Question

You are given vectors A = 5.0i - 6.5j and B = -2.5i + 7.0j. A third vector C lies in the xy-plane. Vector C is perpendicular to vector A and the scalar product of C with B is 15.0. Find the x and y components to vector C.

Answer key shows x component is 8.0, y component is 6.1

I don't know how to work the problem out.

Answer 1

slope of vector A = -6.5/5
so slope of perpendicular = 5/6.5
or
C dot A = 0
5 Cx - 6.5 Cy = 0
Cy/Cx = 5/6.5

if Cx = 6.5 k
then Cy = 5 k

C dot B = 6.5 k (-2.5) + 5 k (7) = 15
so
-16.25 k + 35 k = 15
18.75 k = 15
k = .8
so
Cx = .8 * 6.5
Cy = .8 * 5

Answer 2

Thank you for that. I don't understand the rule for "Cy/Cx = 5/6.5; if Cx = 6.5k, then Cy = 5k." I also realize that your solution doesn't match my book's solution. Here's the step by step from the book that I'm not understanding either, perhaps someone can suggest a rule I can review for this type of problem:

The target variables are the components of C. We are given A and B. We also know A⋅C and B⋅C, and this gives us two equations in the two unknowns Cx and Cy.

A and C are perpendicular, so A⋅C = 0. AxCx + AyCy = 0, which gives 5.0Cx − 6.5Cy = 0.
B⋅C = 15.0, so −3.5Cx + 7.0Cy = 15.0

We have two equations in two unknowns Cx and Cy. Solving gives Cx = 8.0 and Cy = 6.1.

We can check that our result does give us a vector C that satisfies the two equations A⋅C = 0 and B⋅C =15.0.

Answer 3

Just saw the flaw. -2.5 in vector B should be -3.5. This has had me stuck for too long.