1
$\begingroup$

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.

  • 0
    I've been looking at books for something similar for hours now and have found nothing on this. How can i answer this question?2012-02-12

3 Answers 3

0

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.

  • 0
    @mr_e_man I do not remember why I used the wedge. Definitely, I mean vector product. And as you have edited the question feel free to edit answers.2018-08-14
0

$a, b, a\times b$ 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 $r = ua +\cfrac {a\times b}{||a||^2}$

..............................................................................................................

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.

..............................................................................................................

  • 0
    I suggest you change $\wedge$ to $\times$; the wedge product usually means something else which isn't a vector. I don't know whether this is a common alternative notation for the cross product, or a common mistake to identify the cross product with the exterior product.2018-08-12
0

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] $