How do you find the angle between the vectors #u=3i+4j# and #v=-7i+5j#?

1 Answer
Euan S.
Jul 3, 2016

91.33 degrees

Explanation:

We're working in #RR^2# here. The dot product of two vectors given by #vec(u) = ((u_1),(u_2))# and #vec(v) = ((v_1),(v_2))# is given by:

# vec(u)*vec(v) = u_1v_1 + u_2v_2 = |vec(u)||vec(v)|costheta# where theta is the angle between the vectors.

#therefore theta = arccos[(u_1v_1 + u_2v_2)/(|vec(u)||vec(v)|)]#

#|vec(u)| = sqrt(3^2+4^2) = 5#

#|vec(v)| = sqrt((-7)^2 + 5^2) = sqrt(74)#

#theta = arccos[((3)(-7) + (4)(5))/(5sqrt(74))] = 91.33# degrees

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