1. The problem statement, all variables and given/known data 1)Give a unit vector a) in the same direction as v=2i +3j b)Perpendicular to v 2) a) Find a vector perpendicular to the plane z=2+3x-y b)Find a vector parallel to the plane 2. Relevant equations 3. The attempt at a solution 1) a) Method 1: u=v/||v|| so i get u=2/√13i + 3/√13j Method 2: ||u||*||v||cos(θ)=2 arccos(2/√13)=56.31*=θ ||u||=1 so u=cos(56.31)i + sin(56.31)j=2/√13i + 3/√13j Method 3: u.v=||v|| (xi + yj).(2i +3j)=√13 2x + 3y=√13 Here i get stuck. How do i do this and is there any other way? b) Method 1: u.v=||u||*||v||cos(θ)=0 1*√13cos(θ)=0 I get stuck here aswell Method 2: I realized that if i take the angle between v & the x axis I should get the angle between u and the x axis so I can take sin and cos of those angles to get the unit vector perpendicular u=-3/√13i+2/√13j The reason I'm doing all these ways is because I'm curious of the most efficient and proper way to solve the problem. My class is currently covering dot products and I was expecting it all to be as simple as u.v=(xi + yj).(2i +3j)=√13 or (xi + yj).(2i +3j)=0 then I would solve algebraically, but I cant seem to figure out how to do it that way. 2) a) I know that the normal vector is orthogonal to the pane so n=-3i+j+k is there another way to do this? b) n.v=||n||*||v||cos(θ)=0 (-3i+j+k).(xi+yj +zk)=0 (-3x +y +z)=0 I get the equation of a similar plane but not the one they originally gave me. Thanks for your help in advance.