Three Dimensional Geometry

In a plane, two lines that are not parallel must meet. In space, two lines that are not parallel usually do not.

Take the $x$-axis and the line through $(0,0,1)$ parallel to the $y$-axis. Their directions are $\hat{i}$ and $\hat{j}$, so they are certainly not parallel. But every point of the first has $z=0$ and every point of the second has $z=1$, so they share nothing. They are skew, and the shortest distance between them is $1$.

This is the single structural difference between plane and space geometry, and almost every error in the chapter traces back to it. Two lines in space fall into three cases, not two: parallel, intersecting, or skew — and the third is the generic one.

The test is a scalar triple product. Lines $\mathbf{r}={\mathbf{a}}_1+\lambda {\mathbf{b}}_1$ and $\mathbf{r}={\mathbf{a}}_2+\mu {\mathbf{b}}_2$ are coplanar, and therefore either parallel or intersecting, exactly when

$\left[{\mathbf{a}}_2-{\mathbf{a}}_1,{\mathbf{b}}_1,{\mathbf{b}}_2\right]=0,$

and when it is not zero the shortest distance between them is

$d=\frac{|\left({\mathbf{a}}_2-{\mathbf{a}}_1\right)\cdot \left({\mathbf{b}}_1\times {\mathbf{b}}_2\right)|}{|{\mathbf{b}}_1\times {\mathbf{b}}_2|}.$

Setting two Cartesian forms equal and solving for a common point is therefore never a proof of intersection: the equations may simply have no solution, and only the triple product says so in advance.

1. Directions in space

A direction is described either by its direction cosines $l,m,n$, the cosines of the angles it makes with the three axes, or by any proportional triple of direction ratios $a,b,c$. Since the direction cosines are the components of a unit vector,

$l^2+m^2+n^2=1,l=\frac{a}{\sqrt{a^2+b^2+c^2}},$

with the other two similar. Direction ratios are not unique and direction cosines are unique up to an overall sign, which corresponds to traversing the line the other way.

Illustration 1

A line makes angles of $60^{\circ }$ and $45^{\circ }$ with the $x$- and $y$-axes. Find the angle it makes with the $z$-axis.

${\cos }^{2}60^{\circ }+{\cos }^{2}45^{\circ }+n^2=1\Rightarrow \frac{1}{4}+\frac{1}{2}+n^2=1,$

so $n^2=\frac{1}{4}$ and $n=\pm \frac{1}{2}$, giving $60^{\circ }$ or $120^{\circ }$. Both are genuine, because the line has two directions and the question does not choose between them.

2. Lines, and the three ways two of them can sit

A line through $\mathbf{a}$ with direction $\mathbf{b}$ is $\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}$, or in Cartesian form

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}.$

The angle between two lines is the angle between their directions, $\cos \theta =\frac{\left|{\mathbf{b}}_1\cdot {\mathbf{b}}_2\right|}{\left|{\mathbf{b}}_1\right|\left|{\mathbf{b}}_2\right|}$, with the modulus chosen so that the acute angle is reported.

For parallel lines the shortest distance is $\frac{\left|\left({\mathbf{a}}_2-{\mathbf{a}}_1\right)\times \mathbf{b}\right|}{\left|\mathbf{b}\right|}$, since the cross product of the directions vanishes and the earlier formula becomes meaningless.

Illustration 2

Show that $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+1}{3}$ are skew, and find the shortest distance between them.

Here ${\mathbf{a}}_2-{\mathbf{a}}_1=(1,2,-1)$, ${\mathbf{b}}_1=(2,3,4)$ and ${\mathbf{b}}_2=(1,2,3)$. Then

${\mathbf{b}}_1\times {\mathbf{b}}_2=\left(9-8,4-6,4-3\right)=(1,-2,1),$

and $\left({\mathbf{a}}_2-{\mathbf{a}}_1\right)\cdot (1,-2,1)=1-4-1=-4\ne 0$, so the lines are skew. The shortest distance is

$\frac{|-4|}{\sqrt{1+4+1}}=\frac{4}{\sqrt{6}}.$

Had the triple product vanished, the lines would have been coplanar and the same numerator would have given a distance of zero, correctly reporting an intersection.

