You are not logged in.
Pages: 1
Hi,
I am doing 1st year university algebra. I was wondering if anyone could tell me if i have done this question correctly?
Q. Let u and v be perpendicular vectors of lengths 1 and 2 respectively. Evaluate (3u+v) (u-2v).
By expanding the bracket I get
3(u.u)-5(u.v)-2(v.v)
Since u is a unit vector u.u=1 and as they are perpendicular u.v=0
Hence 3(1) - 5(0) - 8 = -5.
Is this correct i am unsure of what to do with v.v as v is of length 2 therefore not a unit vector so I just took the square of 2.
I would be grateful for any help with this.
Thanks ![]()
Offline
hi Natkirky
What you have done is exactly right. Well done!
:)
v.v is size of v times size of v times cosine of 'angle between v and v' (ie cos zero = 1)
Bob
Last edited by Bob (2011-02-08 01:09:13)
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob ![]()
Offline
Hi Bob,
Thanks a lot for your help I have a test on vectors tomorrow and that question was bothering me ![]()
Offline
hi Natkirky
You are very welcome. Good luck with the test!
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob ![]()
Offline
Pages: 1