# Find intersection of two 3D lines

I have two lines $(5,5,4) (10,10,6)$ and $(5,5,5) (10,10,3)$ with same $x$, $y$ and difference in $z$ values.

Please some body tell me how can I find the intersection of these lines.

EDIT: By using the answer given by coffemath I would able to find the intersection point for the above given points. But I'm getting a problem for $(6,8,4) (12,15,4)$ and $(6,8,2) (12,15,6)$. I'm unable to calculate the common point for these points as it is resulting in Zero. Any ideas to resolve this?

Thanks, Kumar.

• For your second problem, the first line has a constant z value of 4. Can you see that? If you plug this into the second line equation you will be well on your way. Commented Jan 5, 2013 at 12:55
• Kumar: I included in my answer below a treatment of the "zero coordinate of direction numbers" case you refer to in your EDIT. In a way the occurrence of a zero makes things easier since one knows immediately one of the variable values. Commented Jan 6, 2013 at 6:36

There are at least two ways to approach such problems, through vectors as such and through coordinates. Some problems are easier one way, and this one is definitely easier via coordinates. Several such solutions have been given. But the more ways you can solve a problem, the better you probably understand the underlying mathematics. So I'll share a vector method.

We can arrive at the solution without the use of coordinates, except as a matter of calculation of the final answer. We also assume that a line is given by a point and a (nonzero) direction vector (in either direction). In the examples given, the lines are defined by two points, but a direction vector may be obtained by taking the difference in the coordinates of the two given points.

Here is the construction. We leave the proof as an exercise, so that the interested reader may benefit by working it out (hint: the Law of Sines is helpful).

Let $\alpha$, $\beta$ be the given lines.

Let $C$, $D$ be points on $\alpha$, $\beta$, resp. If $C$ is on $\beta$ or $D$ is on $\alpha$, we have found the point of intersection. So let us assume they are not.

Let ${\bf e}$ and ${\bf f}$ be direction vectors of $\alpha$, $\beta$, resp. Let ${\bf g} = \vec{CD}$.

Let $h = ||{\bf f} \times {\bf g}||$, $k=||{\bf f} \times {\bf e}||$. If either is $0$, then there is no intersection; otherwise, they are nonzero and proceed.

Let ${\bf l} = {h \over k} {\bf e}$. Then one of $M = C \pm {\bf l}$ is the point of intersection (or depending on the notation you prefer, $\vec{OM} = \vec{OC} \pm {\bf m}$), where the sign may be determined as follows: If ${\bf f} \times {\bf g}$ and ${\bf f} \times {\bf e}$ point in the same direction, then the sign is $+$; otherwise, the sign is $-$.

One can compose all the steps to get a formula: $$M = C \pm {||{\bf f} \times {\vec{CD}}|| \over ||{\bf f} \times {\bf e}||}\,{\bf e}\,.$$

If you want to check, we can, in the second example, take $C = (6,8,4)$, $D = (6,8,2)$ and ${\bf e} = (6,7,0)$ and ${\bf f} = (6,7,4)$. Therefore ${\bf g} = \vec{CD} = (0,0,-2)$.

So ${\bf f} \times {\bf g} = (-14, 12,0)$ and ${\bf f} \times {\bf e} = (-28, 24, 0)$, which point in the same direction; so we take the $+$ sign.

Therefore the point of intersection is $C + {1\over2}{\bf e} = (9,{23\over2},4)$.

• The $13/2$ should be $23/2$. At the point where you find $C+(1/2)e$ the calculation for second coord. is $8+(1/2)7=23/2.$ This then agrees with the other answers for this case. Commented Jan 6, 2013 at 7:29
• @coffeemath Thanks for pointing it out Commented Jan 6, 2013 at 13:49
• What if the lines don't perfectly intersect, will this method still give the "closest" point of intersection between the two lines? Commented Sep 17, 2020 at 7:51
• Also worth to mention that when both lines are collinear there are infinite solutions and therefore event though h and k are zero that doesn't mean there is no intersection Commented Dec 12, 2022 at 11:00

The direction numbers $(a,b,c)$ for a line in space may be obtained from two points on the line by subtracting corresponding coordinates. Note that $(a,b,c)$ may be rescaled by multiplying through by any nonzero constant.

The first line has direction numbers $(5,5,2)$ while the second line has direction numbers $(5,5,-2).$ Once one has direction numbers $(a,b,c)$, one can use either given point of the line to obtain the symmetric form of its equation as $$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}.$$ Note that if one or two of $a,b,c$ are $0$ the equation for that variable is obtained by setting the top to zero. That doesn't happen in your case.

Using the given point $(5,5,4)$ of the first line gives its symmetric equation as $$\frac{x-5}{5}=\frac{y-5}{5}=\frac{z-4}{2}.$$ And using the given point $(5,5,5)$ of the second line gives its symmetric form $$\frac{x-5}{5}=\frac{y-5}{5}=\frac{z-5}{-2}.$$ Now if the point $(x,y,z)$ is on both lines, the equation $$\frac{z-4}{2}=\frac{z-5}{-2}$$ gives $z=9/2$, so that the common value for the fractions is $(9/2-4)/2=1/4$. This value is then used to find $x$ and $y$. In this example the equations are both of the same form $(t-5)/5=1/4$ with solution $t=25/4$. So we may conclude the intersection point is $$(25/4,\ 25/4,\ 9/2).$$

ADDED CASE: The OP has asked about another case, which illustrates what happens when one of the direction numbers of one of the two lines is $0$.