Illustration 3

Find the equation of the plane containing the two lines of the previous illustration, if it exists.

It does not. A plane containing both would make them coplanar, and the non-zero triple product rules that out. Skew lines lie in no common plane, though each lies in a plane parallel to the other — and the distance between those two parallel planes is exactly the shortest distance $\frac{4}{\sqrt{6}}$ just computed.

3. Planes

A plane is fixed by a point on it and a normal direction: $\left(\mathbf{r}-\mathbf{a}\right)\cdot \mathbf{n}=0$, or $ax+by+cz=d$ where $(a,b,c)$ is the normal. Three other forms recur.

The normal form $lx+my+nz=p$ uses direction cosines and the perpendicular distance $p$ from the origin. The intercept form $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ reads off where the plane meets the axes. And a plane through three non-collinear points is found by taking the cross product of two edge vectors as the normal.

Comparing two planes is entirely a comparison of normals. They are parallel when the normals are proportional, identical when the constant terms are in that same proportion as well, and perpendicular when the normals have zero dot product. The corresponding test for three points in space is equally direct: they are collinear when the direction ratios of two of the joining segments are proportional, which is the three-dimensional version of equal slopes.

The angle between two planes is the angle between their normals, and the distance from a point to a plane is

$\frac{|a{x}_1+b{y}_1+c{z}_1-d|}{\sqrt{a^2+b^2+c^2}}.$

Illustration 4

Find the plane through $(1,1,0)$, $(1,2,1)$ and $(-2,2,-1)$.

Two edge vectors are $(0,1,1)$ and $(-3,1,-1)$, whose cross product is

$\left(1\cdot (-1)-1\cdot 1,1\cdot (-3)-0\cdot (-1),0\cdot 1-1\cdot (-3)\right)=(-2,-3,3).$

So the plane is $-2(x-1)-3(y-1)+3z=0$, that is $2x+3y-3z=5$. Substituting each of the three given points returns $5$, which is the check.

The same triple product that tested two lines also tests four points. Points $A,B,C,D$ are coplanar exactly when $\left[\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}\right]=0$, and when it is not zero its sixth part is the volume of the tetrahedron they span:

$V=\frac{1}{6}\left|\left[\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}\right]\right|.$

Illustration 5

Find the volume of the tetrahedron with vertices $(0,0,0)$, $(1,0,0)$, $(0,2,0)$ and $(0,0,3)$, and confirm that the four points are not coplanar.

The three edge vectors from the origin are the rows of a diagonal matrix with entries $1$, $2$ and $3$, so the triple product is $6$ and the volume is $1$.

Because the triple product is non-zero, no plane contains all four, which is the same statement. A zero volume and a coplanarity condition are two readings of one determinant, and questions frequently ask for one while supplying data suited to the other.

4. A line and a plane: sine, not cosine

The angle between a line and a plane is measured from the line to the plane, while the vectors available are the line's direction $\mathbf{b}$ and the plane's normal $\mathbf{n}$ — and those two are separated by the complement of that angle. Hence

$\sin \theta =\frac{|\mathbf{b}\cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}.$

Using a cosine here is the most frequent single error in the chapter, and it is invisible in the arithmetic: the answer is simply the complement of the right one.

Illustration 6

Find the angle between the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{6}$ and the plane $x+y+z=7$.

Here $\mathbf{b}=(2,3,6)$ with $\left|\mathbf{b}\right|=7$, and $\mathbf{n}=(1,1,1)$ with $\left|\mathbf{n}\right|=\sqrt{3}$. Then

$\sin \theta =\frac{|2+3+6|}{7\sqrt{3}}=\frac{11}{7\sqrt{3}},$

so $\theta ={\sin }^{-1}\frac{11}{7\sqrt{3}}\approx 65.4^{\circ }$. Reporting the cosine instead would give about $24.6^{\circ }$ — a plausible-looking answer that is exactly the complement.

Illustration 7

Show that $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ lies in the plane $2x-2y-z+5=0$.

Two conditions are needed, and both must be checked. First, the line's direction must be perpendicular to the normal: $(2,3,4)\cdot (2,-2,-1)=4-6-4=-6\ne 0$.

