vector equation of line

Question

If I write equation as: $r-a=k(b-r)$ ,then equation becomes $r=\frac{(a+kb)}{(1+k)}$.While equation is $r=a+k(b-a)$ Now are these two vector equations different or same. Please assume $r,a,b$ as vectors,i.e. arrow above them

Answer 1

$\mathbf r = \frac {\mathbf a+k\mathbf b}{1+k}\\ \mathbf r = \frac {\mathbf a(1+k)+k(\mathbf b-\mathbf a)}{1+k}\\ \mathbf r = \mathbf a + \frac{k}{1+k} (\mathbf b-\mathbf a)\\ $

Now $k$ is any scalar. And $\frac{k}{1+k}$ is just as easily any scalar.

The equations are eqiuvalent.

Answer 2

As long as the result is a linear combination of two constant vectors then you have a vector equation of a line.

For example the line between two points A and B is

$$\mathbf{r} = (1-t) \mathbf{a} + t\, \mathbf{b} \\ = \mathbf{a} +t\, (\mathbf{b}-\mathbf{a})$$

But also the line from point A with direction $\mathbf{e}$ is

$$\mathbf{r} = \mathbf{a} + s\, \mathbf{e} $$ but if $s = t\, \| \mathbf{b}-\mathbf{a} \|$ then the above is identical to the first expression.

Even if the scalar multiplier of the linear combination is non linear quantity like

$$\mathbf{r} = \mathbf{a} +\frac{u}{1+u}\, (\mathbf{b}-\mathbf{a})$$ you can always make a substitution $t = \frac{u}{1+u}$ to a single value to get back to the equation of the line.

In summary, a line is a linear combination of two points with the general expression

$$ \mathbf{r} = \lambda(t)\, \mathbf{a} + \mu(t)\, \mathbf{b} $$ if $\lambda(t)$ and $\mu(t)$ are scalar function of a single parameter $t$.