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

1. anyone got solutions to these questions. its from the modular books by heinemann, P6.
2. Could you post the question so the people doing different exam boards could maybe help
3. (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.
4. 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).
5. (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).

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