Solutions For Revision Paper 3

p. 174 8c.

OP = 6i – j + 3k OT lies on line l, so OT = 2i + s (i + 3j + 4k) PT = PO + OT = – (6, -1, 3) + (2+s, 3s, 4s) = ( s – 4, 3s + 1, 4s – 3) Since PT is perpendicular to the line, so PT  d = 0 (s – 4, 3s + 1, 4s – 3)  (1, 3, 4) = 0 s – 4 + 9s + 3 + 16s – 12 = 0 26s – 13 = 0 s = 1/2 Therefore OT = 2 1/2 i + 1 1/2 j + 2k (ans) 8 iii

iz 2  2 2 z  2 3  0 z

 (2 2 )  (2 2 ) 2  4(i )(2 3 ) 2i

2 2  8  8 3i 2 2  2 3  2i  2i 2i 2 2  2 3  2i i 2 2i  2 3i  2    2i i 2   2i  3i  1 z

or   2i  3i  1

Now find the square root of

8  8 3i =  2 3  2i 8  8 3i  a  ib 8  8 3i  a 2  b 2  2abi a 2  b 2  8      (1) 2ab  8 3      (2) a  2 3 b  2  2 3  2i

1. Use numerical method to find all the roots of the cubic equations x3 – 2x2 – 2x + 2 = 0, giving your answers correct to 2 decimal places. Let f(x) = x3 – 2x2 – 2x + 2 F(-2) = -8 – 8 + 4 + 2 = - 10 F(-1)= -1 -2 + 2 + 2 = 1

A sign change, root lies between -2 and -1

F(0) = 0 – 0 – 0 + 2 = 2 F(1) = 1 – 2 – 2 + 2 = -1

A sign change, root lies between 0 and 1

F(2) = 8 – 8 – 4 + 2 = -2 F(3) = 27 – 18 – 6 + 2 = 5

A sign change, root lies between 2 and 3

[normally in paper 3, the iteration formula will be given. Since not given, we need to do trial and error, making your own iteration formula.] Let x3 subject x3 = 2x2 + 2x – 2

x  3 2x 2  2x  2 x n 1  3 2 x n2  2 x n  2 U try substitute x = -1.6, -1.4, 0, 0.5, 0.7, 2.4, 2.6 See whether will converge to a root or not. If not then try this formula: x(x2 - 2x - 2) = – 2 Let x2 subject,

x

x3  2x  2 2

or

x

2 2 x x2

Let x subject or

x

x3  2x 2  2 or 2

Make subject x3 +2 = 2x (x+1)

x

x3  2 2( x  1)

Trial and error. See which one converge to the answer. I haven’t tried this yet because I left my scientific calculator in the school. Tomorrow will let you know which one give the answer.