Basic trig really. Find the angle theta that the vector a makes with the x axis. Add 90 degrees and you now have a perpendicular vector. You know the magnitude so just use the angle of the new vector to resolve it into components. The other vector is the same thing except subtract 90 degrees.

PS - I dont know what vector b is doing in there, unless you left out part of the question.

thx..haha such an easy question damn i look stupid..and yeah the b vector was part of another question hehe

...or u can use the dot (scalar) product of the two vectors:

a•c = a[x]*c[x] + a[y]*c[y]

a•c = 2.7*c[x] + 1.0*c[y]

since they are perpendicular, their dot product equals zero

2.7*c[x] + 1.0*c[y] = 0,

c[x] = (-1/2.7)*c[y]

now use the magnitude

c = √(c²[x] + c²[y]),

5.4 = √({(-1/2.7)c}²[y] + c²[y]),

u'll get two answers for c[y], one is for vector c, the other one u can use for vector d.

use c[y] and d[y] to find c[x] and d[x] by substitution.

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i wish my physics course was that easy...

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It has to start somewhere,
It has to start sometime,
What better place then here?
What better time than now?
--Rage Against the Machine

quote:

Originally posted by DrummeRaver86 i wish my physics course was that easy...

Haha yeah, /cringes at physics midterm Friday

Bleh, simple.

Convert the vector into polar coord.

So A = |A|/_thetaA

Then thetaA - 90 to get thetaC

C = |C|/_thetaC = 5.4/_thetaC = 5.4*cos(thetaC)*Cx + 5.4*sin(thetaC)*Cy

Then for D, which is the suplement, just reverse the signs:

D = -(C)

Voila.

I feel so smart now.