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    Hi! I Was Just Wondering If You Had Looked At Jan 2004 Edexcel Paper P3. I Wanted To Ask Help On The Last Question Of The Paper. Thanks
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    Do you mean the vectors question if so here it is:

    L1 : r = i + 3j + 5k + lamda(i + 2j - k)
    L2 : r = -2i + 3j - 4k + mew(2i + j + 4k)

    a) Show L1 and L2 intersect and then give the coordinates of this intersection Q.

    b) Show l1 and l2 are perpendicular

    c) Point P is on L1 with x-coordinate 3. Point R is on L2 with x-coordinate 4 .

    Find in the simplest form the exact area of Triangle PQR.

    I have offered my solution to this on the other thread but would be interested to see other solutions and thoughts. Mensandan
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    (Original post by thecaped)
    Hi! I Was Just Wondering If You Had Looked At Jan 2004 Edexcel Paper P3. I Wanted To Ask Help On The Last Question Of The Paper. Thanks
    Why don't you PM him?

    Or at least put this on the academic subforum.
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    Good point, keep forgetting to do that. Will try harder in future. Good idea about the PM though, cheers dude.
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    Forgive me if I make any mistakes, as I'm tired and doing this as I type. Also, I use the convention a = lambda, b = mew.

    (Original post by mensandan)
    Do you mean the vectors question if so here it is:

    L1 : r = i + 3j + 5k + lamda(i + 2j - k)
    L2 : r = -2i + 3j - 4k + mew(2i + j + 4k)

    a) Show L1 and L2 intersect and then give the coordinates of this intersection Q.
    To find where they intersect, solve the system of linear equations (or just spot appropriate values of a and b by sight):

    a - 2b = -3
    2a - b = 0
    a + 4b = 9

    This yields one solution a = 1, b = 2, and the point of intersection Q = (2, 5, 4).

    (Original post by mensandan)
    b) Show l1 and l2 are perpendicular
    The dot product of their direction vectors is:

    (1, 2, -1).(2, 1, 4) = 2 + 2 - 4= 0, which implies that the angle between the lines has cosine of 0, so they are perpendicular.

    (Original post by mensandan)
    c) Point P is on L1 with x-coordinate 3. Point R is on L2 with x-coordinate 4 .

    Find in the simplest form the exact area of Triangle PQR.
    We want the area of the triangle defined by the three points P, Q and R, where:

    P = (3, 7, 3), Q = (2, 5, 4) and R = (4, 6, 8)

    If we now perform a rigid transformation of the plane (so it does not affect the triangle's shape and thus its area), by translating the plane such that Q is at the origin, we have reduced the problem to finding the area of the triangle defined by the points:

    O (the origin), P' = (1, 2, -1) and R' = (2, 1, 4)

    It's easy to prove that the area of a triangle with one corner at the origin and the other two corners given by position vectors t and s is (1/2)|t x s|, so the area of our triangle is:

    Area = (1/2) | (1, 2, -1) x (2, 1, 4) | = (1/2)Sqrt(126) = (3/2)Sqrt(14)

    As long as I haven't made any calculation errors.

    I hope this helps,
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    (Original post by rahaydenuk)
    Forgive me if I make any mistakes, as I'm tired and doing this as I type. Also, I use the convention a = lambda, b = mew.



    To find where they intersect, solve the system of linear equations (or just spot appropriate values of a and b by sight):

    a - 2b = -3
    2a - b = 0
    a + 4b = 9

    This yields one solution a = 1, b = 2, and the point of intersection Q = (2, 5, 4).



    The dot product of their direction vectors is:

    (1, 2, -1).(2, 1, 4) = 2 + 2 - 4= 0, which implies that the angle between the lines has cosine of 0, so they are perpendicular.



    We want the area of the triangle defined by the three points P, Q and R, where:

    P = (3, 7, 3), Q = (2, 5, 4) and R = (4, 6, 8)

    If we now perform a rigid transformation of the plane (so it does not affect the triangle's shape and thus its area), by translating the plane such that Q is at the origin, we have reduced the problem to finding the area of the triangle defined by the points:

    O (the origin), P' = (1, 2, -1) and R' = (2, 1, 4)

    It's easy to prove that the area of a triangle with one corner at the origin and the other two corners given by position vectors t and s is (1/2)|t x s|, so the area of our triangle is:

    Area = (1/2) | (1, 2, -1) x (2, 1, 4) | = (1/2)Sqrt(126) = (3/2)Sqrt(14)

    As long as I haven't made any calculation errors.

