1. ## Vectors again..

For the two vectors in this figure,
A = 12m W
B = 18m 37 $^\circ$ N of E
a.) Find the magnitude and direction of the vector product $\vec{A} X \vec{B}$
solutions:
$\vec{A} X \vec{B} = AB\sin{\phi} = (-12.0m)(18.0m)\sin{143}$
$\vec{A} X \vec{B} = -129.99$
but im not sure but it also asks for a direction

b.) Find the magnitude and direction $\vec{B} X \vec{A}$

same as part a and the rule says $\vec{B} X \vec{A} = - \vec{A} X \vec{B}$
$129.99$?

For the two vectors in this figure,
A = 12m W
B = 18m 37 $^\circ$ N of E
a.) Find the magnitude and direction of the vector product $\vec{A} X \vec{B}$
solutions:
$\vec{A} X \vec{B} = AB\sin{\phi} = (-12.0m)(18.0m)\sin{143}$
$\vec{A} X \vec{B} = -129.99$
but im not sure but it also asks for a direction

b.) Find the magnitude and direction $\vec{B} X \vec{A}$

same as part a and the rule says $\vec{B} X \vec{A} = - \vec{A} X \vec{B}$
$129.99$?

For a) in the formula $\vec{A} \times \vec{B}$ we have that
$|\vec{A} \times \vec{B}| = | \vec{A} || \vec{B} | sin( \theta_{AB})$

So your working is incorrect. The negative sign in front of A (which really depends on your coordinate system anyway) is not there.

There are two ways to get the direction. One is the "right hand rule" (which I'll suggest you do a web search on) or simply work out the determinant:
$\vec{A} \times \vec{B} = \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{array} \right |$

Since your vectors are both in the xy plane (most people would choose +x in the E and +y in the N directions) your cross product will end up in the z direction. (I get the - z direction, to be precise.)

-Dan

3. my mistake is that a resultant vector has no negative...

so A is 129.99

B is -129.99

my mistake is that a resultant vector has no negative...

so A is 129.99

B is -129.99
A vector is neither positive nor negative. It has a "size" (in whatever units it has) in a certain direction.

In this case, choosing +x to be E and +y to be N:
$\vec{A} \times \vec{B} = \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & -12 & 0 \\ 18cos(37) & 18 sin(37) & 0 \end{array} \right |$ $= \hat{k}(0 - (-12)(18 cos(37))) = 172.505 \hat{k}$

Which is $172.505 ~ m^2$ in the +z direction. (Perpendicular to both E and N.)

And as you correctly noted:
$\vec{B} \times \vec{A} = - \vec{A} \times \vec{B} = -172.505 \hat{k}$
or $172.505 ~ m^2$ in the -z direction.

-Dan

5. thanks!!!