What is the angle between #<-8,2,8># and #< 2,-3,5>#?

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Answer 1 · 2016-10-19T09:56:51

Angle between #< -8,2,8># and #< 2,-3,5># is #75.28^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 #< -8,2,8># and #< 2,-3,5># is given by

#costheta=((-8)xx2+2xx(-3)+8xx5)/(sqrt((-8)^2+2^2+8^2)xxsqrt((2)^2+(-3)^2+5^2))#

= #(-16-6+40)/(sqrt(64+4+64)xxsqrt(4+9+25))#

= #18/(sqrt132xxsqrt38)#

= #18/(11.4891xx6.1644)#

= #0.25415#

and #theta=75.28^o#