What is the angle between #<-4,3,-8 ># and #< 6,-3,8 >#?

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Answer 1 · 2016-10-19T10:07:07

Angle between #< -4,3,-8># and #< 6,-3,8># is #170.01^o#

Angle between two vectors #vecu=a_1hati+b_1hatj+c_1hatk# or #< a_1,b_1,c_1>#

and #vecv=a_2hati+b_2hatj+c_2hatk# or #< a_2,b_2,c_2># is given by

#costheta=((vecu*vecv))/((|vecu|*|vecv|))#,

where #vecu*vecv=a_1a_2+b_1b_2+c_1c_2#

and #|vecu|# or #|vecv|# are magnitudes of vectors #vecu# or #vecv# and here they are

#sqrt(a_1^2+b_1^2+c_1^2)# and #sqrt(a_2^2+b_2^2+c_2^2)#

Hence angle between #< -4,3,-8># and #< 6,-3,8># is given by

#costheta=((-4)xx6+3xx(-3)+(-8)xx8)/(sqrt((-4)^2+3^2+(-8)^2)xxsqrt(6^2+(-3)^2+8^2))#

= #(-24-9-64)/(sqrt(16+9+64)xxsqrt(36+9+64))#

= #-97/(sqrt89xxsqrt109)#

= #-97/(9.43398xx10.44031)#

= #-0.9848#

and #theta=170.01^o#