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P6 Edexcel - vectors rev. exe. qn 53(b) & (d) and qn 61 (c) watch

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    anyone got solutions to these questions. its from the modular books by heinemann, P6.
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    Could you post the question so the people doing different exam boards could maybe help
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    (Original post by geekypoo)
    Could you post the question so the people doing different exam boards could maybe help
    all right here goes. by the ways i need qn 53(b) and (d) in addition to qn 61(c), once again.

    qn 53) The points A(3,0,0), B (0,2,-1) and C(2,0,1) have position vectors a,b and c with respect to a fixed origin O. the line L has equation (r-a)*b = 0. The plane II contains L and the point C
    (a) find ac*ob
    (b) hence or otherwise show that an equation of II is 2x+y+2z = 6.
    (c) Find the perpendicular distance of II from O.
    the point R is the reflection of O in II
    (d) find the position vector of R.

    qn 61) The plane II passes through A (3,-5,-1), B(-1,5,7) and C(2,-3,0).
    FIND
    (a) ac*bc
    (b)hence or otherwise, find an equation, in the form r.n=p of the plane II.
    (c) the perpendicular from the point (2,3,-2) to II meets the plane at P. Find the co ordinates of P.
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    53.
    (b)
    II contains the points A, B and C. A vector perpendicular to the plane is ACxOB=(-2, -1, -2). So:
    r . n = a . n
    (x, y, z) . (-2, -1, -2) = (2, 0, 1) . (-2, -1, -2)
    -2x - y - 2z = -4 - 2
    2x + y + 2z = 6, as required.

    (d)
    R lies on the line that passes through O and is perpendicular to II. And equation for such a line is: r = t(2, 1, 2), the direction vector is n which we found in part (a). So R=(2t, t, 2t).

    The second piece of information we have is that the distance of O from II is 2 units long. Hence the distance from II to R is also 2 units long. So the distance of R from O is 4 units long, i.e. |OR|=4. That is:
    |OR| = |2t, t, 2t|
    = sqrt[4t² + t² + 4t²]
    = sqrt[9t²]
    = 3t = 4
    => t = 4/3

    Hence:
    R = (8/3, 4/3, 8/3)

    61.
    (b) r.(-6, -4, 2) = 0
    (c) An equation of the perpendicular is:
    r = (2, 3, -2) + t(-6, -4, 2), since it passes through (2, 3, -2) and it's parallel to ACxBC.

    Hence P has coords (2-6t, 3-4t, -2+2t). Since P lies on II, then:
    (2-6t, 3-4t, -2+2t) . (-6, -4, 2) = 0 (from the equation of the plane)
    -6(2-6t) - 4(3-4t) + 2(-2+2t) = 0
    t = 1/2

    So... P has coords (2-3, 3-2, -2+1) = (-1, 1, -1).
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    (Original post by dvs)
    53.
    (b)
    II contains the points A, B and C. A vector perpendicular to the plane is ACxOB=(-2, -1, -2). So:
    r . n = a . n
    (x, y, z) . (-2, -1, -2) = (2, 0, 1) . (-2, -1, -2)
    -2x - y - 2z = -4 - 2
    2x + y + 2z = 6, as required.

    (d)
    R lies on the line that passes through O and is perpendicular to II. And equation for such a line is: r = t(2, 1, 2), the direction vector is n which we found in part (a). So R=(2t, t, 2t).

    The second piece of information we have is that the distance of O from II is 2 units long. Hence the distance from II to R is also 2 units long. So the distance of R from O is 4 units long, i.e. |OR|=4. That is:
    |OR| = |2t, t, 2t|
    = sqrt[4t² + t² + 4t²]
    = sqrt[9t²]
    = 3t = 4
    => t = 4/3

    Hence:
    R = (8/3, 4/3, 8/3)

    61.
    (b) r.(-6, -4, 2) = 0
    (c) An equation of the perpendicular is:
    r = (2, 3, -2) + t(-6, -4, 2), since it passes through (2, 3, -2) and it's parallel to ACxBC.

    Hence P has coords (2-6t, 3-4t, -2+2t). Since P lies on II, then:
    (2-6t, 3-4t, -2+2t) . (-6, -4, 2) = 0 (from the equation of the plane)
    -6(2-6t) - 4(3-4t) + 2(-2+2t) = 0
    t = 1/2

    So... P has coords (2-3, 3-2, -2+1) = (-1, 1, -1).
    thanks loads. :rock:
 
 
 
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