Calculating the Product of Vectors A, B and C

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Homework Statement



The vectors A, B and C have components Ax = 3, Ay = -2, Az = 2, Bx = 0, By = 0, Bz = 4, Cx = 2, Cy = -3, Cz = 0. Calculate the A X (B + C) ??




Homework Equations





The Attempt at a Solution

 
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hey hypsm - where are you stuck?

evaluate B+C first
 
I am ding this way: A X (B + C)=AXB+AXC = AXBsin(x) + AXCsin(x); how I can get angle of x degree ?
 
Don't you know how to calculate a cross product using the components of the two vectors? BTW, lanedance's suggestion will save you some calculating, since the addition of two vectors first is much simpler than taking two cross products and then adding.
 
hpysm said:
I am ding this way: A X (B + C)=AXB+AXC = AXBsin(x) + AXCsin(x); how I can get angle of x degree ?

that doesn't quite make sense... AXB is a vector and the magnitude is given by
|AXB| = |A|.|B|sin(x).n

where n is a unit vector perpendicular to both A & B, and x is the angle between them.

the angle between A and B, will not in general be the same as that between A and C, neither will thr normal vector.

Hence I would do it the way first suggested... (evaluate B+C first)

however, if you want to keep going your way it should be
A X (B + C)=A X B+ AX C = n|A|B||sin(x) + m|A||C|sin(y)

one good way to find the angel is using the dot product..

if you were just looking to find the magnitude of the vector this would all simplify
 
If \vec{A}= A_x\vec{i}+ A_y\vec{j}+ A_z\vec{k} and \vec{B}= B_x\vec{i}+ B_y\vec{j}+ B_z\vec{k} then, symbolically,
\vec{A}\times\vec{B}= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{array}\right|

You really need to know that before you can do this problem.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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