Three-Dimensional Geometry

Every line in three-dimensional space has a definite direction, and we describe this direction using two closely related ideas: direction cosines and direction ratios.

If a line makes angles α, β, γ with the positive directions of the x-axis, y-axis and z-axis respectively, then cos α, cos β, cos γ are called the direction cosines (DCs) of the line, usually written l, m, n. A line has two opposite directions, so it has two sets of direction cosines that differ only in sign. For any line, the direction cosines always satisfy the fundamental identity l² + m² + n² = 1.

Numbers a, b, c that are proportional to l, m, n (that is, a = kl, b = km, c = kn for some non-zero constant k) are called the direction ratios (DRs) of the line. Unlike direction cosines, direction ratios are not unique — any non-zero multiple of a valid set of DRs is still a valid set of DRs for the same line. Given DRs a, b, c, the direction cosines are recovered by dividing each by the magnitude √(a²+b²+c²).

If a line passes through two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), its direction ratios are simply x₂-x₁, y₂-y₁, z₂-z₁, and its direction cosines are obtained by dividing each of these by the distance PQ.

Worked Example 1. Find the direction cosines of the line joining P(1, 2, 3) and Q(4, 5, 6).

Direction ratios: (4-1, 5-2, 6-3) = (3, 3, 3). Magnitude = √(3²+3²+3²) = √27 = 3√3. So the direction cosines are (3/3√3, 3/3√3, 3/3√3) = (1/√3, 1/√3, 1/√3). Check: (1/√3)² + (1/√3)² + (1/√3)² = 1/3 + 1/3 + 1/3 = 1.

Worked Example 2. A line makes angles 90°, 135°, 45° with the x, y and z axes respectively. Find its direction cosines.

l = cos 90° = 0, m = cos 135° = -1/√2, n = cos 45° = 1/√2. Check: 0² + (-1/√2)² + (1/√2)² = 0 + 1/2 + 1/2 = 1.

To pin down a line completely we need a fixed point on it together with its direction. This idea leads to two standard forms for the equation of a line in space: through a given point parallel to a given vector, and through two given points.

Line through a point, parallel to a given vector. Let the line pass through the point A with position vector a, and be parallel to a vector b. Then any point R on the line has position vector r = a + λb, where λ is a real parameter (the vector equation). If A is (x₁,y₁,z₁) and b has direction ratios a, b, c, the same line in Cartesian form is (x-x₁)/a = (y-y₁)/b = (z-z₁)/c.

Line through two points. If the line passes through two points A and B with position vectors a and b, its direction vector is b - a, so the vector equation is r = a + λ(b - a). In Cartesian form, with A(x₁,y₁,z₁) and B(x₂,y₂,z₂), the equation is (x-x₁)/(x₂-x₁) = (y-y₁)/(y₂-y₁) = (z-z₁)/(z₂-z₁).

Worked Example 1. Find the vector and Cartesian equations of the line through (5, 2, -4) and parallel to the vector 3i + 2j - 8k.

Vector form: r = (5i + 2j - 4k) + λ(3i + 2j - 8k). Cartesian form: (x-5)/3 = (y-2)/2 = (z+4)/-8.

Worked Example 2. Find the Cartesian equation of the line through (-2, 4, -5) and (1, 2, 3).

Direction ratios: (1-(-2), 2-4, 3-(-5)) = (3, -2, 8). Equation: (x+2)/3 = (y-4)/-2 = (z+5)/8.

When two lines are given in space (either intersecting or skew — non-intersecting and non-parallel), two natural questions arise: what angle do they make, and if they don't meet, how far apart are they?

Angle between two lines. If two lines have direction vectors b₁ and b₂, the acute angle θ between them satisfies cos θ = |b₁·b₂| / (|b₁||b₂|). In terms of direction ratios (a₁,b₁,c₁) and (a₂,b₂,c₂), cos θ = |a₁a₂+b₁b₂+c₁c₂| / (√(a₁²+b₁²+c₁²)·√(a₂²+b₂²+c₂²)). Taking the absolute value ensures we always get the acute angle. The lines are perpendicular when a₁a₂+b₁b₂+c₁c₂=0, and parallel when a₁/a₂=b₁/b₂=c₁/c₂.

Worked Example 1. Find the angle between two lines whose direction ratios are (2, 2, 1) and (4, 1, 8).

cos θ = |2(4)+2(1)+1(8)| / (√(4+4+1)·√(16+1+64)) = |8+2+8| / (3·√81) = 18/(3·9) = 18/27 = 2/3. So θ = cos^-1(2/3).

