How to find vector equation of a plane given a line and a point? I have the following information:
a.b = 0 where . represents the scalar product. 
The plane contains the line r x a = b where r is a general point on the line and x represents the cross product (vector product). 
A point, y, lies on the plane.
How do I go about finding the equation of the plane involving r,a,y,b? Any hints/pointers would be great.
Also could someone roughly explain the intuition behind a line in E^3 being of the form r x a = b? I can't quite get my head around it :( 
Many thanks,
 A: Consider the orthogonal basis $\{\vec{a}, \vec{b}, \vec{a} \times \vec{b}\}$ in $E^3$.
We know that $\vec{a} \cdot \vec{b}=0$, $\vec{a} \cdot (\vec{a} \times \vec{b})=0$ and $\vec{b} \cdot (\vec{a} \times \vec{b})=0$.
Let 
$$\vec{r}= r_1 \vec{a}+r_2 \vec{b}+r_3 (\vec{a} \times \vec{b})\quad(1)$$
as $\vec{r} \times \vec{a} = \vec{b}$, let's apply the cross product by $\vec{a}$ on both sides:
$$(\vec{r} \times \vec{a}) \times \vec{a} = \vec{b}\times \vec{a} \quad (2)$$
Recall that:
$$(\vec{u} \times \vec{v}) \times \vec{w} = -(\vec{v}\cdot \vec{w}) \vec{u} + (\vec{u}\cdot \vec{w}) \vec{v} \quad(3) $$
Substituting $(1)$ in $(2)$ and using $(3)$, it follows:
$$-(\vec{a}\cdot \vec{a}) \vec{r} + (\vec{r}\cdot \vec{a}) \vec{a}= \vec{b} \times \vec{a} \Rightarrow $$
$$-(\vec{a}\cdot \vec{a})( r_1 \vec{a}+r_2 \vec{b}+r_3 (\vec{a} \times \vec{b})) + ((r_1 \vec{a}+r_2 \vec{b}+r_3 (\vec{a} \times \vec{b}))\cdot \vec{a}) \vec{a}= \vec{b} \times \vec{a} \Rightarrow $$
$$-|\vec{a}|^2r_2 \vec{b} - |\vec{a}|^2r_3(\vec{a} \times \vec{b})= \vec{b} \times \vec{a} \Rightarrow $$
$$\Rightarrow \left \{
\begin{array}{l}
r_2= 0 \\
r_3 = \frac{1}{|\vec{a}|^2}\\
\end{array}
\right.
$$
Therefore
$$\vec{r}=r_1 \vec{a}+\frac{1}{|\vec{a}|^2}(\vec{a} \times \vec{b}) \quad (4)$$
As we can see, $(4)$ is a line equation, since $r_1$ ($r_1 \in \mathbb R$) is arbitrary.
Let $Y_0$ a point of line $(4)$ such that $r_1=0$, we conclude that:
$$Y_0=\frac{1}{|\vec{a}|^2}(\vec{a} \times \vec{b}) \quad (5)$$
Let's now find out the plane equation that contains the line $(4)$ and the point $Y$ ($Y$ is not in line $(4)$).
Let $P$ a point of the plane, $\lambda$ and $\mu$ real numbers, the plane equation is given by:
$$P=Y_0 + \lambda(Y-Y_0)+ \mu \vec{a} \quad (6)$$
Substituting $(5)$ in $(6)$, we get:
$$P=\frac{1}{|\vec{a}|^2}(\vec{a} \times \vec{b})  + \lambda(Y-\frac{1}{|\vec{a}|^2}(\vec{a} \times \vec{b}) )+ \mu \vec{a} \quad (7)$$
