man0005's specialist question thread :)

[man0005]

Will examiners test us on the semi circle theorem?

[man0005]

Let a = 3i - 6j + 4k
and b = 2i + j - 2k

c is the vector component of a perpendicular to b
d is the vector resolute of c in the direction of a

hence show that | a | | d | = | c | ^2

cant seem to get it :/

[evaever]

the result is true for any non-parallel a, b

[jane1234]

Hmm... there has to be an easier way than this:
But basically I just worked out the two required components (or resolutes) c and d and then found the magnitudes of a, c and d and found that |a|*|d| = |c|^2

c = a - (unit(b) . a)*(unit(b)) ---> this is just the formula for finding a perpendicular component in direction of b.

c = 3i - 6j + 4k - ((2/3i + 1/3j - 2/3k) . 3i - 6j + 4k)*(2/3i + 1/3j - 2/3k)
c = 3i - 6j + 4k - (-8/3)*(2/3i + 1/3j - 2/3k)
c = 43/9i - 46/9j + 20/9k

...|c| = (sqrt(485))/3
...|c|^2 = 485/9

d = (unit(a) . c)*(unit(a)) ---> formula for finding parallel component in direction of a

d = (3/sqrt(61)i - 6/sqrt(61)j + 4/sqrt(61)k . 43/9i - 46/9j + 20/9k)*(3/sqrt(61)i - 6/sqrt(61)j + 4/sqrt(61)k)
d = ((485*sqrt(61))/549)*3/sqrt(61)i - 6/sqrt(61)j + 4/sqrt(61)k
d = 485/183i - 970/183j + 1940/549k

... |d| = (485*sqrt(61))/549
and |a| = sqrt(61)

... |a|*|d| = 485/9
                = |c|^2 (as required)
 

[man0005]

ah yeah thats right silly mistake by me thanks

i have another one ><

            O


A                       D

OAD forms a triangle like so with B the midpoint of AD and C the midpoint of OD
OA - a
OB - b
OC - c

E is a point on OA produced such that OE = 4AE. If CB = k AE, find the value of k
i think the answer may be wrong..so if anyone could tell me what they get, it would be great

[brightsky]

1/2 OD + 1/2 DA = 1/2(OD+DA) = CB
OD + DA = OA, so 1/2 OA = CB
since OE = 4AE and OE - AE = OA
then 4AE - AE = 3AE = OA
so 1/2 (3AE) = CB
3/2 AE = CB

[man0005]

thanks! i got that too, but the answers said 2.5 :/

i have another question
how do you prove if something is "concurrent"? :S
what does it mean

[man0005]

How would you do 1c) ?
:S