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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

Fernando Revilla
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Koolman
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    No, this only shows that they lie on a plane. What you've calculated is the triple product, which measures the volume of a parallelepiped formed by the three vectors. There are more ways to have zero volume than to have all the points collinear. – Michael Burr Jan 20 '17 at 12:20
  • Take two of the points and construct the line through those two. Then show that the third point also lies on that line. – Mathematician 42 Jan 20 '17 at 12:26

2 Answers2

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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.

  • Can you explain it more – Koolman Jan 20 '17 at 12:35
  • @koolman We have $\overrightarrow{AB} =(-1, 10, 7) $ and $\overrightarrow{AC} =(1, -10, -7) $. We can see that they are scalar multiples of each other so $A,B,C $ are collinear. –  Jan 20 '17 at 12:45
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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.

bubba
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