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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 $B$.

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  • $\begingroup$ 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

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

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