the book carefully and then do lots of exam q's -standard stuff really
 
 
 
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