Consider two skew lines in 3D space. M₁: x-1 1 = 2-y 1 = z-5 1 and M₂: x+3 1 = y-7 2 = z+4 1 Let L₁ be the line of shortest distance (common perpendicular) between M₁ and M₂ If L₂ is a line parallel to the vector b = j + k , Then the acute angle between the lines L₁ and L₂ is:

Step-by-step solution

The equation of line M₁ can be rewritten in standard form as: x-1 1 = y-2 -1 = z-5 1 The direction vector of M₁ is v ₁ = i - j + k . The direction vector of M₂ is v ₂ = i + 2 j + k . The line L₁ is the common perpendicular to M₁ and M₂ . Its direction vector d ₁ is given by the cross product of v ₁ and v ₂ : d ₁ = v ₁ v ₂ = vmatrix i & j & k 1 & -1 & 1 1 & 2 & 1 vmatrix d ₁ = i (-1 - 2) - j (1 - 1) + k (2 - (-1)) = -3 i + 3 k We can take the simpler direction vector for L₁ as d ₁ = - i + k . The line L₂ is parallel