Question

Let u and v be vectors in ${\mathrm{ℝ}}^3$. Prove Lagrange’s identity, Theorem 10.30:

role="math" localid="1649924794767" ${||u\times v||}^2={||u||}^2{||v||}^2-{(u·v)}^2$

Short answer

Hence, prove that${||u\times v||}^2={||u||}^2{||v||}^2-{(u·v)}^2$.

Step-by-step solution

Step 1. Given Information

Let u and v be vectors in ${\mathrm{ℝ}}^3$. Prove Lagrange’s identity, Theorem 10.30:

${||u\times v||}^2={||u||}^2{||v||}^2-{(u·v)}^2$

Step 2. We have to prove u×v2=u2v2−(u·v)2

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

Firstly finding the value of$$\begin{gathered} {(u·v)}^2={(u_1{v}_1+u_2{v}_2+u_3{v}_3)}^2 \\ {(u·v)}^2={{\left(u_1{v}_1\right)}^2+{\left(u_2{v}_2\right)}^2+\left(u_3{v}_3\right)}^2+2·u_1{v}_1·u_2{v}_2+2·u_2{v}_2·u_3{v}_3+2·u_3{v}_3·u_1{v}_1 \\ {(u·v)}^2=u_1^2{v}_1^2+u_2^2{v}_2^2+u_3^2{v}_3^2+2u_1{v}_1{u}_2{v}_2+2u_2{v}_2{u}_3{v}_3+2u_3{v}_3{u}_1{v}_1 \end{gathered}$$

Step 3. Now finding the value of u×v2

$$\begin{gathered} u\times v=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ {\mathrm{u}}_1& {\mathrm{u}}_2& {\mathrm{u}}_3\\ {\mathrm{v}}_1& {\mathrm{v}}_2& {\mathrm{v}}_3\end{array}\right] \\ \mathrm{u}\times \mathrm{v}=\mathrm{i}\left({\mathrm{u}}_2{\mathrm{v}}_3-{\mathrm{u}}_3{\mathrm{v}}_2\right)-\mathrm{j}\left({\mathrm{u}}_1{\mathrm{v}}_3-{\mathrm{u}}_3{\mathrm{v}}_1\right)+\mathrm{k}\left({\mathrm{u}}_1{\mathrm{v}}_2-{\mathrm{u}}_2{\mathrm{v}}_1\right) \\ \mathrm{u}\times \mathrm{v}=i({\mathrm{u}}_2{\mathrm{v}}_3-{\mathrm{u}}_3{\mathrm{v}}_2)-j({\mathrm{u}}_1{\mathrm{v}}_3-{\mathrm{u}}_3{\mathrm{v}}_1)+k({\mathrm{u}}_1{\mathrm{v}}_2-{\mathrm{u}}_2{\mathrm{v}}_1) \\ \mathrm{u}\times \mathrm{v}=({\mathrm{u}}_2{\mathrm{v}}_3-{\mathrm{u}}_3{\mathrm{v}}_2,{\mathrm{u}}_1{\mathrm{v}}_3-{\mathrm{u}}_3{\mathrm{v}}_1,{\mathrm{u}}_1{\mathrm{v}}_2-{\mathrm{u}}_2{\mathrm{v}}_1) \\ {||\mathrm{u}\times \mathrm{v}||}^2={({\mathrm{u}}_2{\mathrm{v}}_3-{\mathrm{u}}_3{\mathrm{v}}_2,{\mathrm{u}}_1{\mathrm{v}}_3-{\mathrm{u}}_3{\mathrm{v}}_1,{\mathrm{u}}_1{\mathrm{v}}_2-{\mathrm{u}}_2{\mathrm{v}}_1)}^2 \\ {||\mathrm{u}\times \mathrm{v}||}^2={{({\mathrm{u}}_2{\mathrm{v}}_3-{\mathrm{u}}_3{\mathrm{v}}_2)}^2+{({\mathrm{u}}_1{\mathrm{v}}_3-{\mathrm{u}}_3{\mathrm{v}}_1)}^2+({\mathrm{u}}_1{\mathrm{v}}_2-{\mathrm{u}}_2{\mathrm{v}}_1)}^2 \\ {||\mathrm{u}\times \mathrm{v}||}^2={\mathrm{u}}_2^2{\mathrm{v}}_3^2+{\mathrm{u}}_3^2{\mathrm{v}}_2^2-2{\mathrm{u}}_2{\mathrm{v}}_3{\mathrm{u}}_3{\mathrm{v}}_2+{\mathrm{u}}_1^2{\mathrm{v}}_3^2+{\mathrm{u}}_3^2{\mathrm{v}}_1^2-2{\mathrm{u}}_1{\mathrm{v}}_3{\mathrm{u}}_3{\mathrm{v}}_1+{\mathrm{u}}_1^2{\mathrm{v}}_2^2+{\mathrm{u}}_2^2+{\mathrm{v}}_1^2-2{\mathrm{u}}_1{\mathrm{v}}_2-{\mathrm{u}}_2{\mathrm{v}}_1 \\ {||\mathrm{u}\times \mathrm{v}||}^2={\mathrm{u}}_1^2{\mathrm{v}}_2^2+{\mathrm{u}}_1^2{\mathrm{v}}_3^2+{\mathrm{u}}_2^2{\mathrm{v}}_1^2+{\mathrm{u}}_2^2{\mathrm{v}}_3^2+{\mathrm{u}}_3^2{\mathrm{v}}_1^2+{\mathrm{u}}_3^2{\mathrm{v}}_2^2-2{\mathrm{u}}_1{\mathrm{v}}_1{\mathrm{u}}_2{\mathrm{v}}_2-2{\mathrm{u}}_2{\mathrm{v}}_2{\mathrm{u}}_3{\mathrm{v}}_3-2{\mathrm{u}}_3{\mathrm{v}}_3{\mathrm{u}}_1{\mathrm{v}}_1 \end{gathered}$$

