Equation of a line in 3D

A straight line in three dimensions is fixed the moment you know two things about it: one point it passes through, and the direction it runs in. That single fact — a point plus a direction — is all the algebra in this article builds on. Starting from the vector equation r⃗=a⃗+λb⃗\vec r = \vec a + \lambda \vec b, we expand it component by component to get the familiar Cartesian symmetric form, learn to write it directly when two points are given instead of a point-and-direction, and practise checking whether two differently-written equations secretly describe the same line. This topic sits right after direction cosines and direction ratios in the CBSE Class 12 Three-Dimensional Geometry chapter, and every later result in the chapter — the angle between two lines, the shortest distance between two skew lines — is built directly on top of the equation you write here.

The core insight

A line through a point with position vector a⃗\vec a, parallel to a vector b⃗\vec b, is the set of all points r⃗=a⃗+λb⃗\vec r = \vec a + \lambda \vec b for real λ\lambda, which expands component-wise into the Cartesian symmetric form x−x1a=y−y1b=z−z1c\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}.

The core idea

A line is not a list of points you write down one by one — it is generated by a rule. Fix one point on the line and one direction along it, and every other point on the line is reachable by walking some multiple of that direction, starting from the fixed point.

Vector form. If a⃗\vec a is the position vector of a known point on the line and b⃗\vec b is any vector parallel to the line, then the position vector r⃗\vec r of a general point on the line satisfies

r⃗=a⃗+λb⃗,λ∈R\vec{r} = \vec{a} + \lambda \vec{b}, \qquad \lambda \in \mathbb{R}

where: r⃗=xı^+yȷ^+zk^\vec r = x\hat\imath+y\hat\jmath+z\hat k is the position vector of the running point, a⃗\vec a is the position vector of a fixed point on the line, b⃗\vec b is a direction vector of the line (its components are direction ratios — it need not be a unit vector), and λ\lambda is a free real parameter. As λ\lambda sweeps through every real number, r⃗\vec r sweeps out every point on the line; λ=0\lambda = 0 gives back the starting point a⃗\vec a.

Cartesian (symmetric) form. Writing a⃗=x1ı^+y1ȷ^+z1k^\vec a = x_1\hat\imath+y_1\hat\jmath+z_1\hat k and b⃗=aı^+bȷ^+ck^\vec b = a\hat\imath+b\hat\jmath+c\hat k, the vector equation splits into three scalar equations x=x1+λax=x_1+\lambda a, y=y1+λby=y_1+\lambda b, z=z1+λcz=z_1+\lambda c. Solving each for λ\lambda and setting the three expressions equal gives the symmetric form used throughout the chapter:

x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}

where: (x1,y1,z1)(x_1,y_1,z_1) is a known point on the line and a,b,ca,b,c are its direction ratios (any nonzero scalar multiple of them describes the same line). The three fractions are forced equal because all three equal the same λ\lambda.

Line through two points. Given two points P(x1,y1,z1)P(x_1,y_1,z_1) and Q(x2,y2,z2)Q(x_2,y_2,z_2) on the line, there is no direction vector handed to you — you build one as b⃗=PQ→\vec b = \overrightarrow{PQ}, i.e. direction ratios (x2−x1, y2−y1, z2−z1)(x_2-x_1,\ y_2-y_1,\ z_2-z_1), and anchor the equation at either point:

r⃗=a⃗+λ(b⃗−a⃗),x−x1x2−x1=y−y1y2−y1=z−z1z2−z1\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}), \qquad \frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}

This is exactly the same construction used earlier in the chapter to get direction cosines from two points — nothing new is being invented here. Two equations that look nothing alike on paper can still be the same line: the point used to anchor the equation is not unique (any point on the line will do), and neither are the direction ratios (any nonzero scalar multiple works). The standard check is to confirm the direction ratios of both equations are proportional, and that a point from one equation satisfies the other.

