If the lines L₁: x-1 -3 = y-2 2k = z-3 2 , L₂: x-1 3k = y-5 1 = z-6 -5 are perpendicular to each other, then the equation of a plane containing the line L₁ and parallel to the line L₂ for the value of k that satisfies this condition is ...

Step-by-step solution

The direction ratios of the line L₁ are d₁ = -3, 2k, 2 . The direction ratios of the line L₂ are d₂ = 3k, 1, -5 . Since the lines L₁ and L₂ are perpendicular, their dot product is zero: d₁ d₂ = 0 (-3)(3k) + (2k)(1) + (2)(-5) = 0 -9k + 2k - 10 = 0 -7k = 10 k = - 10 7 Substituting k = - 10 7 into the direction ratios and scaling by 7 to avoid fractions, we get: v₁ = 7 d₁ = -21, -20, 14 v₂ = 7 d₂ = -30, 7, -35 The plane contains L₁ and is parallel to L₂ , so its normal vector n is perpendicular to both v₁ and v₂ : n =