# Can skew lines share a normal vector? And how would I derive the equation for the new plane?

Can skew lines share a normal vector?

I know that if I take the cross product of two vectors then I will be able to find the normal vector to the plane created by $a\times b$.

And if I take the cross product of the direction vectors from the skew lines and it is non zero, then I will get the normal vector to a plane created by the direction vectors.

And from there I could come up with an equation for the plane created by those two vectors?

How would I do that?

• For any pair of non-coplanar lines, there exists a unique line which is perpendicular to both. Commented Feb 13, 2017 at 16:28

If you have a pair of skew lines with direction vectors ${\bf a}$ and ${\bf b}$, then since they are skew, their direction vectors are not parallel. Non-parallel vectors will always yield a nonzero cross product. So ${\bf n} = {\bf a} \times {\bf b}$ will (for skew lines) always be a nonzero vector.

So just as with any nonzero vector, you can use ${\bf n}$ as a normal for a plane. But to determine a plane you need a normal vector and a point.

If you choose a point on one of the skew lines, you'll get a plane containing that line and the other skew line will be parallel to your plane.

If you choose a point not on either of the skew lines, you'll get a plane which is parallel to both of the lines but contains neither line.

Let the first of the skew lines be parameterized as $$\vec{a}(t) = \vec{\alpha}t + \vec{\sigma}$$ and the second skew line be $$\vec{b}(t) = \vec{\beta}t + \vec{\rho}$$ Then unless $\vec{\alpha} \times \vec{\beta} = 0$ -- and for true skew lines that cross product cannot be zero -- we can construct the following line which will meet the first skew line when $s=0$ and the second when $s=1$: $$\vec{c}(s) =\frac{(\vec{\rho}-\vec{\sigma})\cdot(\vec{\alpha} \times \vec{\beta})}{|\vec{\alpha} \times \vec{\beta}|^2 } (\vec{\alpha} \times \vec{\beta}) s + \vec{\sigma}$$ The line parameterized as $\vec{c}(s)$ will be orthogonal to both the skew lines, and will meet each of those lines.

BTW, for any two lines in 3-space, there is a third line that meets both the two lines and forms a right angle at the meeting point. If the two lines are parallel, rather than skew, the third line can be any of the family of lines joining the two parallel lines and at right angels to one (hence both) of them. If the lines meet, then the desired line is the line through the meeting point, normal to the plane determined by the two lines.

• The perpendicular line should meet $\boldsymbol{a}$ at $$\boldsymbol{\sigma}+ \frac{(\boldsymbol{\sigma}-\boldsymbol{\rho}) \cdot (\boldsymbol{\alpha} \times \boldsymbol{\beta}) \times \boldsymbol{\beta}} {(\boldsymbol{\alpha} \times \boldsymbol{\beta})^2} \; \boldsymbol{\alpha}$$ and $\boldsymbol{b}$ at $$\boldsymbol{\rho}+ \frac{(\boldsymbol{\sigma}-\boldsymbol{\rho}) \cdot (\boldsymbol{\alpha} \times \boldsymbol{\beta}) \times \boldsymbol{\alpha}} {(\boldsymbol{\alpha} \times \boldsymbol{\beta})^2} \; \boldsymbol{\beta}$$ Commented Feb 13, 2017 at 20:56