Since this already fails, the line is not parallel to the plane and certainly does not lie in it — it crosses it at a single point. Had the dot product vanished, the second condition would still be required: some point of the line must satisfy the plane's equation, since otherwise the line is parallel to the plane and misses it entirely.

5. Feet, images and distances

The foot of the perpendicular from $(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$ comes from the same relation as in two dimensions, with one more coordinate:

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-\frac{a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2},$

and doubling the right-hand side gives the image.

The distance from a point to a line has no such formula and is computed with a cross product: for the line $\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}$ and the point $P$,

$d=\frac{|\left(\mathbf{p}-\mathbf{a}\right)\times \mathbf{b}|}{|\mathbf{b}|},$

which is the area of a parallelogram divided by its base.

Two parallel planes, written with the same normal, are $ax+by+cz=d_1$ and $ax+by+cz=d_2$, and the distance between them is $\frac{\left|d_1-d_2\right|}{\sqrt{a^2+b^2+c^2}}$. The coefficients must be scaled to match before the difference is taken: for $2x-y+2z=5$ and $4x-2y+4z=3$ the second must first be halved to $2x-y+2z=\frac{3}{2}$, after which the distance is $\frac{\left|5-\frac{3}{2}\right|}{3}=\frac{7}{6}$.

Illustration 8

Find the image of $(1,3,4)$ in the plane $2x-y+z+3=0$.

Here $a{x}_1+b{y}_1+c{z}_1+d=2-3+4+3=6$ and $a^2+b^2+c^2=6$, so for the image the common ratio is $-2\cdot \frac{6}{6}=-2$. Then

$x=1+2(-2)=-3,y=3-1(-2)=5,z=4+1(-2)=2,$

giving $(-3,5,2)$. Checking, the midpoint $(-1,4,3)$ satisfies $-2-4+3+3=0$, as it must.

Illustration 9

Find the distance of $(2,3,-1)$ from the line $\frac{x-1}{1}=\frac{y+1}{2}=\frac{z}{2}$.

Take $\mathbf{a}=(1,-1,0)$ and $\mathbf{b}=(1,2,2)$, so $\mathbf{p}-\mathbf{a}=(1,4,-1)$. Then

$\left(\mathbf{p}-\mathbf{a}\right)\times \mathbf{b}=\left(4\cdot 2-(-1)\cdot 2,(-1)\cdot 1-1\cdot 2,1\cdot 2-4\cdot 1\right)=(10,-3,-2),$

whose magnitude is $\sqrt{100+9+4}=\sqrt{113}$. Dividing by $\left|\mathbf{b}\right|=3$ gives $\frac{\sqrt{113}}{3}$.

6. The line where two planes meet

Two non-parallel planes meet in a line, and that line is found without solving anything simultaneously. Its direction must be perpendicular to both normals, so it is ${\mathbf{n}}_1\times {\mathbf{n}}_2$; a point on it is obtained by setting one coordinate to a convenient value, usually zero, and solving the two remaining equations in two unknowns.

Illustration 10

Find the line of intersection of $x+y+z=1$ and $2x+3y+4z=5$.

The direction is

${\mathbf{n}}_1\times {\mathbf{n}}_2=(1,1,1)\times (2,3,4)=\left(4-3,2-4,3-2\right)=(1,-2,1).$

For a point, set $z=0$ and solve $x+y=1$ with $2x+3y=5$, giving $y=3$ and $x=-2$. So the line is

$\frac{x+2}{1}=\frac{y-3}{-2}=\frac{z}{1}.$

Substituting $(-2,3,0)$ into both planes returns $1$ and $5$, which confirms the point, and the direction is perpendicular to both normals by construction.

If the chosen coordinate happens to give an inconsistent pair, the line simply does not meet that coordinate plane, and a different coordinate should be fixed instead.

7. Families of planes

Every plane through the line of intersection of $P_1=0$ and $P_2=0$ has the form $P_1+\lambda P_2=0$, exactly as for lines in two dimensions. The line itself never has to be found.

Illustration 11

Find the plane through the line of intersection of $x+y+z=1$ and $2x+3y+4z=5$ that is perpendicular to $x-y+z=0$.

