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.

1 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
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