Question

If u and v are nonzero vectors in ${\mathrm{ℝ}}^3$, why do the equations role="math" localid="1649263352081" $u·(u\times v)=0$ and $v·(u\times v)=0$ tell us that the cross product is orthogonal to both u and v?

Short answer

We prove that$u·(u\times v)=0\mathrm{and}v·(u\times v)=0$tell us that the cross product is orthogonal to both u and v.

Step-by-step solution

Step 1. Given Information 

If u and v are nonzero vectors in ${\mathrm{ℝ}}^3$, why do the equations $u·(u\times v)=0$ and $v·(u\times v)=0$ tell us that the cross product is orthogonal to both u andv?

Step 2. u and v are nonzero vectors in ℝ3.

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

The determinant of a 3 × 3 matrix is

$u\times v=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ {\mathrm{u}}_1& {\mathrm{u}}_2& {\mathrm{u}}_3\\ {\mathrm{v}}_1& {\mathrm{v}}_2& {\mathrm{v}}_3\end{array}\right]$

Step 3. Then, by the definition of the cross product,

$u\times v=(u_2{v}_3-u_3{v}_2,u_3{v}_1-u_1{v}_3,u_1{v}_2-u_2{v}_1)$

Now proving the equation $u·(u\times v)=0$

$$\begin{gathered} u·(u\times v)=(u_1,u_2,u_3)·(u_2{v}_3-u_3{v}_2,u_3{v}_1-u_1{v}_3,u_1{v}_2-u_2{v}_1) \\ u·(u\times v)=u_1(u_2{v}_3-u_3{v}_2)+u_2(u_3{v}_1-u_1{v}_3)+u_3(u_1{v}_2-u_2{v}_1) \\ u·(u\times v)=0 \end{gathered}$$

Step 4. Now proving the equationv·(u×v)=0.

$$\begin{gathered} v·(u\times v)=(v_1,v_2,v_3)·(u_2{v}_3-u_3{v}_2,u_3{v}_1-u_1{v}_3,u_1{v}_2-u_2{v}_1) \\ v·(u\times v)=v_1(u_2{v}_3-u_3{v}_2)+v_2(u_3{v}_1-u_1{v}_3)+v_3(u_1{v}_2-u_2{v}_1) \\ v·(u\times v)=0 \end{gathered}$$