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.


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 $B$.

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

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.