You are given vectors A⃗ = 5.2 i^− 6.2 j^ and B⃗ = - 3.0 i^+ 6.9 j^. A third vector C⃗ lies in the xy-plane. Vector C⃗ is perpendicular to vector A⃗ and the scalar product of C⃗ with B⃗ is 15.0.

Question

You are given vectors A⃗ = 5.2 i^− 6.2 j^ and B⃗ = - 3.0 i^+ 6.9 j^. A third vector C⃗  lies in the xy-plane. Vector C⃗  is perpendicular to vector A⃗  and the scalar product of C⃗  with B⃗  is 15.0.

oobleck · Answer

If C = ai+bj, then since C⊥A, 5.2a - 6.2b = 0 Since B•C = 15, -3a+6.9b = 15 Solve for a and b, and we find that C = 5.38i + 4.51j