A problem with cross product of two vectors

Question

Problem:

Let $u$ and $v$ be vectors in $\mathbb{R}^3$. Can $u \times v$ be a non-zero scalar multiple of $u$?

Answer:

Recall that: $$ u \times v = (u_2 v_3 - u_3 v_2)i + (u_3v_1 - u_1 v_3)j + (u_1 v_2 - u_2 v_1)k $$ Since $u \times v$ is perpendicular to $u$ and $v$ then I believe the statement is false. I will now attempt a more rigours proof.

Suppose the statement is true. We have a $K$ such that $u \times v = Ku$ where $u$, $v$ and $K$ are non-zero. Then $$(u_2 v_3 - u_3 v_2)i + (u_3v_1 - u_1 v_3)j + (u_1 v_2 - u_2 v_1)k = K(u_1,u_2,u_3)$$ that is $$ \begin{cases} u_2 v_3 - u_3 v_2 = K u_1 \\ u_3 v_1 - u_1 v_3 = K u_2 \\ u_1 v_2 - u_2 v_1 = K u_3 \\ \end{cases}$$ Now I am struck. Am on the right track?

Answer 1

Wirsing this as another comment is too cumbersome. So I write an answer. Using \begin{cases} u_2 v_3 - u_3 v_2 = K u_1 \\ u_3 v_1 - u_1 v_3 = K u_2 \\ u_1 v_2 - u_2 v_1 = K u_3 \\ \end{cases} we multiply the first equation by $u_1$, the second by $u_2$ and the last one by $u_3$ and add these the equations. Then the righthand side equals $K(u_1^2+u_2^2+u_3^2)$. The lefthand side equals $(u_2 v_3 - u_3 v_2) u_1+(u_3 v_1 - u_1 v_3)u_2+(u_1 v_2 - u_2 v_1 )u_3 $, which turns out to be 0. So also $K(u_1^2+u_2^2+u_3^2)=0$. Since $K\not=0$ by assumption we have $u_1^2+u_2^2+u_3^2=0$. This implies that $u_1=u_2=u_3=0$.

Answer 2

The cross-product of two vectors are orthogonal to these vectors. See this link. If, $u\times v=Ku$ where $K\neq 0$, then $$u\cdot Ku=u\cdot(u\times v)$$ $$K||u||^2=0$$ $$||u||=0$$ $$u=0.$$ Therefore, $u\times v$ can be a non-zero scalar multiple of $u$ iff $u$ is zero vector.