Show that the points are collinear

Question

Show that the points $$\mathbf a-2\mathbf b+3\mathbf c ,\quad 2\mathbf a+3\mathbf b-4\mathbf c ,\quad-7\mathbf b+10\mathbf c$$ are collinear. Where $\mathbf a ,\mathbf b,\mathbf c$ are vectors .

I thought as

Let the points be $k ,l ,m$ then

$(k\times l)\cdot m=0$

Can I prove it like this

Answer 1

The simplest way:

If $A,B,C$ are collinear, then $AB$ and $AC$ are proportional.

In your case, $$(-1,10,7)=\lambda(1,-10,-7).$$

Obviously, yes they are collinear.

Answer 2

Let's denote the three points by $\mathbf{k}$, $\mathbf{l}$, $\mathbf{m}$, as you suggested. Then let $\mathbf{u} = \mathbf{l} - \mathbf{k}$, and $\mathbf{v} = \mathbf{k} - \mathbf{m}$. Then the three points $\mathbf{k}$, $\mathbf{l}$, $\mathbf{m}$ are collinear if and only if $\mathbf{u}$ and $\mathbf{v}$ are parallel. To show that they are parallel, you can check that $\mathbf{u} \times \mathbf{v} = \mathbf{0}$, or you can show that $\mathbf{u}$ is a scalar multiple of $\mathbf{v}$.

A couple of people have suggested that $\mathbf{k}$, $\mathbf{l}$, $\mathbf{m}$ are collinear if and only if $(\mathbf{k} \times \mathbf{l}) \cdot \mathbf{m} = 0$. This is not correct. The triple product (or the corresponding determinant) will be zero iff the position vectors of the three points are coplanar. Collinear implies coplanar, of course, but the converse is not true. So, this condition is necessary bu not sufficient for collinearity.