OSU MATH 1172 Final Exam with Solutions

(T/F) if u and v are not ‖, |scal_v_u||u| - True. Geometrically: scal_v_u is the magnitude of the ‖ component of u in the direction of v; the only way that this magnitude can be equal to the magnitude of the vector u itself is if u is in the same direction as v. (T/F) if u and v are not ‖, |scal_v_u||scal_u_v| - Not enough information given. Geometrically: scal_v_u is the magnitude of the ‖ component of u in the direction of v. Vice versa for scal_u_v. If the magnitude of u and v are equal, scal_v_u and scal_u_v are equal. But if their magnitudes differ, scal_v_u and scal_u_v are not equal. Since we do not know their magnitudes we cannot know which one is larger. (T/F) If u and v are orthogonal, proj_v_u=0 - True. Geometrically, if u and v are orthogonal, then the ‖ component of u along v should be 0! It won't cast a "shadow" (T/F) scal_u_u=0 - False. Geometrically, the scalar projection of a vector onto itself should just be the magnitude of the original vector! (T/F) If u is not ‖ to v, (u×v)×u=0 - False. Since |u×v|=|u||v|sinθ, the cross product of nonzero vectors can only be zero if sinθ=0, which occurs when θ=0,π, meaning that the vectors must be parallel. But, u×v is ⊥ to u since the cross product of any two nonzero vectors must be perpendicular to both of the original vectors. Hence,(u×v)×u≠0 (T/F) If u is not ‖ to v, then proj_u_v is parallel to v - False. By definition proj_u_v is ‖ to u (T/F) u-(proj_v_u) is ⊥ to v - True. Geometrically: proj_v_u is the component of u that is parallel to v, so u-(proj_v_u) is ⊥ to v (T/F) (u-(proj_v_u))⊗v=0 - True. This is just the previous question restated! If two vectors are perpendicular, their dot product must be 0! (T/F) If u and v are ‖, then u⊗v=0 - False. u⊗v=|u||v|cosθ. Since u and v are ‖, θ=0,π. Thus cosθ≠0 and since neither v or u is 0, u⊗v≠0 (T/F) If u and v are ⊥, then u⊗v=0 - True. u⊗v=|u||v|cosθ. Since u and v are ⊥, θ=90°. Thus cosθ=0 and u⊗v=0 (T/F) If u and v are ‖, then u×v=0 - True. u×v=|u||v|sinθ. Since u and v are ‖, θ=0,π. Thus sinθ=0 and u×v=0 (T/F) If u and v are ⊥, then u×v=0 - False. u×v=|u||v|sinθ. Since u and v are ⊥, θ=90°. Thus sinθ≠0 and since neither v or u is 0, u×v≠0 (scalar/vector) u⊗v - Scalar. A dot product is defined between two vectors, and returns a scalar. (scalar/vector) u×v - Vector. A cross product is defined between two vectors and returns a third vector (with the property that the third vector is perpendicular to the other two!) (scalar/vector) (u+v)v - Undefined. While adding vectors returns a vector. However, "products" of vector such as u(v) or u/v are not defined. (scalar/vector) (u×v)×u - Vector. A cross product is defined between two vectors and returns a third vector (with the property that the third vector is perpendicular to the other two!)