Calculate the angle between the vectors [-1,1,2] and [1, -2, 2].?

Question

I used the formula costheta=u*v/IIuII*IIvII, and came up with the answer of 35.86 degrees, but the book is saying this answer is incorrect. Does anyone know what I'm doing wrong?

Answer 1

u*v = (-1*1) + (1*-2) + (2*2) = -1 -2 + 4 = 1
|u| = sqrt(-1^2 + 1^2 + 2^2) = sqrt(6)
|v| = sqrt(1^2 + -2^2 + 2^2) = 3

theta = cos^-1((u*v)/(|u|*|v|))
theta = 82.2

Answer 2

Let the angle between the vectors = A

So, cos A = ( -1, 1, 2 ) . ( 1, -2, 2 ) / [ l ( -1, 1, 2 ) l l ( 1, -2, 2 ) l ]
= ( -1 -2 +4 ) / sqrt[ (-1)^2 + (1)^2 + (2)^2 ] sqrt [ (1)^2 + (-2)^2 + (2)^2 ]
= 1 / 3 sqrt ( 6 )

So, A = arc cos (1/3sqrt(6)) = 82.2 degrees

Answer 3

The formula seems correct. Let us see:
|-1,1,2| = sqr[1 + 1 + 4] = sqr[6]
|1,-2,2| = sqr[1 + 4 + 4] = sqr[9] = 3
vector dot u.v = [-1,1,2].[1,-2,2] = -1 + (-2) + 4 = 1
cos t = 1/{3 sqr[6]} = 0.1361
t = arccos(0.1361) = ~ 82.3 degrees x pi/180 degrees = .457 pi

Answer 4

[-1,1,2] • [1,-2,2] = -1 - 2 + 4 = 1
||[-1,1,2]|| = √(1+1+4) = √6
||[1,-2,2]|| = √(1+4+4) = √9 = 3
so cos Θ = 1/(3√6) = 0.136
Θ = 82.18°