Shortest distance between two lines. Two lines in space that are neither parallel nor intersecting are called skew lines, and the shortest distance between them is measured along their common perpendicular. For lines r = a₁+λb₁ and r = a₂+μb₂, this shortest distance is d = |(a₂-a₁)·(b₁×b₂)| / |b₁×b₂|. If the lines intersect, this distance is zero. For two parallel lines r = a₁+λb and r = a₂+μb (same direction vector b), the distance between them is d = |b×(a₂-a₁)| / |b|.

Worked Example 2. Find the shortest distance between the lines r = (i+j) + λ(2i-j+k) and r = (2i+j-k) + μ(3i-5j+2k).

Here a₁=(1,1,0), b₁=(2,-1,1), a₂=(2,1,-1), b₂=(3,-5,2), so a₂-a₁=(1,0,-1).

b₁×b₂ = ((-1)(2)-(1)(-5), (1)(3)-(2)(2), (2)(-5)-(-1)(3)) = (3, -1, -7). So |b₁×b₂| = √(9+1+49) = √59.

(a₂-a₁)·(b₁×b₂) = (1)(3)+(0)(-1)+(-1)(-7) = 3+0+7 = 10. Hence the shortest distance d = 10/√59 units.

A plane is a flat two-dimensional surface extending infinitely in space, and just like a line, it can be described in several equivalent ways depending on what information is given: a point and a normal direction, three points on the plane, its intercepts on the axes, or as a combination of two other planes.

Normal form. If n̂ is the unit vector normal (perpendicular) to a plane and d is the plane's perpendicular distance from the origin, every point on the plane has position vector r satisfying r·n̂ = d. In Cartesian form, if l, m, n are the direction cosines of the normal, the equation is lx + my + nz = d.

Plane through a point, perpendicular to a given vector. If the plane passes through a point A (position vector a) and is perpendicular to a vector n, then for any point R on the plane, the vector AR is perpendicular to n, giving (r - a)·n = 0. If n has direction ratios a, b, c and A is (x₁,y₁,z₁), the Cartesian form is a(x-x₁) + b(y-y₁) + c(z-z₁) = 0.

Worked Example 1. Find the equation of the plane through (1, 0, -2) with normal direction ratios (1, -2, 1).

1(x-1) - 2(y-0) + 1(z+2) = 0 ⟹ x - 1 - 2y + z + 2 = 0 ⟹ x - 2y + z + 1 = 0.

Plane through three non-collinear points. If A(x₁,y₁,z₁), B(x₂,y₂,z₂), C(x₃,y₃,z₃) lie on the plane, the vectors AB and AC both lie in the plane, so their cross product AB × AC gives a normal to it.

Worked Example 2. Find the equation of the plane through (1,1,0), (1,2,1), (-2,2,-1).

Take A(1,1,0). Then AB = (0,1,1) and AC = (-3,1,-1). Normal = AB × AC = (1·(-1)-1·1, 1·(-3)-0·(-1), 0·1-1·(-3)) = (-2, -3, 3). The plane through A with this normal is -2(x-1) - 3(y-1) + 3(z-0) = 0, which simplifies to 2x + 3y - 3z = 5.

Intercept form. If a plane cuts the x, y, z axes at distances a, b, c from the origin, its equation is x/a + y/b + z/c = 1.

Worked Example 3. Express 2x - 3y + 4z = 12 in intercept form.

Dividing throughout by 12: x/6 - y/4 + z/3 = 1, so the intercepts are a=6, b=-4, c=3.

Plane through the intersection of two planes. Any plane passing through the line of intersection of a₁x+b₁y+c₁z=d₁ and a₂x+b₂y+c₂z=d₂ can be written as (a₁x+b₁y+c₁z-d₁) + λ(a₂x+b₂y+c₂z-d₂) = 0, with λ fixed using an extra condition such as another point the plane must pass through.

Worked Example 4. Find the plane through the intersection of x+y+z=1 and 2x+3y+4z=5, also passing through (1,1,1).

Family: (x+y+z-1) + λ(2x+3y+4z-5) = 0. Substituting (1,1,1): (1+1+1-1) + λ(2+3+4-5) = 2 + 4λ = 0, so λ = -1/2. Multiplying by 2 and simplifying: 2(x+y+z-1) - (2x+3y+4z-5) = 0 ⟹ -y-2z+3 = 0, i.e. y + 2z = 3.

Planes can be compared with each other, and lines can be compared against planes, using the same dot-product idea used for lines: the angle between two directions is governed by the angle between their normal or direction vectors.

