Given line be :
vector r = 4i - j + λ(i + 2j - 2k) ...(i)
and vector r = i - j + 2k - μ(2i + 4j - 4k) ... (ii)
The first line is parallel to the vector b1 = i + 2j - 2k and second line is parallel to the vector b2 = 2i + 4j - 4k. If θ be the angle between the lines, then θ is also the angle between the vector b1 & b2.
So, cos θ = vector(b1.b2/|b1| |b2|)
= (1.2 + 2.4 + (-2)(-4))/|(12 + 22 + (-2)2| √|22 + 42 + (-4)2| = 18/√9√36
= 18/3 x 6 = 1
∴ θ = 0
Hence, angle between the lines is 0
