Question 2.145: By using Eqs. (2.23) and (2.34), show that U.(V × W) = | Ux......
By using Eqs. (2.23) and (2.34), show that
U.(V\times W)= \left|\begin{matrix} U_x & U_{y} & U_z \\ V_x & V_y & V_z \\ W_x & W_y & W_z \end{matrix} \right|Final Answer
One strategy is to expand the determinant in terms of its components, take the dot product, and then collapse the expansion. Eq. (2.23) is an expansion of the dot product: Eq. (2.23): U · V = U_X V_X + U_Y V_Y + U_Z V_Z. Eq. (2.34) is the determinant representation of the cross product:
\text { Eq. (2.34) } U \times V=\left|\begin{array}{ccc} i & j & k \\U_X & U_Y & U_Z \\V_X & V_Y & V_Z\end{array}\right|For notational convenience, write P = (U × V). Expand the determinant about its first row:
P = i \left|\begin{array}{ll}U_Y & U_Z \\V_Y & V_Z\end{array}\right|- j \left|\begin{array}{ll}U_X & U_Z \\V_X & V_Z\end{array}\right|+ k \left|\begin{array}{cc}U_X & U_Z \\V_X & V_Z\end{array}\right|Since the two-by-two determinants are scalars, this can be written in the form: P = iP_X + jP_Y + kP_Z where the scalars P_X,~P_Y,~and~P_Z are the two-by-two determinants. Apply Eq. (2.23) to the dot product of a vector Q with P. Thus Q · P = Q_X P_X + Q_Y P_Y + Q_Z P_Z. Substitute P_X,~P_Y,~and~P_Z into this dot product
Q \cdot P =Q_X\left|\begin{array}{ll}U_Y & U_Z \\V_Y & V_Z\end{array}\right|-Q_Y\left|\begin{array}{ll}U_X & U_Z \\V_X & V_Z\end{array}\right|+Q_z\left|\begin{array}{cc}U_X & U_Z \\V_X & V_Z\end{array}\right|But this expression can be collapsed into a three-by-three determinant directly, thus: