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

1 Answer
Oct 25, 2016

THe angle is #126.9º#

Explanation:

The angle is obtained by using the definition of the dot product
#vecu.vecv=∣vecu∣*∣vecv∣costheta#
where #theta# is the angle between the vectors #(vecu,vecv)#

So #costheta=(vecu.vecv)/(∣vecu∣*∣vecv∣)#

Here #vecu=〈-1,-2,8〉#

and #vecv=〈-2,6,-4〉#

The dot product is given by #vecu.vecv=〈u_1,u_2,u_3〉〈v_1,v_2,v_3〉=u_1v_1+u_2v_2+u_3v_3#
So #vecu.vecv=2-6-32=-36#

#∣vecu∣=sqrt(u_1^2+u_2^2+u_3^2)=sqrt(1+4+64)=sqrt69#

#∣vecv∣=sqrt(v_1^2+v_2^2+v_3^2)=sqrt(4+36+16)=sqrt56#

So #costheta=-36/(sqrt69.sqrt56)=-0.601#
#theta=126.9º#