Vector problem using scalar (dot) and vector (cross) product??

Let vectors A = (2, -1, 1), B = (3, 0, 5), and C = (1, 4, -2), where (x, y, z) are the components of the vectors along x [hat], y [hat], and z [hat] respectively. Calculate:





A * (B x C).


That is, vector A times the cross product between vector B and vector C.





I already know that the cross product between vector B and vector C is |B| * |C| * sin (theta) = 26... I know that I can't just multiply 26 by the magnitude of vector A because when I submitted it, it was wrong. Help please? Thanks in advance!

Vector problem using scalar (dot) and vector (cross) product??
The cross product BxC is a vector. 26 is it's magnitude, but you must also figure out what direction it has.


Then, take your previous answer and multiply it by the cosine of the angle between the vectors A and BxC.





Alternate method:





First, get the actual coordinates of the cross product BxC:


x-component = By*Cz - Bz*Cy


y-component = Bz*Cx - Bx*Cz


z-component = Bx*Cy - By*Cx





If we call this vector "D", then the dot product will be


A.(BxC) = A.D = Ax*Dx + Ay*Dy + Az*Dz


where Dx = x-component calculated above


= By*Cz - Bz*Cy,


and similarly for Dy and Dz.





Hope this helps.
Reply:A*(B x C) is the triple product. You forgot something in


B x C = |B||C| sin(theta). It's


B x C = a[hat]|B||C| sin(theta), where a is a unit vector perpendicular to B and C.


However, that is not the best way of doing this problem. The cross product of B and C can be calculated by finding the determinate of the matrix:


| x[hat] j[hat] k[hat]|


|3 0 5|


|1 4 -2|


where x[hat] = (1,0,0), y[hat] = (0,1,0), z[hat] = (0,0,1)


But since this is a triple product you don't need to do that either (but it's important for you to know).


Just find the determinant of the above, but instead of i, j, k use the components of A; that is, just place the vectors on top of each other to form a matrix.


| 2 -1 1 |


|3 0 5 |


|1 4 -2|


..and find the determinant, which gives you a number, a scaler, which is the triple product.


OK, maybe your matrix stuff is a little rusty. Just in case it is, it goes like this: The triple product A*(B x C) is


2(0*-2 - 5*4) - -1(3*-2 - 5*1) + 1(3*4 - 0*1) =


2*(-20) + (-11) + 12 = -40 + -11 + 12 = -39


But don't take my word for it, check it. I do make mistakes.