5.0Cx − 6.5Cy = 0
−3.5Cx + 7.0Cy = 15.0
using determinates D= 35-22.75 = 12.25
then x= +15*6.5 divided by 12.25
cx= 7.96
cy= 75/12.25=6.12
A.C= (5*7.96-6.5*6.12)= zero
B.C= (-3.5*7.96+7*6.12)= 15
You don't have to use determinants, but it is pretty fast on a 2x2
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.
Here's the step by step from the book that I'm having a hard time wrapping around, perhaps someone can help me solve and 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.
(It's the "Solving" that has me scratching my head...)
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.
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Fantastic. I realized I've been doing this with the -2.5 instead of -3.5 as the x component of vector B, and of course it didn't make any sense. Thank you for the response!
1 answer