What is the angle between #<5,8,2 ># and #< 4,6,-1 >#?

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Answer 1 · 2016-06-26T08:54:12

Angle between vectors #< 5, 8, 2 ># and #< 4, 6, -1 ># is #28.682^o#

The angle between vectors #A=< a_1, a_2, a_3 ># and #B=< b_1, b_2, b_3 ># is given by

#A*B=|A||B|costheta#, where #A*B# is the dot product and #|A|# and #|B|# are absolute value of the two vectors.

As #A=< 5, 8, 2 ># and #B=< 4, 6, -1 >#, hence

#A*B=5*4+8*6+2*(-1)=20+48-2=66#

#|A|=sqrt(5^2+8^2+2^2)=sqrt(25+64+4)=sqrt93#

#|B|=sqrt(4^2+6^2+(-1)^2)=sqrt(16+36+1)=sqrt53#

Hence #costheta=66/(sqrt93xxsqrt53)=66/(9.64365xx7.28011)#

= #66/75.231=0.8773#

and #theta=28.682^o#