Step 4. Now finding the value of u2v2

$$\begin{gathered} {||u||}^2{||v||}^2=(u_1^2,u_2^2,u_3^2)·(v_1^2,v_2^2,v_3^2) \\ {||u||}^2{||v||}^2=u_1^2(v_1^2,v_2^2,v_3^2)+u_2^2(v_1^2,v_2^2,v_3^2)+u_3^2(v_1^2,v_2^2,v_3^2) \\ {||u||}^2{||v||}^2=u_1^2{v}_1^2+u_1^2{v}_2^2+u_1^2{v}_3^2+u_2^2{v}_1^2+u_2^2{v}_2^2+u_2^2{v}_3^2+u_3^2{v}_1^2+u_3^2{v}_2^2+u_3^2{v}_3^2 \end{gathered}$$

Step 5. Now finding the value of u2v2−(u·v)2

$$\begin{gathered} {||u||}^2{||v||}^2-{(u·v)}^2=u_1^2{v}_1^2+u_1^2{v}_2^2+u_1^2{v}_3^2+u_2^2{v}_1^2+u_2^2{v}_2^2+u_2^2{v}_3^2+u_3^2{v}_1^2+u_3^2{v}_2^2+u_3^2{v}_3^2 \\ -(u_1^2{v}_1^2+u_2^2{v}_2^2+u_3^2{v}_3^2+2u_1{v}_1{u}_2{v}_2+2u_2{v}_2{u}_3{v}_3+2u_3{v}_3{u}_1{v}_1) \\ {||u||}^2{||v||}^2-{(u·v)}^2=u_1^2{v}_1^2+u_1^2{v}_2^2+u_1^2{v}_3^2+u_2^2{v}_1^2+u_2^2{v}_2^2+u_2^2{v}_3^2+u_3^2{v}_1^2+u_3^2{v}_2^2+u_3^2{v}_3^2 \\ -u_1^2{v}_1^2-u_2^2{v}_2^2-u_3^2{v}_3^2-2u_1{v}_1{u}_2{v}_2-2u_2{v}_2{u}_3{v}_3-2u_3{v}_3{u}_1{v}_1 \\ {||u||}^2{||v||}^2-{(u·v)}^2=u_1^2{v}_2^2+u_1^2{v}_3^2+u_2^2{v}_1^2+u_2^2{v}_3^2+u_3^2{v}_1^2+u_3^2{v}_2^2 \\ -2u_1{v}_1{u}_2{v}_2-2u_2{v}_2{u}_3{v}_3-2u_3{v}_3{u}_1{v}_1 \\ {||u||}^2{||v||}^2-{(u·v)}^2={||\mathrm{u}\times \mathrm{v}||}^2 \end{gathered}$$