1. The problem statement, all variables and given/known data Strain tensor E = [0.02 0 0; 0 0.01 0; 0 0 0.03] (no shear strains). A triangle consists of points A, B, and C, each on axis X1, X2, and X3 respectively. The lengths OA = OB = OC, and D is the midpoint of AC. The direction cosines of AC are (1/sqrt(2), 0, -1/sqrt(2)) and those of BD are (-1/sqrt(6), sqrt(2)/sqrt(3),-1/sqrt(6)). Find: A) The elongation of AC B) The change of initial right angle BDA 2. Relevant equations A^2 + B^2 = C^2 for right triangle 3. The attempt at a solution Part A -- Drawing the triangle OAC (in the plane X1-X2), we see that the initial angles CAO and ACO are 45 degrees, while COA is a right angle. The deformation at point A is AO(0.02) while the deformation at point C is CO(0.03). Since AO=CO, we can write the change as: A'C' - AC = sqrt( (1.02AO)^2 + (1.03AO)^2 ) - sqrt ( 2 AO^2 ) = 0.0354 AO Part B -- Here I am a bit stuck. It seems to me that it should still be a right angle, but this is probably not true. I know I should have to use the direction cosines for BD. Any advice or hints?