3D - 2

The 3D vector r has magnitude 1 and makes 60˚ angles with unit vectors i and j?

(a) Write r as a column vector. (b) State the angle between r and the unit vector k.

Public Comments

1. Suppose r = [a, b, c] or the same as a column vector. Then r dot i = [a,b,c] dot [1,0,0] = a = cos60 = 1/2. Similarly, r dot j = [a,b,c] dot [0,1,0] = b = cos60 = 1/2. Thus r = [1/2, 1/2, c]. But r dot r = 1 since r is a unit vector. Hence (1/2)^2 + (1/2)^2 + c^2 = 1, so c^2 = 1-1/4-1/4=1/2. So c=1/sqrt(2). Thus r = [1/2, 1/2, 1/sqrt(2)]. Now just write it as a column vector. The angle between r and k is arccos(r dot k) = arccos(1/sqrt(2)) = 45 degrees. Consider also using -1/sqrt(2) for the third component of r.
2. v = (a, b, c) v.i = |v||i|cos(∡v,i) = 1(1)(1/2) = 1/2 = (a, b, c).(1, 0, 0) = a v.j = |v||j|cos(∡v,j) = 1(1)(1/2) = 1/2 = (a, b, c).(0, 1, 0) = b |v| = 1 --> a^2 + b^2 + c^2 = 1 --> c^2 + 1/4 + 1/4 = 1 --> c = √2/2 or -√2/2 so r = (1/2 , 1/2, √2/2) or (1/2, 1/2, -√2/2) angle between r and the unit vector k = ArcCos(√2/2) or ArcCos(-√2/2) = 45˚ or 135˚