Two Beams That Never Meet Three Dimensional Geometry

Class 11 studied coordinate geometry using nothing but Cartesian coordinates and separate x, y, z bookkeeping; the previous chapter then built up a whole toolkit of vectors - addition, dot products, cross products - without yet pointing it at lines and distances. This chapter aims both tools at the same target: a line drifting through three-dimensional space. A directed line, one with an arrow on it, makes three angles with the positive x, y and z axes in turn, called its direction angles and usually written α, β, γ. The cosines of these three angles, written l, m, n, are called the direction cosines of the line, and they always satisfy l^2 + m^2 + n^2 = 1, no matter which way the line points. Flip the arrow around to point the opposite way, and each direction angle is replaced by its supplement (180° minus the original), which flips the sign of every cosine - the same line, described in reverse, has direction cosines -l, -m, -n instead. Everything that follows - equations of lines, angles between pairs of lines, and the shortest distance between them - is really just vector algebra, applied directly to geometry in space.

Direction cosines pin down a line's exact tilt in space, independent of where the line happens to sit. For a line making angles 90°, 60° and 30° with the positive x, y and z axes respectively, the direction cosines are simply l = cos90° = 0, m = cos60° = 1/2, and n = cos30° = √3/2 - and checking the identity, 0^2 + (1/2)^2 + (√3/2)^2 = 0 + 1/4 + 3/4 = 1 exactly, as it must. Because a line extends in two opposite directions, it technically owns two sets of direction cosines, one the negative of the other; to make the description unique, a line is treated as a directed line with one chosen positive sense. When a line does not pass through the origin, its direction cosines are still found the same way - imagine sliding a parallel copy of the line through the origin, since parallel lines anywhere in space always share identical direction cosines regardless of where they actually sit.

Direction ratios are any three numbers proportional to a line's direction cosines - a rougher, unscaled description that is often far more convenient to write down. If a, b, c are direction ratios and l, m, n are the true direction cosines, then a = kl, b = km, c = kn for some constant k, and it turns out k = ±1/√(a^2+b^2+c^2), so l = ±a/√(a^2+b^2+c^2), and similarly for m and n. Take direction ratios 2, -1, -2: the magnitude √(2^2+(-1)^2+(-2)^2) = √9 = 3, so the direction cosines are 2/3, -1/3, -2/3, and checking, (2/3)^2 + (-1/3)^2 + (-2/3)^2 = 4/9+1/9+4/9 = 1 exactly. Unlike direction cosines, which are unique up to an overall sign, a line has infinitely many valid sets of direction ratios: (2,-1,-2), (4,-2,-4) and (-6,3,6) all describe the exact same direction, since any nonzero multiple ka, kb, kc works equally well.

Given two points P(x1,y1,z1) and Q(x2,y2,z2), the direction ratios of the line joining them are just the coordinate differences: x2-x1, y2-y1, z2-z1. Dividing each by the distance PQ = √[(x2-x1)^2+(y2-y1)^2+(z2-z1)^2] turns these ratios into genuine direction cosines. For P(-2,4,-5) and Q(1,2,3): the differences are 3, -2, 8, and PQ = √(3^2+(-2)^2+8^2) = √(9+4+64) = √77, giving direction cosines 3/√77, -2/√77, 8/√77. This same idea gives a clean test for three points lying on one straight line: compute the direction ratios of AB and of BC, and if they turn out proportional to each other, then AB and BC point along the same (or exactly opposite) direction, and since B is common to both, all three points must be collinear. Checking A(2,3,-4), B(1,-2,3), C(3,8,-11): direction ratios of AB are -1,-5,7, and of BC are 2,10,-14, which is exactly -2 times (-1,-5,7) - proportional, confirming A, B, C lie on one line.

A line in space is completely fixed either by one point on it plus its direction, or by any two points on it. In the first case, if A is a point with position vector a, and the line runs parallel to a direction vector b, then any point P on the line, with position vector r, satisfies r = a + λb for some real number λ - as λ sweeps through every real number, P sweeps through every point of the line. If instead the line is given by two points A and B with position vectors a and b, the direction vector is simply b - a, and the equation becomes r = a + λ(b-a). Take the point (5,2,-4) with direction vector 3i + 2j - 8k: the vector equation is r = (5i+2j-4k) + λ(3i+2j-8k), a single compact statement standing in for three separate coordinate equations at once.

