# find magnitude of a vector

• May 15th 2009, 12:03 AM
s_ingram
find magnitude of a vector
Hi folks!

I joined today to find the answer to a vector question that has been driving me nuts. You know how it is! So here goes:

The two vectors a and b are of equal magnitude k (k not 0) and the angle between a and b is 60 degrees. If c = 3a - b and d = 2a - 10b

i) show that c and d are perpendicular
ii) find magnitudes of c and d in terms of k.

i) is OK. c.d = 0 for perpendicular vectors so evaluate (3a - b).(2a - 10b). This comes to 0 since a.b = a.b. cos 60. = k squared / 2 and a squared = b squared = k squared. (Sorry about the notation folks! - I'll figure out the superscripts later)

My problem is the second part. Can anyone help?

regards and thanks
• May 15th 2009, 02:52 AM
HallsofIvy
Quote:

Originally Posted by s_ingram
Hi folks!

I joined today to find the answer to a vector question that has been driving me nuts. You know how it is! So here goes:

The two vectors a and b are of equal magnitude k (k not 0) and the angle between a and b is 60 degrees. If c = 3a - b and d = 2a - 10b

i) show that c and d are perpendicular
ii) find magnitudes of c and d in terms of k.

i) is OK. c.d = 0 for perpendicular vectors so evaluate (3a - b).(2a - 10b). This comes to 0 since a.b = a.b. cos 60. = k squared / 2 and a squared = b squared = k squared. (Sorry about the notation folks! - I'll figure out the superscripts later)

My problem is the second part. Can anyone help?

regards and thanks

The magnitude of c is $\sqrt{c\cdot c}$. Now, $c\cdot c= (3a- b)\cdot(3a- b)= 9 a\cdot a- 6 a\cdot b+ b\cdot b$. Of course, $a\cdot a= b\cdot b= k^2$ any you calculated $a\cdot b$ before. Do the same with the magnitude of d.
• May 15th 2009, 04:03 AM
s_ingram
It's always easy when someone else gives you the answer! I was faffing about looking at the equalateral triangle made by vectors a and b, then I tried expressing c and d in terms of i and j (vectors at right angles) when all I had to do was mod c = SQRT c.c! Thanks to you HallsofIvy.