Write the family as $\left(x+y+z-1\right)+\lambda \left(2x+3y+4z-5\right)=0$, whose normal is $\left(1+2\lambda ,1+3\lambda ,1+4\lambda \right)$. Perpendicularity to the third plane means the normals are perpendicular:

$\left(1+2\lambda \right)-\left(1+3\lambda \right)+\left(1+4\lambda \right)=0\Rightarrow 1+3\lambda =0,$

so $\lambda =-\frac{1}{3}$. Substituting and clearing fractions gives $x-2z+2=0$, whose normal $(1,0,-2)$ is indeed perpendicular to $(1,-1,1)$.

Illustration 12

Find where the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ meets the plane $x-y+z=5$.

Write the general point of the line as $\left(2+3t,-1+4t,2+12t\right)$ and substitute:

$\left(2+3t\right)-\left(-1+4t\right)+\left(2+12t\right)=5\Rightarrow 5+11t=5,$

so $t=0$ and the point is $(2,-1,2)$. Parametrising the line and substituting is always the route; solving the Cartesian pair simultaneously with the plane is longer and no more reliable.

Illustration 13

Find the equation of the perpendicular from $(1,0,2)$ to the line $\frac{x+1}{3}=\frac{y-2}{-2}=\frac{z+1}{-1}$.

Let the foot be $\left(-1+3t,2-2t,-1-t\right)$. The vector from the given point to the foot is $\left(-2+3t,2-2t,-3-t\right)$, and it must be perpendicular to the direction $(3,-2,-1)$:

$3(-2+3t)-2(2-2t)-1(-3-t)=0\Rightarrow 14t-7=0,$

so $t=\frac{1}{2}$ and the foot is $\left(\frac{1}{2},1,-\frac{3}{2}\right)$. The perpendicular is the line joining $(1,0,2)$ to that foot, with direction $\left(-\frac{1}{2},1,-\frac{7}{2}\right)$, or $(1,-2,7)$ after scaling.

Illustration 14

Find the plane containing the line $\frac{x-1}{2}=\frac{y}{3}=\frac{z+1}{4}$ and the point $(2,1,3)$.

The plane contains the line's point $(1,0,-1)$ and its direction $(2,3,4)$, and also the vector from that point to $(2,1,3)$, namely $(1,1,4)$. Its normal is therefore

$(2,3,4)\times (1,1,4)=\left(12-4,4-8,2-3\right)=(8,-4,-1).$

So the plane is $8(x-1)-4y-(z+1)=0$, that is $8x-4y-z=9$. Substituting $(1,0,-1)$ gives $8+1=9$ and substituting $(2,1,3)$ gives $16-4-3=9$, confirming both.

A plane is always built the same way: assemble two independent directions lying in it, cross them for the normal, and use any known point. Whether the data arrives as three points, a line and a point, or two intersecting lines makes no difference to the method.

Summary

In space, two non-parallel lines need not meet. The scalar triple product $\left[{\mathbf{a}}_2-{\mathbf{a}}_1,{\mathbf{b}}_1,{\mathbf{b}}_2\right]$ decides: zero means coplanar and therefore parallel or intersecting, and non-zero means skew, with the same expression divided by $\left|{\mathbf{b}}_1\times {\mathbf{b}}_2\right|$ giving the shortest distance. Attempting to solve for a common point is not a test, because failure to solve can mean either skewness or an algebraic slip.

Directions are carried by direction cosines, whose squares sum to one, or by any proportional set of direction ratios. The angle between two lines is the angle between directions; the angle between two planes is the angle between normals; but the angle between a line and a plane needs a sine, because the available vectors are separated by its complement.

A line lies in a plane only if two conditions hold: its direction is perpendicular to the normal, and one of its points satisfies the plane. Checking only the first leaves open the case of a line parallel to the plane and missing it.

The foot of a perpendicular to a plane and the image of a point come from one relation, with the image using twice the ratio. The distance from a point to a line has no analogous formula and is computed as a cross product divided by the direction's length.

Every plane through the intersection of two planes is $P_1+\lambda P_2=0$, so that line never needs to be found, and every question about a line meeting a plane is answered by parametrising the line and substituting.