    I hope this helps,
    simplay the best mathematician i know.
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    (Original post by rahaydenuk)
    We want the area of the triangle defined by the three points P, Q and R, where:

    P = (3, 7, 3), Q = (2, 5, 4) and R = (4, 6, 8)

    If we now perform a rigid transformation of the plane (so it does not affect the triangle's shape and thus its area), by translating the plane such that Q is at the origin, we have reduced the problem to finding the area of the triangle defined by the points:

    O (the origin), P' = (1, 2, -1) and R' = (2, 1, 4)

    It's easy to prove that the area of a triangle with one corner at the origin and the other two corners given by position vectors t and s is (1/2)|t x s|, so the area of our triangle is:

    Area = (1/2) | (1, 2, -1) x (2, 1, 4) | = (1/2)Sqrt(126) = (3/2)Sqrt(14)
    Of course your method is completely sound, as long as you know about cross/vector products and determinants; but I don't think those come along until P6!

    Therefore for the benefit of those who did not have knowledge of P6 methods in their P3 exam I will provide an alternative solution for this part of the question.

    We know that the lines L1 and L2 are perpendicular, and that the points P (3,7,3) and R (4,6,8) lie on lines L1 and L2 respectively. We also know that Q (2,5,4) is the point of intersection between these two lines. Hence we are dealing with a right angled triangle and can calculate the area by using the GCSE formula 1/2 * base * height. We can calculate the length of the base and height by pythagoras. Thus...

    height = sqrt((3-2)^2+(7-5)^2+(3-4)^2) = sqrt6
    base = sqrt((2-4)^2+(5-6)^2+(4-8)^2) = sqrt21

    => Area = 1/2 * sqrt6 * sqrt21 = 1/2 * sqrt126 = 3/2 * sqrt14
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    The question about solving the partial fraction differential equation, what was the final answer, please?
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    (Original post by thecaped)
    The question about solving the partial fraction differential equation, what was the final answer, please?
    The Partial Fraction was:

    y = 4/(2x-3) - 3/(x+1)

    Integrating you get:

    Lny = 2Ln(2x-3) - 3Ln(x+1) + c

    Now it said y=4 when x=2 so:

    ln4 = ln 1 - ln 27 + c

    So C = ln 4 + ln 27 = ln 108

    Final answer

    ln y = 2ln(2x-3) - 3ln(x+1) + ln108
    ln y = 108 ((2x-3)^2/(x+1)^3)
    Or
    y=108((2x-3)^2/(x+1)^3)
    This seems to be correct though I think I messed up the constant part in the exam. Never mind.
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    (Original post by mensandan)
    The Partial Fraction was:

    y = 4/(2x-3) - 3/(x+1)

    Integrating you get:

    Lny = 2Ln(2x-3) - 3Ln(x+1) + c

    Now it said y=4 when x=2 so:

    ln4 = ln 1 - ln 27 + c

    So C = ln 4 + ln 27 = ln 108

    Final answer

    ln y = 2ln(2x-3) - 3ln(x+1) + ln108
    ln y = 108 ((2x-3)^2/(x+1)^3)
    Or
    y=108((2x-3)^2/(x+1)^3)
    This seems to be correct though I think I messed up theconstant part in the exam. Never mind.
    wow. another good mathematician on the forum....
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    (Original post by Darkness)
    Of course your method is completely sound, as long as you know about cross/vector products and determinants; but I don't think those come along until P6!
    I didn't use determinants in that solution. :confused: However, I think you're right about cross products not coming up until P6. My apologies.
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    (Original post by rahaydenuk)
    I didn't use determinants in that solution. :confused:
    Well generally when working out the cross product of two vectors one writes it out in determinant form (although of course there are other methods).
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    (Original post by bono)
    simplay the best mathematician i know.
    lol..........yea..
    but the method as pretty long and intimidating !!!
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    (Original post by bono)
    wow. another good mathematician on the forum....
    I remember them helping out whenever there were maths problems for the june exams too.
 
 
 
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