Chapter 9.7, Problem 57AYU

Prove for vectors $u\text{ and }v$ that

${\Vert u\times v\Vert }^2\text{ = }{\Vert u\Vert }^2{\Vert v\Vert }^2-{\left(u\cdot v\right)}^2$

[Hint: Proceed as in the proof of property (4), computing first the left side and then the right side.]

To determine

To prove: For vectors $u$ and $v$ that

${\Vert u \times v\Vert }^2 = {\Vert u\Vert }^2{\Vert v\Vert }^2 - {\left(\text{u}\cdot \text{v}\right)}^2$

Answer to Problem 57AYU

${\Vert u \times v\Vert }^2 = {\Vert u\Vert }^2{\Vert v\Vert }^2 - {\left(u\cdot v\right)}^2$

Explanation of Solution

Proof:

Let $\text{u} = {\text{a}}_{\text{1}}\text{i} + {\text{b}}_{\text{1}}\text{j} + {\text{c}}_{\text{1}}\text{k}$ and $\text{v} = {\text{a}}_{\text{2}}\text{i} + {\text{b}}_{\text{2}}\text{j} + {\text{c}}_{\text{2}}\text{k}$, then

$u\times v= \left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ {\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\end{array}\right|=\left(b_1{c}_2–b_2{c}_1\right)i–\left(a_1{c}_2-a_2{c}_1\right)j+\left(a_1{b}_2–a_2{b}_1\right)k$

${\Vert u\times v\Vert }^2=b_1{}^2{c}_2{}^2–2b_1{b}_2{c}_1{c}_2+b_2{}^2{c}_1{}^2+a_1{}^2{c}_2{}^2–2a_1{a}_2{c}_1{c}_2+a_2{}^2{c}_1{}^2+a_1{}^2{b}_2{}^2–2a_1{a}_2{b}_1{b}_2+a_2{}^2{b}_1{}^2$

${\Vert u\Vert }^2=a_1{}^2+b_1{}^2+c_1{}^2$

${\Vert v\Vert }^2=a_2{}^2+b_2{}^2+c_2{}^2$

${\Vert u\Vert }^2{\Vert v\Vert }^2=\left(a_1{}^2+b_1{}^2+c_1{}^2\right)\left(a_2{}^2+b_2{}^2+c_2{}^2\right)$

$=a_1{}^2{a}_2{}^2+ a_1{}^2{b}_2{}^2+ a_1{}^2{c}_2{}^2 +a_2{}^2{b}_1{}^2+ b_1{}^2{b}_2{}^2+b_1{}^2{c}_2{}^2+a_2{}^2{c}_1{}^2+b_2{}^2{c}_1{}^2+c_1{}^2{c}_2{}^2$

${\left(u.v\right)}^2={\left(a_1{a}_2+b_1{b}_2+c_1{c}_2\right)}^2$

$=a_1{}^2{a}_2{}^2+b_1{}^2{b}_2{}^2+c_1{}^2{c}_2{}^2+ 2a_1{a}_2{b}_1{b}_2+ 2b_1{b}_2{c}_1{c}_2+ 2a_1{a}_2{c}_1{c}_2$

${\Vert u\Vert }^{2 }{\Vert v\Vert }^{ 2}–{\left(u.v\right)}^2= a_1{}^2{b}_2{}^2+ a_1{}^2{c}_2{}^2 +a_2{}^2{b}_1{}^2 +b_1{}^2{c}_2{}^2+a_2{}^2{c}_1{}^2+b_2{}^2{c}_1{}^2-2a_1{a}_2{b}_1{b}_2-2b_1{b}_2{c}_1{c}_2 -2a_1{a}_2{c}_1{c}_2$

which equals to ${\Vert u\times v\Vert }^2$

Hence ${\Vert u \times v\Vert }^2 = {\Vert u\Vert }^2{\Vert v\Vert }^2 - {\left(u\cdot v\right)}^2$