SOLUTION: Evening All. Struggling on this vectors problem. i)If a and b are perpendicular, then by writing a= ¦a¦ and b= ¦b¦ simplify the expression (a-2b).(3a+5b) ii)Show that t

Algebra ->  Length-and-distance -> SOLUTION: Evening All. Struggling on this vectors problem. i)If a and b are perpendicular, then by writing a= ¦a¦ and b= ¦b¦ simplify the expression (a-2b).(3a+5b) ii)Show that t      Log On


   

Question 365047: Evening All.
Struggling on this vectors problem.
i)If a and b are perpendicular, then by writing a= ¦a¦ and b= ¦b¦ simplify the expression (a-2b).(3a+5b)
ii)Show that the vectors a=i+2j and b=4i-2j are perpendicular
iii) Find the angle theta in degrees between the vectors p=i-j-k and q=2i+j+2k
Many Thanks
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
Struggling on this vectors problem.
i)If a and b are perpendicular, then by writing a= ¦a¦ and b= ¦b¦ simplify the expression (a-2b).(3a+5b)
Sorry I don't understand "writing a= ¦a¦ and b= ¦b¦".  If you
can explain what that means I can help you.  Here are the others:

ii)Show that the vectors a=i+2j and b=4i-2j are perpendicular 
Just show that their dot product (scalar product) is 0:

(i+2j)·(4i-2j) = (1)(4)+(2)(-2) = 4-4 = 0

iii) Find the angle theta in degrees between the vectors p=i-j-k and q=2i+j+2k 
Use the formula 

           u·v 
cos(@) = ------ =
         |u||v|


 =

%282-1-2%29%2F%28%28sqrt%281%2B1%2B1%29%29sqrt%284%2B1%2B4%29%29 =

%28-1%29%2F%28%28sqrt%283%29%29sqrt%289%29%29 = -1%2F%28%28sqrt%283%29%29%283%29%29 = -1%2F%283sqrt%283%29%29 = %28-1%2F%283sqrt%283%29%29%29%28sqrt%283%29%2Fsqrt%283%29%29 = -sqrt%283%29%2F%28%283%29%283%29%29 = -sqrt%283%29%2F9

So @ = 101.1° approximately

Edwin