What is the angle between #<2,7,-1 ># and #< 3,2,-8 >#?

Question

1835 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-28T08:35:38

#alpha=89.61^o#

#"find :" A*B#

#A=<2,7,-1>#

#A_x=2#
#A_y=7#
#A_z=-1#

#B=<3,2,-8>#

#B_x=3#
#B_y=2#
#B_z=-8#

#A*B=A_x*B_x+A_y*B_y+A_z*B_z#

#A*B=2*3+7*2+((-1)*(-8))#
#A*B=6+14+8#
#A*B=28" ;dot product for A and B"#

#"find :"||A||" magnitude of A"#

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

#||A||=sqrt(2^2+7^2+(-1)^2)#

#||A||=sqrt(4+49+1)" ;" ||A||=sqrt54#

#"find :"||B||" magnitude of B"#

#||B||=sqrt(B_x^2+B_y^2+B_z^2)#

#||B||=sqrt(3^2+2^2+(-8)^2)=sqrt(9+4+64)#

#||B||=sqrt77#

#cos alpha=(A*B)/(||A||*||B||)#

#cos alpha=(28)/(sqrt54*sqrt77)#

#cos alpha=28/sqrt(54*77)#

#cos alpha=28/4158#

#cos alpha=0.0067340067#

#alpha=89.61^o#