Question

Find the angle of the following pairs of vectors
1: 3i-2j+k and 4i+3j-2k
2: i-j+k and 2i+3j- k

Answer 1

To find the angle between two vectors, we can use the dot product formula:

cos(theta) = (a.b) / (||a|| ||b||)

1: For vectors 3i-2j+k and 4i+3j-2k:

a = 3i-2j+k
b = 4i+3j-2k

a.b = (3*4) + (-2*3) + (1*-2) = 12 - 6 - 2 = 4
||a|| = sqrt(3^2 + (-2)^2 + 1^2) = sqrt(9 + 4 + 1) = sqrt(14)
||b|| = sqrt(4^2 + 3^2 + (-2)^2) = sqrt(16 + 9 + 4) = sqrt(29)

cos(theta) = 4 / (sqrt(14) * sqrt(29))
theta = arccos(4 / (sqrt(14) * sqrt(29)))

theta ≈ 52.12 degrees

2: For vectors i-j+k and 2i+3j-k:

a = i-j+k
b = 2i+3j-k

a.b = (1*2) + (-1*3) + (1*-1) = 2 - 3 - 1 = -2
||a|| = sqrt(1^2 + (-1)^2 + 1^2) = sqrt(1 + 1 + 1) = sqrt(3)
||b|| = sqrt(2^2 + 3^2 + (-1)^2) = sqrt(4 + 9 + 1) = sqrt(14)

cos(theta) = -2 / (sqrt(3) * sqrt(14))
theta = arccos(-2 / (sqrt(3) * sqrt(14)))

theta ≈ 118.54 degrees