The perimeter a square whose two sides have equations x-1 2 = y+2 3 = z-3 4 and x 2 = y-1 3 = z+1 4 is

Step-by-step solution

Identical direction ratios 2 , 3 , 4 establish the lines as parallel, implying they are opposite sides of the square. A point on L₁ is P₁(1, -2, 3) and on L₂ is P₂(0, 1, -1) . The connecting vector is P₁P₂ = (-1, 3, -4) . The side length s is the perpendicular distance between the parallel lines. Using the direction vector d = (2, 3, 4) , compute the cross product: P₁P₂ d = vmatrix i & j & k -1 & 3 & -4 2 & 3 & 4 vmatrix = 24 i - 4 j - 9 k = (24, -4, -9) | P₁P₂ d | = 24^2 + (-4)^2 + (-9)^2 = 673 | d | = 2^2 + 3^2 +