How do you find the angle between the vectors #u=-6i-3j# and #v=-8k+4j#? Precalculus Dot Product of Vectors Angle between Vectors 1 Answer Ratnaker Mehta Sep 3, 2016 #approx 101.537^@# Explanation: Let #theta=/_(vec u, vec v)#. #" Then, "vec u*vec v=||vec u||||vec v||cos theta, ...............(star)#. Here, #vec u =-6i-3j=(-6,-3,0), and, vec v=(0,4,-8)# #:. vec u*vec v=0-12+0=-12#, and, #||vec u||=sqrt(36+9)=sqrt45=3sqrt5, and, ||vec v||=4sqrt5#. Hence, by #(star)#, we have, #cos theta=-12/((3sqrt5)(4sqrt5))=-1/5#. #:. theta=arc (cos (-0.2)=pi-arc cos (0.2)~~(180-78.463)^@# #:. theta~~101.537^@# Answer link Related questions How do I calculate the angle between two vectors? How do I find the sine of the angle between two vectors? How do I find the angle between two vectors using the law of cosines? What are common mistakes students make with angles between vectors? How do I find the angle between a vector and the x-axis? What is the dot product of two vectors that are parallel? How do I find out if vectors are parallel? What is the dot product of two vectors that are perpendicular? How do I find the angle between vectors #<3, 0># and #<5, 5>#? Consider the following vectors: v = 3i + 4j, w = 4i + 3j, how do you find the angle between v and w? See all questions in Angle between Vectors Impact of this question 4476 views around the world You can reuse this answer Creative Commons License