B bubb Member Joined Oct 3, 2014 Messages 31 Gender Female HSC 2013 Feb 14, 2015 #1 Vectors & are defined in terms of , & as: = + ( + ) − (2 + ) = + (3 + 2) − (5 − 3) If = then show that: |1.25 + 5 + 6| = √46
Vectors & are defined in terms of , & as: = + ( + ) − (2 + ) = + (3 + 2) − (5 − 3) If = then show that: |1.25 + 5 + 6| = √46
A Ambility Active Member Joined Dec 22, 2014 Messages 336 Gender Male HSC 2016 Feb 14, 2015 #2 Umm, what? I think your latex is playing up.
L LostInHSC New Member Joined Dec 14, 2014 Messages 22 Gender Undisclosed HSC 2015 Feb 14, 2015 #3 You are in the Mathematics (Extension 1) forum. Mathematics (Extension 2): http://community.boredofstudies.org/14/mathematics-extension-2/
You are in the Mathematics (Extension 1) forum. Mathematics (Extension 2): http://community.boredofstudies.org/14/mathematics-extension-2/
F FrankXie Active Member Joined Oct 17, 2014 Messages 330 Location Parramatta, NSW Gender Male HSC N/A Uni Grad 2004 Feb 14, 2015 #4 two vectors are same if and only if all components are same. a=b, so z=1, 3y+2=x+y, 5x-3=2x+y, solving simutaneously we have x=4/5, y=-3/5, z=1, so the magnitude of 1.25xi+5yj+6zk=i-3j+6k is \sqrt{1^2+(-3)^2+6^2}=\sqrt{46}.
two vectors are same if and only if all components are same. a=b, so z=1, 3y+2=x+y, 5x-3=2x+y, solving simutaneously we have x=4/5, y=-3/5, z=1, so the magnitude of 1.25xi+5yj+6zk=i-3j+6k is \sqrt{1^2+(-3)^2+6^2}=\sqrt{46}.
