Recent content by smakhtar

So I got (x^2)=(A^2)+(B^2)-2AB(cos180 - theta) and 78(x^2)=(A^2)+(B^2)-2ABcos(theta) I rearranged for x in the first equation, and subbed that into the second, made B=A because they have the same magnitude. I canceled out the A's, and I got to 78(2+theta)=1-cos(theta). How do I solve for theta...

Homework Statement Vectors A and B have the same magnitude. |A+B| is 78 times greater than |A-B|, for this to happen, what must be the angle between vectors A and B? Homework Equations 78((A^2) + (B^2)-2ABcos(theta))=((A^2)+(B^2)+2ABcos(theta)) Made it by using dot product. The...

I did.

Hey, I finally figured it out. Thanks for the help. I did BC x AC and I got it right, which is the same as |BC||AC|sin (theta) right?

So it would be A= |CB|(|CA|sin (theta) all over 2?

I am assuming that you are referring to the coordinates mentioned above. I think it would be (AB x BC) (Dot) CA.

The magnitude of cross product equals the magnitude of |CB||CA|sin (theta).

The length of the sides.

And the finders and palm would refer to the x and y axis.

CA and CB are vectors. It geometrically represents the right hand rule where the thumb represents the z axis.

It has to be done using Cartesian vectors not geometric.

Homework Statement A triangle has verticies A(-2,1,3), B(7,8,-4), and C(5,0,2). Determine the area of the triange ABC. The correct answer is 35.9 square units. Homework Equations Has to be done by using dot product and/or cross product. Dot product: a(dot) b= |a||b|cos(theta) Cross...

I already got it all figured out. Thanks for trying to help.

Can you please reply and give me a solution as soon as possible? I actually have to hand it in tomorrow.

This is the updated one.