If #A = <3 ,2 ,5 >#, #B = <5 ,-7 ,8 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Jun 25, 2016

#theta=87.12#

Explanation:

#"Strategy...."#

#1 diamond" find " C=A-B #
#2 diamond" find || A|| (magnitude of A)" #
#3 diamond" find ||C|| (magnitude of C)"#
#4 diamond" find A.C (dot product of A and C)"#
#5 diamond" use A.C=||A||.||C||. cos" theta " (dot product formula)"#

#1diamond" ......................................................"#

#A_x=3" ; "A_y=2" ; "A_z=5#

#B_x=5" ; "B_5=-7" ; "B_z=8#

# C= < (A_x-B_x),(A_y-B_y),(A_z-B_z) >#

#C= < (3-5),(2+7),(5-8)#

#"we obtain :"#
#C= < -2,9,-3>#

#2diamond" ........................................................"#

#||A||=sqrt(A_x^2+A_y^2+A_z^2)=sqrt(3^2+2^2+5^2)=sqrt38#

#3diamond" .........................................................."#

#||C||=sqrt(c_x^2+C_y^2+C_z^2)=sqrt((-2)^2+9^2+(-3)^2)#

#||C||=sqrt(4+81+9)=sqrt94#

#4diamond" ..........................................................." #

#A*C=A_x*C_x+A_y*C_y+A_z*C_z#

#A*C=3*(-2)+2*9+5*(-3)=-6+18-15#

#A*C=-3#

#5diamond" ............................................................"#

#-3=sqrt38*sqrt94*cos theta#

#cos theta=-3/(sqrt(38*94))#

#cos theta=-3/(59.766)#

#cos theta=0.0501957635#

#theta=87.12#