Simultaneous Vector Equations How do I solve the simultaneous vector equations for $r$
$$r \times a = b, \qquad r \cdot c = \alpha $$ 
given that $a\cdot b=0$ and $a$ is not equal to $0$?
I am required to distinguish between the cases $a\cdot c$ is not equal to $0$ and $a\cdot c=0$ and give a geometrical interpretation.
 A: I'll try. 
Using the property of triple product $c\cdot (r \times a) = r \cdot (a \times c) = c\cdot b$.
So there are 
$$r\cdot b = 0$$
$$r\cdot(a \times c) = c\cdot b $$
$$r\cdot c = \alpha $$
If $a \nparallel c$ vectors {$c, a \times c, b$} is basis $R^3$.
Applying Gram–Schmidt process we will have orthonormal basis: $e_1 = \frac{c}{|c|}, e_2 = \frac{a \times c}{|a \times c|}, e_3 = b - (b\cdot e_1)e_1 - (b\cdot e_2)e_2$.
Final $r = \alpha e_1 + \frac{(c\cdot b)}{|a \times c|}e_2 + (-\frac{(b\cdot c)}{|c|^2}\alpha  -\frac{(b\cdot a \times c)}{|a \times c|^2}(b\cdot c))e_3$
I just took appropriate basis.
A: $a, b, a\times b$ form a basis for $\Re^3$ hence $ r = ua+vb+w(a\times b)$ for scalars 
$u,v,w$
Sub in given equations to get
$$ 0 + v(b\times a) + w( (a\cdotp a) b - (a\cdotp b)a = b  $$
$$ u(a\cdotp c) + v(b\cdotp c) + w((a\times b) \cdotp c ) = \alpha $$
Using $a\cdotp b = 0 $ the first equation gives  v=0 and $ w = \cfrac{1}{||a||^2} $
The second equation is then $ u(a\cdotp c) = \alpha - \cfrac {(a\times b) \cdotp c}{||a||^2} $
If $a\cdotp c \ne 0 $ then $ r = \cfrac {\left(\alpha - \cfrac {(a\times b) \cdotp c}{||a||^2} \right)}{c\cdotp a} a 
+ \cfrac {a\times b}{||a||^2}$
If $a\cdotp c=0$ and $\alpha \ne \cfrac {(a\times b) \cdotp c}{||a||^2} $ then there is no solution for $r$
If $a\cdotp c=0$ and $\alpha = \cfrac {(a\times b) \cdotp c}{||a||^2} $ then u is arbitrary and $r = ua +\cfrac {a\times b}{||a||^2}$ 
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This is the intersection of a plane with normal $c$ and a line with direction $a$.
If $a\cdotp c\ne 0$ they meet at a unique point. 
If $a\cdotp c =0$ then they do not meet unless the line is entirely contained in the plane. 
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A: Taking the scalar product with $b$ and exchanging the factors of the triple product cyclic we get
$$
b^2 
= (r\times a) \cdot b 
= (a\times b) \cdot r
$$
which is a plane with normal vector $n=a\times b$ and distance $b^2/\lVert n \rVert$ to the origin.
Similar we get
$$
b\cdot c 
= (r\times a) \cdot c
= (a\times c) \cdot r
$$
The second given equation
$$
c\cdot r = \alpha
$$
already is a plane equation.
So we have three planes, or three linear equations of a linear system in three unknowns.
$$
\left[
\begin{array}{ccc|c}
a_2 b_3 - a_3 b_2 & a_3 b_2 - a_2 b_3 &
a_1 b_2 - a_2 b_1 & b\cdot b \\
a_2 c_3 - a_3 c_2 & a_3 c_2 - a_2 c_3 &
a_1 c_2 - a_2 c_1 & b\cdot c \\
c_1 & c_2 & c_3 & \alpha 
\end{array}
\right]
$$
