# Find the x and y component of vector c

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.

## Answers and Replies

Hurkyl
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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?

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

But why does it appear in your equations?

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

Hurkyl
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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

?

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
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Science Advisor
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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)