asteroidfodder
New member
- Joined
- Jul 23, 2009
- Messages
- 6
Hi, I'd appreciate some help with this problem. Some background, this is a problem from an A-level pure maths book, a chapter on vector dot product. I couldn't find an exact fit; I hope that this is a reasonable area to post this to.
The vectors a and b are of equal length magnitude k (k != 0) and the angle between a and b is 60 degrees.
if c = 3a - b and d= 2a - 10 b,
(a) show that c and d are perpendicular vectors
(b) find the magnitudes of c and d in terms of k.
I got (a) as follows:
the dot product of two perpendicular vectors = 0, so
(3a-b) . (2a-10b) = 0
6a^2 - 2(a.b) + 10b^2 -30(a.b) = 0
Since the angle between a and b is 60 degrees and cos 60 = 1/2
a . b = |a| * |b| * cos 60 = 1/2 k^2
substitute k
6k^2 - 2 * (1/2k^2)+10k^2 -30(1/2k^2) = 0
6k^2 - k^2 + 10k^2 - 15k^2 = 0
which is correct.
My problem is part (b). I have a nasty feeling that I'm missing (or have forgotten) something really basic, so a hint rather than a full solution would be great.
Thanks.
The vectors a and b are of equal length magnitude k (k != 0) and the angle between a and b is 60 degrees.
if c = 3a - b and d= 2a - 10 b,
(a) show that c and d are perpendicular vectors
(b) find the magnitudes of c and d in terms of k.
I got (a) as follows:
the dot product of two perpendicular vectors = 0, so
(3a-b) . (2a-10b) = 0
6a^2 - 2(a.b) + 10b^2 -30(a.b) = 0
Since the angle between a and b is 60 degrees and cos 60 = 1/2
a . b = |a| * |b| * cos 60 = 1/2 k^2
substitute k
6k^2 - 2 * (1/2k^2)+10k^2 -30(1/2k^2) = 0
6k^2 - k^2 + 10k^2 - 15k^2 = 0
which is correct.
My problem is part (b). I have a nasty feeling that I'm missing (or have forgotten) something really basic, so a hint rather than a full solution would be great.
Thanks.