What is the angle between #<7,2,-4 > # and #< -3,6,-2>#?

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Answer 1 · 2017-07-21T16:02:25

The angle is #=91^@#

Let #veca=<7,2,-4>#

#vecb=<-3,6,-2>#

The angle is calculated with the dot product.

#veca.vecb=||veca||*||vecb||*costheta#

Where #theta# is the angle between the #2# vectors

The dot product is

#veca.vecb= <7,2,-4> .< -3,6,-2> =-21+12+8=-1#

The modulus of #veca# is

#=||veca||=||<7,2,-4>|| = sqrt(49+4+16)=sqrt69#

The modulus of #vecb# is

#=||vecb||=||<-3,6,-2>|| = sqrt(9+36+4)=sqrt49#

The angle is

#theta=cos^-1((veca.vecb)/(||veca||*||vecb||))=cos^-1(-1/(sqrt69sqrt49))#

#=cos^-1(-0.017)=91^@#