Quote:
|
Originally Posted by event_staff_9
ill answer half-havocs first since there are no calculations involved; i hope im not too late for spankyou; i had just left when you posted :/
anyways, yeah the bases must be the same
correct: x^5 / x ^3 = x ^ (5-3) = x^2
incorrect: x^5 / y ^ 3 does not equal (x/y)^2
working on yours spankyou
edit: okay spankyou, to do a triple scalar product, you'll need to find a determinant of the following matrix of A, B and C:
THIS WAY MAY NOT WORK ALL THE TIME
A(1,3,2), B(3,-1,6), C(5,2,0), and D(3,6,-4)
[1 3 2
3 -1 6
5 2 0]
which is 100
note the numbers of the 3 x 3 matrix above is:
[A1 A2 A3
B1 B2 B3
C1 C2 C3]
next try, A B and D
answer: 100
lets try A, C, and D
answer: 100
how about B, C, and D now
answer: 100
so b/c you had to use triple scalar products and the question asked whether all the points lied in the same plane, and since all the answers are equal its safe to assume that they do :D
THIS WAY WORKS ALL THE TIME
heres a cooler way using a different formula:
A dot (B x C) (x = cross product) and B dot (C x D)
first you need to make three vectors. AB we'll call vector X = [ 2 -4 4] BC we'll call Y, [2 3 -6] and CD we'll call Z [-2 4 4]
so lets do X dot (Y x Z) first:
B x C = [12 20 14] dot it with A and you get 0! so A B and C are coplanar
so lets try Y dot (X x Z) X Z is [0 0 0] so the dot product must be zero! yipee! we found that A B and C lie in the same plane and B C and D also do therefore all four lie on the same plane!
|
very very interesting
you see I did vector AB, AC, and AD instead
so I was off right at the start
I turned in the HW at 3 today.. 10 minutes before you submitted yours
so that one I got wrong, but it's all right, thank you for the explaination, are you in college or something?