vectors

[ios]

You are given vectors a = 4.50i-7.00j and b = -3.50i+7.40j. 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 10.0.

now i figured that since vector a is in quad 4, then to be perpendicular, vector c would have to be in quad 1 or 3, but im not sure which one. i got that to find the scalar product, its 10 = -3.5(Xc)+7.4(1.556Yc) & arctan(32.7352)=Xc/1.556Yc. but then i kind of reach a dead end and i can't figure out what to do. the question asks to find the x and y component of vector c.

[Hurkyl]

10 = -3.5(Xc)+7.4(1.556Yc) & arctan(32.7352)=Xc/1.556Yc

Can't you solve that system of equations for Xc and Yc?

P.S. what are Xc and Yc, and where did the 32.7352 and the 1.556 come from?

[ios]

Xc is the x component of vector c, and Yc is the y component of vector c. i'm just not sure if my math was right or not. 32.5372 is the angle if you take tan of 4.5/7. and 1.556 is the ratio of x to y (vector a).

[Hurkyl]

1.556 is the ratio of x to y (vector a).

But why does it appear in your equations?

[ios]

don't i need that for vector c to be perpendicular to vector a?

[Hurkyl]

I get the feeling you're mixing fragments of equations together without really knowing why.

You have (correctly) determined that Xc / Yc = 14/9 (= 1.556).

But why did you put the 1.556 into the two equations

10 = -3.5(Xc)+7.4(1.556Yc) & arctan(32.7352)=Xc/1.556Yc

?

[ios]

don't i need to maintain that x to y ratio for the angle of vector c to be perpendicular to vector a? i thought that was the only way to keep the angle so that added together, the angle between vector c and vector a equals 90.

[Hurkyl]

Remember that the definition of dot product is:

"The product of the x coordinates plus the product of the y coordinates".

The y coordinates of b and c are 7.40 and Yc respectively. (not 1.556 Yc)