Posted by **lina** on Sunday, August 30, 2009 at 10:45pm.

The concept i get, but somehow i just can't execute this problem, please help me!

You are given vectors A = 5.5 6.2 and B = - 3.7 7.4 . 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 19.0.

I need x and y components for vector C

- physics [vectors] -
**Damon**, Monday, August 31, 2009 at 6:09am
A = 5.5 i + 6.2 j

B = -3.7i + 7.4 j

C = x i + y j

A dot C = |A| |C| cos theta

if perpendicular theta is 90 degrees and cos theta = 0

so

5.5 x + 6.2 y = 0

B dot C = 19

so

-3.7 x + 7.4 y = 19.0

solve those two equations for x and y

- physics [vectors] -
**DN**, Tuesday, November 17, 2015 at 12:28pm
5.5(_3.7x+7.4y=19)

3.7(5.5X+2.2y=0)

(20.35x+22.94y=0)

+ (_20.35x+40.70y=384.65)

Y=384.68/63.64

X=_(22.94(384.68/63.64)/20.35)

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