Writing out r = xi+yj+zk, a = x1i+y1j+z1k and b = ai+bj+ck, then matching up the i, j and k parts of r = a+λb gives three parametric equations at once: x = x1+λa, y = y1+λb, z = z1+λc. Solving each for λ and setting them equal eliminates the parameter entirely, producing the Cartesian equation of the line: (x-x1)/a = (y-y1)/b = (z-z1)/c. Continuing the earlier example, the point (5,2,-4) with direction ratios 3,2,-8 gives the Cartesian equation (x-5)/3 = (y-2)/2 = (z+4)/-8. This single chain of equal fractions is really three equations in disguise, and it says something very concrete: as you move along the line, the x, y and z coordinates all change in the same fixed proportion - 3 parts of x-change for every 2 parts of y-change and -8 parts of z-change.

Given the direction ratios of two lines, a1,b1,c1 and a2,b2,c2, the angle θ between them (always taken as the acute angle) satisfies cosθ = |a1a2+b1b2+c1c2| / [√(a1^2+b1^2+c1^2) × √(a2^2+b2^2+c2^2)] - the absolute value bars guarantee the answer always comes out between 0° and 90°, even though two crossing lines technically form a pair of supplementary angles. Two lines are perpendicular exactly when a1a2+b1b2+c1c2 = 0, and parallel exactly when a1/a2 = b1/b2 = c1/c2. For lines with direction ratios 3,5,4 and 1,1,2: the sum is 3(1)+5(1)+4(2) = 3+5+8 = 16, while √(9+25+16) = √50 = 5√2 and √(1+1+4) = √6, so cosθ = 16/(5√2×√6) = 16/(5√12) = 8√3/15, giving θ = cos^-1(8√3/15). The same formula, written with vectors b1 and b2 instead of raw direction ratios, is simply cosθ = |b1·b2| / (|b1||b2|) - the dot-product-over-magnitudes idea from the vectors chapter, now doing geometric work.

In a flat two-dimensional world, any two distinct lines either cross at exactly one point or run parallel forever, never meeting. Three-dimensional space allows for a genuinely third option: two lines that are neither parallel nor ever intersecting, called skew lines. Picture a rectangular room: the diagonal line running across the ceiling and the diagonal line running down one wall are skew - they point in different directions, so they are not parallel, yet they never actually cross, because they sit at different heights and positions within the room. Skew lines are, by definition, not coplanar: there is no single flat plane containing both of them at once. When two lines do intersect, the shortest distance between them is trivially zero; when two lines are parallel, the shortest distance is just the ordinary perpendicular distance from any point on one line to the other. Skew lines need a genuinely new idea: the one line segment that is perpendicular to both lines simultaneously, called the common perpendicular, whose length is defined as the shortest distance between the two skew lines.

For skew lines r = a1+λb1 and r = a2+μb2, the common perpendicular's direction is found from the cross product b1×b2, since a vector perpendicular to both b1 and b2 is exactly what a cross product produces. The shortest distance is then d = |(b1×b2)·(a2-a1)| / |b1×b2| - the dot product projects the vector joining any point on each line onto the unit vector along that common perpendicular. Take lines through a1 = i+j with direction b1 = 2i-j+k, and through a2 = 2i+j-k with direction b2 = 3i-5j+2k. First, b1×b2 = 3i-j-7k, so |b1×b2| = √(9+1+49) = √59. Next, a2-a1 = i-k, and the dot product (b1×b2)·(a2-a1) = 3(1)+(-1)(0)+(-7)(-1) = 3+0+7 = 10. So the shortest distance is d = 10/√59 - a single positive number measuring exactly how close these two beams ever get, even though extended forever in both directions they will never actually touch.

When two lines share the exact same direction vector b - meaning they are parallel - they are automatically coplanar, and the shortest distance between them collapses to the perpendicular distance from any point on one line to the other. Writing the lines as r = a1+λb and r = a2+μb, this distance works out to d = |b×(a2-a1)| / |b|, using a single cross product rather than the cross-then-dot recipe needed for skew lines. Take a1 = i+2j-4k, a2 = 3i+3j-5k, both lines sharing direction b = 2i+3j+6k: then a2-a1 = 2i+j-k, and b×(a2-a1) = -9i+14j-4k, with magnitude √(81+196+16) = √293. Since |b| = √(4+9+36) = √49 = 7, the distance between the two parallel lines is d = √293/7. Notice the pattern across the whole chapter: every distance question, whether between two skew lines or two parallel ones, reduces to building the right perpendicular vector with a cross product, then measuring its length.