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.
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
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!
To solve the system of equations -
1. Start with the equations:
5.0Cx - 6.5Cy = 0
-2.5Cx + 7.0Cy = 15.0
2. Multiply the first equation by 2.5 and the second equation by 5 to eliminate the coefficients of Cx.
12.5Cx - 16.25Cy = 0
-12.5Cx + 35.0Cy = 75.0
3. Add the two equations together, canceling out the Cx terms:
-16.25Cy + 35.0Cy = 75.0
18.75Cy = 75.0
4. Solve for Cy:
Cy = 75.0 / 18.75
Cy = 4.0
5. Substitute the value of Cy back into one of the original equations to solve for Cx:
-2.5Cx + 7.0(4.0) = 15.0
-2.5Cx + 28.0 = 15.0
-2.5Cx = 15.0 - 28.0
-2.5Cx = -13.0
Cx = -13.0 / -2.5
Cx = 5.2
6. Therefore, the x-component of vector C is Cx = 5.2 and the y-component is Cy = 4.0.
To check the result, substitute these values of Cx and Cy back into the original equations:
A⋅C = 5.0(5.2) - 6.5(4.0) = 0 (approximately)
B⋅C = -2.5(5.2) + 7.0(4.0) = 15.0
The result satisfies both equations, confirming that the values of Cx = 5.2 and Cy = 4.0 are correct.
To solve for the components of vector C, we can use the two equations given:
Equation 1: 5.0Cx - 6.5Cy = 0
Equation 2: -2.5Cx + 7.0Cy = 15.0
We can solve this system of linear equations using any method we prefer. One common method is substitution, where we solve one equation for one variable and substitute it into the other equation.
Let's solve Equation 1 for Cx:
5.0Cx - 6.5Cy = 0
5.0Cx = 6.5Cy
Cx = (6.5/5.0)Cy
Cx = 1.3Cy
Now, substitute this value of Cx into Equation 2:
-2.5(1.3Cy) + 7.0Cy = 15.0
-3.25Cy + 7.0Cy = 15.0
3.75Cy = 15.0
Cy = 15.0 / 3.75
Cy = 4.0
Now, substitute this value of Cy back into the equation for Cx:
Cx = 1.3Cy
Cx = 1.3(4.0)
Cx = 5.2
Therefore, the x-component of vector C (Cx) is 5.2 and the y-component (Cy) is 4.0.