Answer to Question #114831 in Analytic Geometry for Reginald Thebe

Question

Suppose u and v are nonzero vectors in 3-space, where u=(u1,u2, u3) and v=(v1,v2,v3). Prove that u × v is perpendicular to both u and v by making use of the dot product

Answer 1

We have $u\times v=\begin{vmatrix} u_1&u_2&u_3\\ v_1&v_2&v_3\\ e_1&e_2&e_3 \end{vmatrix}=$

$=\left(\begin{vmatrix} u_2&u_3\\ v_2&v_3\\ \end{vmatrix},-\begin{vmatrix} u_1&u_3\\ v_1&v_3\\ \end{vmatrix},\begin{vmatrix} u_1&u_2\\ v_1&v_2\\ \end{vmatrix}\right)$

Then $(u,u\times v)=u_1\begin{vmatrix} u_2&u_3\\ v_2&v_3\\ \end{vmatrix}-u_2\begin{vmatrix} u_1&u_3\\ v_1&v_3\\ \end{vmatrix}+u_3\begin{vmatrix} u_1&u_2\\ v_1&v_2\\ \end{vmatrix}=$

$=\begin{vmatrix} u_1&u_2&u_3\\ v_1&v_2&v_3\\ u_1&u_2&u_3 \end{vmatrix}=0$ and $(v,u\times v)=v_1\begin{vmatrix} u_2&u_3\\ v_2&v_3\\ \end{vmatrix}-v_2\begin{vmatrix} u_1&u_3\\ v_1&v_3\\ \end{vmatrix}+v_3\begin{vmatrix} u_1&u_2\\ v_1&v_2\\ \end{vmatrix}=$

$=\begin{vmatrix} u_1&u_2&u_3\\ v_1&v_2&v_3\\ v_1&v_2&v_3 \end{vmatrix}=0$