Line 1: points $(6,8,4),\ (12,15,4);$ directions $(6,7,0)$, "equation" $$\frac{x-6}{6}=\frac{y-8}{7}=\frac{z-4}{0},$$ where I put equation in quotes because of the division by zero, and as noted the zero denominator of the last fraction means $z=4$ (so $z$ is constant on line 1).

Line 2: points $(6,8,2),\ (12,15,6);$ directions $(6,7,4)$, equation $$\frac{x-6}{6}=\frac{y-8}{8}=\frac{z-2}{4}.$$ Now since we know $z=4$ from line 1 equation, we can use $z=4$ in $(z-2)/4$ of line 2 equation, to get the common fraction value of $(4-2)/4=1/2$. Then from either line, $(x-6)/6=1/2$ so $x=9$, and $(y-8)/7=1/2$ so $y=23/2.$ So for these lines the intersection point is $(9,\ 23/2,\ 4).$

It should be pointed out that two lines in space generally do not intersect, they can be parallel or "skew". This would come out as some contradictory values in the above mechanical procedure.

\begin{align} \color{#C00000}{((10,10,6)-(5,5,4))}\times\color{#00A000}{((10,10,3)-(5,5,5))} &=\color{#C00000}{(5,5,2)}\times\color{#00A000}{(5,5,-2)}\\ &=\color{#0000FF}{(-20,20,0)} \end{align} is perpendicular to both lines; therefore, $\color{#0000FF}{(-20,20,0)}\cdot u$ is constant along each line. If this constant is not the same for each line, the lines do not intersect. In this case, the constant for each line is $0$, so the lines intersect. $$\color{#C00000}{(5,5,2)}\times\color{#0000FF}{(-20,20,0)}=\color{#E06800}{(-40,-40,200)}$$ is perpendicular to the first line; therefore, $\color{#E06800}{(-40,-40,200)}\cdot u$ is constant along the first line. In this case, that constant is $400$. The general point along the second line is $$\color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t$$ To compute the point of intersection, find the $t$ so that $$\color{#E06800}{(-40,-40,200)}\cdot(\color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t)=400\\ 600-800t=400\\ t=1/4$$ Plugging $t=1/4$ into the formula for a point along the second line, yields the point of intersection: $$\color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}\cdot1/4=\left(\frac{25}{4},\frac{25}{4},\frac{9}{2}\right)$$

The second example \begin{align} \color{#C00000}{((12,15,4)-(6,8,4))}\times\color{#00A000}{((12,15,6)-(6,8,2))} &=\color{#C00000}{(6,7,0)}\times\color{#00A000}{(6,7,4)}\\ &=\color{#0000FF}{(28,-24,0)} \end{align} is perpendicular to both lines; therefore, $\color{#0000FF}{(28,-24,0)}\cdot u$ is constant along each line. If this constant is not the same for each line, the lines do not intersect. In this case, the constant for each line is $-24$, so the lines intersect. $$\color{#C00000}{(6,7,0)}\times\color{#0000FF}{(28,-24,0)}=\color{#E06800}{(0,0,-340)}$$ is perpendicular to the first line; therefore, $\color{#E06800}{(0,0,-340)}\cdot u$ is constant along the first line. In this case, that constant is $-1360$. The general point along the second line is $$\color{#00A000}{(6,8,2)}+\color{#00A000}{(6,7,4)}t$$ To compute the point of intersection, find the $t$ so that $$\color{#E06800}{(0,0,-340)}\cdot(\color{#00A000}{(6,8,2)}+\color{#00A000}{(6,7,4)}t)=-1360\\ -680-1360t=-1360\\ t=1/2$$ Plugging $t=1/2$ into the formula for a point along the second line, yields the point of intersection: $$\color{#00A000}{(6,8,2)}+\color{#00A000}{(6,7,4)}\cdot1/2=\left(9,\frac{23}{2},4\right)$$

• Very nice, the color helps get it into the head! Commented Dec 5, 2014 at 12:16
• How did you get the constant values? the 400, 0 and -24, -1360 in the examples? Commented Jan 8, 2018 at 7:47
• $\color{#C00}{0}$: since $(-20,20,0)\cdot u$ is constant along each of the first two lines, we just plug in a point from each line: $(-20,20,0)\cdot(5,5,4)=0$ and $(-20,20,0)\cdot(5,5,5)=0$. Since these two are the same, the lines intersect. $\color{#C00}{400}$: since $(-40,-40,200)\cdot u$ is constant along the first line, we again just plug in a point from that line: $(-40,-40,200)\cdot(5,5,4)=400$. The same goes for $\color{#C00}{-24}$ and $\color{#C00}{-1360}$ from the second example.
– robjohn
Commented Jan 8, 2018 at 12:22

The two lines can be represented as $(5,5,4)+s[(10,10,6)-(5,5,4)]$ and $(5,5,5)+t[(10,10,3)-(5,5,5)]$. To find their intersection means to find $s$ and $t$ such that they are equal.

• Thanks for the reply. When i tested this with some other points i am not getting correct values. for example if you take the lines (6,8,4)(12,15,4) and (6,8,2)(12,15,6). Please expline is there any other conserns are there to solve this problem, This will helps a lot for me. Thanks, Kumar. Commented Jan 5, 2013 at 11:12
• @kumar Apply the same method to your new example, you should solve $(6,8,4) + s(6,7,0) = (6,8,2) + t(6,7,4)$ and the answer is $s = t = 1/2$, i.e. the intersection point is $(9,\,23/2,\,4)$. I see no problems here. What do you mean by not getting correct values? Commented Jan 5, 2013 at 15:11