Angle between two planes. The angle between two planes equals the angle between their normal vectors n₁ and n₂: cos θ = |n₁·n₂| / (|n₁||n₂|). In terms of the Cartesian coefficients, for planes a₁x+b₁y+c₁z=d₁ and a₂x+b₂y+c₂z=d₂, cos θ = |a₁a₂+b₁b₂+c₁c₂| / (√(a₁²+b₁²+c₁²)·√(a₂²+b₂²+c₂²)). Two planes are perpendicular when a₁a₂+b₁b₂+c₁c₂=0, and parallel when their normals are proportional.

Worked Example 1. Find the angle between the planes 2x+y-2z=5 and 3x-6y-2z=7.

n₁=(2,1,-2), n₂=(3,-6,-2). n₁·n₂ = 6-6+4 = 4. |n₁| = √(4+1+4) = 3. |n₂| = √(9+36+4) = 7. cos θ = 4/(3·7) = 4/21, so θ = cos^-1(4/21).

Angle between a line and a plane. If a line has direction vector b and a plane has normal n, the angle φ between the line and the plane (not the line and the normal) satisfies sin φ = |b·n| / (|b||n|). Sine, not cosine, appears here because φ is measured from the plane itself, which is perpendicular to n.

Worked Example 2. Find the angle between the line (x+1)/2 = y/3 = (z-3)/6 and the plane 10x+2y-11z=3.

Line direction b=(2,3,6); plane normal n=(10,2,-11). b·n = 20+6-66 = -40, so |b·n|=40. |b| = √(4+9+36)=7. |n| = √(100+4+121) = √225 = 15. sin φ = 40/(7·15) = 40/105 = 8/21. So φ = sin^-1(8/21).

Coplanarity of two lines. Two lines r = a₁+λb₁ and r = a₂+μb₂ are coplanar (they either intersect or are parallel, and lie in one plane) exactly when the scalar triple product (a₂-a₁)·(b₁×b₂) = 0 — the same expression that appears in the numerator of the shortest-distance formula. If it is not zero, the lines are skew.

Worked Example 3. Show that the lines (x-5)/4 = (y-7)/4 = (z+3)/-5 and (x-8)/7 = (y-4)/1 = (z-5)/3 intersect.

Points: (5,7,-3) and (8,4,5); direction ratios: (4,4,-5) and (7,1,3). The connecting vector is (8-5, 4-7, 5-(-3)) = (3,-3,8). The coplanarity determinant is:

| 3 -3 8 |
| 4 4 -5 |
| 7 1 3 |

Expanding: 3(4·3-(-5)·1) - (-3)(4·3-(-5)·7) + 8(4·1-4·7) = 3(12+5) + 3(12+35) + 8(4-28) = 3(17) + 3(47) + 8(-24) = 51 + 141 - 192 = 0. Since the determinant is zero, the lines are coplanar — in fact they intersect (at the point (1,3,2)).

Once a plane's equation is known, a very common practical question is: how far is a given point from the plane? This closes the loop on the chapter's core toolkit and also lets us compare two parallel planes.

Distance of a point from a plane. For a plane written in normal form r·n = d (n not necessarily a unit vector) and a point with position vector a, the perpendicular distance from the point to the plane is D = |a·n - d| / |n|. In Cartesian form, for the plane Ax+By+Cz+D=0 and point (x₁,y₁,z₁), the distance is D = |Ax₁+By₁+Cz₁+D| / √(A²+B²+C²).

Worked Example 1. Find the distance of the point (2, 5, -3) from the plane r·(6i-3j+2k) = 4.

Here a=(2,5,-3), n=(6,-3,2), d=4. a·n = 6(2) + (-3)(5) + 2(-3) = 12 - 15 - 6 = -9. So D = |-9-4| / √(36+9+4) = |-13|/√49 = 13/7 units.

Distance between two parallel planes. Since parallel planes share the same normal direction, if they are written as r·n = d₁ and r·n = d₂, the distance between them is D = |d₁-d₂| / |n|.

Worked Example 2. Find the distance between the parallel planes 2x-2y+z+3=0 and 4x-4y+2z+5=0.

Rewrite the second plane with the same normal as the first by dividing by 2: 2x-2y+z+5/2=0. Both planes now have normal (2,-2,1) with |n| = √(4+4+1)=3, and constants -3 and -5/2 in the r·n=d form. D = |(-3)-(-5/2)|/3 = |-1/2|/3 = 1/6 units.