The equation of a line passing through the point ( -1,2,3 ) and perpendicular to the lines x 2 = y-1 -3 = z+2 -2 and x+3 -1 = y+3 2 = z-1 3 is

Step-by-step solution

The line passes through P(-1, 2, 3) and is perpendicular to both L₁ and L₂. The direction vector of L₁ is d₁ = (2, -3, -2) and for L₂ it is d₂ = (-1, 2, 3). Since the required line is perpendicular to both, its direction vector is the cross product d = d₁ d₂ . d = vmatrix i & j & k 2 & -3 & -2 -1 & 2 & 3 vmatrix = i ((-3)(3) - (-2)(2)) - j ((2)(3) - (-2)(-1)) + k ((2)(2) - (-3)(-1)) Computing yields d = -5 i - 4 j + 1 k = (-5, -4, 1). A line through (-1, 2, 3) with direction (-5, -4, 1) has symmetric form x + 1 -5