The Scalar and Cross Product of Vectors
Given two vectors:
[tex]\begin{gathered} \underline{r_1}=(a,b,c) \\ \underline{r_2}=(d,e,f) \end{gathered}[/tex]
The scalar product is defined as:
[tex]\underline{r_1}\cdot\underline{r_2}=ad+be+cf[/tex]
The cross product is the result of computing the following determinant:
[tex]\underline{r_1}\times\underline{r_2}=\begin{bmatrix}i & j & {k} \\ {a} & {b} & {c} \\ {d} & {e} & {f}\end{bmatrix}[/tex]
Where i, j, and k are the unit vectors in each of the directions x, y, and z, respectively.
This concept will be applied to the following physics problem.
Given a force F= (2, 3, 0) and the distance vector d = (4, 0, 0), the torque is defined by:
[tex]\tau=r\times F[/tex]
Calculating:
[tex]\tau=(4,0,0)\times(2,3,0)[/tex][tex]\tau=\begin{bmatrix}{i} & {j} & {k} \\ {4} & {0} & {0} \\ {2} & {3} & {0}\end{bmatrix}[/tex]
Calculating the determinant:
[tex]\begin{gathered} \tau=0i+12k+0j-(0k+0j+0i) \\ \tau=0i+0j+12k \end{gathered}[/tex]
Expressing in vector form τ = (0, 0, 12) <= should use angle brackets
The magnitude of the torque is:
[tex]\begin{gathered} |\tau|=\sqrt[]{0^2+0^2+12^2} \\ |\tau|=\sqrt[]{144} \\ |\tau|=12 \end{gathered}[/tex]
The power P is equal to the scalar product of the torque by the angular velocity w. We are given the angular velocity w = (3, 3, 2), thus:
[tex]\begin{gathered} P=(0,0,12)\cdot(3,3,2) \\ P=0\cdot3+0\cdot3+12\cdot2 \\ P=24 \end{gathered}[/tex]
P = 24
