Let u and v be vectors in ${\mathrm{ℝ}}^3$ and let c be a scalar. Prove that $c(u\times v)=\left(cu\right)\times v=u\times \left(cv\right)$. (This is Theorem 10.28).

Let u and v be vectors in ${\mathrm{ℝ}}^3$ and let c be a scalar. Prove that $c(u\times v)=\left(cu\right)\times v=u\times \left(cv\right)$.

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

Firstly finding the value of $c(u\times v)$

role="math" localid="1649918741570" $$\begin{gathered} c(u\times v)=c\left\{\right(u_1,u_2,u_3)\times (v_1,v_2,v_3\left)\right\} \\ c(u\times v)=c\left[\begin{array}{ccc}i& j& k\\ u_1& u_2& u_3\\ v_1& v_2& v_3\end{array}\right] \\ c(u\times v)=c\left(i\left|\begin{array}{cc}{u}_2& u_3\\ v_2& v_3\end{array}\right|-j\left|\begin{array}{cc}{u}_1& u_3\\ v_1& v_3\end{array}\right|+k\left|\begin{array}{cc}{u}_1& u_2\\ v_1& v_2\end{array}\right|\right) \\ c(u\times v)=c\left[i(u_2{v}_3-u_3{v}_2)-j(u_1{v}_3-u_3{v}_1)+k(u_1{v}_2-u_2{v}_1)\right] \\ c(u\times v)=c(u_2{v}_3-u_3{v}_2,-u_1{v}_3+u_3{v}_1,u_1{v}_2-u_2{v}_1) \\ c(u\times v)=(c{u}_2{v}_3-c{u}_3{v}_2,-c{u}_1{v}_3+c{u}_3{v}_1,c{u}_1{v}_2-c{u}_2{v}_1) \end{gathered}$$

$$\begin{gathered} \left(cu\right)\times v=(c{u}_1,c{u}_2,c{u}_3)\times (v_1,v_2,v_3) \\ \left(cu\right)\times v=\left[\begin{array}{ccc}i& j& k\\ c{u}_1& c{u}_2& c{u}_3\\ v_1& v_2& v_3\end{array}\right] \\ \left(cu\right)\times v=i\left|\begin{array}{cc}c{u}_2& c{u}_3\\ v_2& v_3\end{array}\right|-j\left|\begin{array}{cc}c{u}_1& c{u}_3\\ v_1& v_3\end{array}\right|+k\left|\begin{array}{cc}c{u}_1& c{u}_2\\ v_1& v_2\end{array}\right| \\ \left(cu\right)\times v=i(c{u}_2{v}_3-c{u}_3{v}_2)-j(c{u}_1{v}_3-c{u}_3{v}_1)+k(c{u}_1{v}_2-c{u}_2{v}_1) \\ \left(cu\right)\times v=(c{u}_2{v}_3-c{u}_3{v}_2,-c{u}_1{v}_3+c{u}_3{v}_1,c{u}_1{v}_2-c{u}_2{v}_1) \end{gathered}$$

$$\begin{gathered} u\times \left(cv\right)=(u_1,u_2,u_3)\times (c{v}_1,c{v}_2,c{v}_3) \\ u\times \left(cv\right)=\left[\begin{array}{ccc}i& j& k\\ u_1& u_2& u_3\\ c{v}_1& c{v}_2& c{v}_3\end{array}\right] \\ u\times \left(cv\right)=i\left|\begin{array}{cc}{u}_2& u_3\\ c{v}_2& c{v}_3\end{array}\right|-j\left|\begin{array}{cc}{u}_1& u_3\\ c{v}_1& c{v}_3\end{array}\right|+k\left|\begin{array}{cc}{u}_1& u_2\\ c{v}_1& c{v}_2\end{array}\right| \\ u\times \left(cv\right)=i(c{u}_2{v}_3-c{u}_3{v}_2)-j(c{u}_1{v}_3-c{u}_3{v}_1)+k(c{u}_1{v}_2-c{u}_2{v}_1) \\ u\times \left(cv\right)=(c{u}_2{v}_3-c{u}_3{v}_2,-c{u}_1{v}_3+c{u}_3{v}_1,c{u}_1{v}_2-c{u}_2{v}_1) \end{gathered}$$

Hence, prove that$\mathrm{c}(\mathrm{u}\times \mathrm{v})=\left(\mathrm{cu}\right)\times \mathrm{v}=\mathrm{u}\times \left(\mathrm{cv}\right)$