Question

Let u and v be vectors in ${\mathrm{ℝ}}^3$. Prove that $v·(u\times v)=0$. (This is Theorem 10.31(b).)

Short answer

Hence, we prove that$v·(u\times v)=0$

Step-by-step solution

Step 1. Given Information

Let u and v be vectors in ${\mathrm{ℝ}}^3$. Prove that role="math" localid="1649924080419" $v·(u\times v)=0$.

Step 2. We have to prove v·(u×v)=0

Let $u=(u_1,u_2,u_3)\mathrm{and}v=(v_1,v_2,v_3)$

Now the determinant is

$$\begin{gathered} v·(u\times v)=\det \left[\begin{array}{ccc}{\mathrm{v}}_1& {\mathrm{v}}_2& {\mathrm{v}}_3\\ {\mathrm{u}}_1& {\mathrm{u}}_2& {\mathrm{u}}_3\\ {\mathrm{v}}_1& {\mathrm{v}}_2& {\mathrm{v}}_3\end{array}\right] \\ v·(u\times v)={\mathrm{v}}_1\left|\begin{array}{cc}{\mathrm{u}}_2& {\mathrm{u}}_3\\ {\mathrm{v}}_2& {\mathrm{v}}_3\end{array}\right|-{\mathrm{v}}_2\left|\begin{array}{cc}{\mathrm{u}}_1& {\mathrm{u}}_3\\ {\mathrm{v}}_1& {\mathrm{v}}_3\end{array}\right|+{\mathrm{v}}_3\left|\begin{array}{cc}{\mathrm{u}}_1& {\mathrm{u}}_2\\ {\mathrm{v}}_1& {\mathrm{v}}_2\end{array}\right| \\ \mathrm{v}·(\mathrm{u}\times \mathrm{v})={\mathrm{v}}_1\left({\mathrm{u}}_2{\mathrm{v}}_3-{\mathrm{u}}_3{\mathrm{v}}_2\right)-{\mathrm{v}}_2\left({\mathrm{u}}_1{\mathrm{v}}_3-{\mathrm{u}}_3{\mathrm{v}}_1\right)+{\mathrm{v}}_3\left({\mathrm{u}}_1{\mathrm{v}}_2-{\mathrm{u}}_2{\mathrm{v}}_1\right) \\ \mathrm{v}·(\mathrm{u}\times \mathrm{v})={\mathrm{v}}_1{\mathrm{u}}_2{\mathrm{v}}_3-{\mathrm{v}}_1{\mathrm{u}}_3{\mathrm{v}}_2-{\mathrm{v}}_2{\mathrm{u}}_1{\mathrm{v}}_3+{\mathrm{v}}_2{\mathrm{u}}_3{\mathrm{v}}_1+{\mathrm{v}}_3{\mathrm{u}}_1{\mathrm{v}}_2-{\mathrm{v}}_3{\mathrm{u}}_2{\mathrm{v}}_1 \\ \mathrm{v}·(\mathrm{u}\times \mathrm{v})=0 \end{gathered}$$