Point of intersection of the lines $\vec{r}\times\vec{a}=\vec{b}\times\vec{a}$ and $\vec{r}\times\vec{b}=\vec{a}\times\vec{b}$

Question

If $\vec{a}=<1,1,0>,\vec{b}=<2,0,-1>$ then find the point of intersection of the lines $\vec{r}\times\vec{a}=\vec{b}\times\vec{a}$ and $\vec{r}\times\vec{b}=\vec{a}\times\vec{b}$

$$ \vec{a}\times(\vec{r}-\vec{b})=\vec{0}\implies\vec{a}||(\vec{r}-\vec{b})\implies\vec{r}-\vec{b}=\lambda\vec{a}\\ \vec{b}\times(\vec{r}-\vec{a})=\vec{0}\implies\vec{b}||(\vec{r}-\vec{a})\implies\vec{r}-\vec{a}=\mu\vec{b}\\ $$ But I do not have any hint of how to find the point intersection of the given lines ?

May be because I am unable to understand how can lines be defined in terms of cross products.

Answer 1

I'll simplify the notation.

As you say, you need $r=\lambda a+b=a+\mu b$.

It strikes me that $r=a+b$ fits the bill.

Answer 2

In other words, $r=b+\lambda a$ and $r=a+\mu b.$ Setting them equal and rearranging gives $$(1-\mu)b=b-\mu b=a -\lambda a=(1-\lambda)a.$$ Clearly one solution is $\lambda=\mu=1.$ Otherwise, write the equation as $$b=\frac{1-\lambda}{1-\mu}a,$$ which says that $a$ and $b$ are parallel. But they are not since clearly $a\cdot b=2\ne \pm\sqrt{10}.$

Thus we have the only solution $$\lambda=\mu=1.$$