# Intersecting Points on a Line in $ℝ^3$

Suppose you have a line in $$ℝ^3$$ that contains points $$A(a,b,c)$$ and $$B(d,e,f)$$.

Is it then true that the midpoint of $$A$$ and $$B$$ (say $$C$$) also lies on that line?

In other words, does it follow that the line also intersects point $$C(\frac{a+d}{2},\frac{b+e}{2},\frac{c+f}{2})$$?

Any help would be greatly appreciated.

CONTEXT:

I have an assignment question where you're given the equations of two lines $$L_1$$ and $$L_2$$ in $$ℝ^3$$ that contain unknown points $$A$$ and $$B$$ respectively, and you have to find the equation of a line that contains both points and is perpendicular to $$L_1$$ and $$L_2$$.

So far I have found the cross product of direction vectors of $$L_1$$ and $$L_2$$, which I will use as the direction vector of my new line.

I was then wondering if, by letting $$C(\frac{a+d}{2},\frac{b+e}{2},\frac{c+f}{2})$$ lie on my new line, this would then satisfy the requirement that the line contains $$A$$ and $$B$$.

• Yes, all of the points $\lambda A + (1-\lambda)B$ (including $\frac 12 A + \frac12 B$) lie on $\overline {AB}$ Sep 12, 2019 at 5:21

It is intuitively clear. To show this, you may take the cross product of the vectors $$\vec {CA}$$ and $$\vec{CB}.$$ That should vanish.