SOLUTION: Hi All, Having difficulty with this particular problem on a task sheet, any help would be greatly appreciated. Given a=i+3j-k and b=2i+j-k, find a.b and the cosine of t

Algebra ->  Length-and-distance -> SOLUTION: Hi All, Having difficulty with this particular problem on a task sheet, any help would be greatly appreciated. Given a=i+3j-k and b=2i+j-k, find a.b and the cosine of t      Log On


   

Question 364466: Hi All,
Having difficulty with this particular problem on a task sheet, any help would be greatly appreciated.
Given a=i+3j-k and b=2i+j-k, find a.b and the cosine of the angle Theta between the vectors
b) Given a= 3i+j-2k and b= -6i-2j+4k find a.b and the angle theta between vectors
Finaly c) For what value of alpha will the vectors a+5i+2j+3k and b=2i+alphaj+alphak be perpendicular?
Many thanks in advance for any help.
Mike
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
b) Use the dot product.
a=(3,1,-2)
b=(-6,-2,4)
a%2Ab=3%28-6%29%2B1%28-2%29%2B%28-2%29%284%29=-18-2-8=-28
a%2Ab=abs%28a%29abs%28b%29cos%28theta%29
abs%28a%29=sqrt%283%5E2%2B1%5E2%2B%28-2%29%5E2%29=sqrt%289%2B1%2B4%29=sqrt%2814%29

-28=sqrt%2814%29%282%2Asqrt%2814%29%29cos%28theta%29
cos%28theta%29=-28%2F%282%2A14%29
cos%28theta%29=-1
highlight%28theta=180%29
.
.
.
c)
Perpendicular means theta=90.
cos%2890%29=0
So then a%2Ab=0
a=(5,2,3)
b=(2,alpha,alpha)
a%2Ab=5%2A2%2B2%2Aalpha%2B3%2Aalpha=10%2B5%2Aalpha
10%2B5%2Aalpha=0
5%2Aalpha=-10
highlight%28alpha=-2%29