Figure
A point A with a direction arrow b and three marked points on the line at different values of the parameter lambda, showing the line generated by walking multiples of b from A
r = a + λb — walking along the line from A A (λ = 0) b (direction) λ = 1 λ = 2 λ = -1 Each step of size b along the line corresponds to one more unit of λ, in either direction.
Derivation ADeriving the Cartesian symmetric form from the vector equation

The goal is to show, starting only from the vector equation r⃗=a⃗+λb⃗\vec r = \vec a + \lambda \vec b and the definition of vector equality (two vectors are equal exactly when all their corresponding components are equal), that the same line is described by the symmetric Cartesian equations x−x1a=y−y1b=z−z1c\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}, so that a student can move between the two forms in either direction without memorising a separate rule.

Step 1. Write every vector in the equation in component form. Let r⃗=xı^+yȷ^+zk^\vec r = x\hat\imath+y\hat\jmath+z\hat k be the position vector of a general point (x,y,z)(x,y,z) on the line, let the known point be a⃗=x1ı^+y1ȷ^+z1k^\vec a = x_1\hat\imath+y_1\hat\jmath+z_1\hat k, and let the given direction vector be b⃗=aı^+bȷ^+ck^\vec b = a\hat\imath+b\hat\jmath+c\hat k, where a,b,ca,b,c are the direction ratios of the line.

Step 2. Substitute into the vector equation. r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b becomes xı^+yȷ^+zk^=(x1+λa)ı^+(y1+λb)ȷ^+(z1+λc)k^x\hat\imath+y\hat\jmath+z\hat k = (x_1+\lambda a)\hat\imath + (y_1+\lambda b)\hat\jmath + (z_1+\lambda c)\hat k.

Step 3. Equate corresponding components. Two vectors are equal only if their ı^\hat\imath, ȷ^\hat\jmath and k^\hat k parts match separately, so this single vector equation is really three scalar equations at once: x=x1+λax = x_1+\lambda a, y=y1+λby=y_1+\lambda b, and z=z1+λcz=z_1+\lambda c.

Step 4. Solve each scalar equation for λ\lambda, since it is the same parameter in all three. λ=x−x1a\lambda = \dfrac{x-x_1}{a}, λ=y−y1b\lambda=\dfrac{y-y_1}{b}, and λ=z−z1c\lambda=\dfrac{z-z_1}{c} (assuming a,b,ca,b,c are all nonzero).

Step 5. Set the three expressions equal, since all three equal the same λ\lambda for the same point. This gives x−x1a=y−y1b=z−z1c\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}, the Cartesian symmetric form — recovering exactly the equation quoted above, purely by unpacking components and eliminating the shared parameter λ\lambda.

Worked example

Problem. Find the vector and Cartesian equations of the line passing through the point (1,2,−4)(1, 2, -4) and parallel to the vector 2ı^+3ȷ^+6k^2\hat\imath + 3\hat\jmath + 6\hat k. Then check whether the point (5,5,2)(5, 5, 2) lies on this line.

Step 1. Identify a⃗\vec a and b⃗\vec b. The given point gives a⃗=ı^+2ȷ^−4k^\vec a = \hat\imath + 2\hat\jmath - 4\hat k, and the given direction gives b⃗=2ı^+3ȷ^+6k^\vec b = 2\hat\imath + 3\hat\jmath + 6\hat k.

Step 2. Write the vector equation directly by substitution:

r⃗=(ı^+2ȷ^−4k^)+λ(2ı^+3ȷ^+6k^)\vec r = (\hat\imath + 2\hat\jmath - 4\hat k) + \lambda(2\hat\imath + 3\hat\jmath + 6\hat k)

Step 3. Read off the point (x1,y1,z1)=(1,2,−4)(x_1,y_1,z_1)=(1,2,-4) and direction ratios a=2,b=3,c=6a=2,b=3,c=6, and substitute into the symmetric form derived above:

x−12=y−23=z+46\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}

Step 4. To check (5,5,2)(5,5,2), substitute x=5,y=5,z=2x=5,y=5,z=2 into each fraction: 5−12=2\frac{5-1}{2}=2, 5−23=1\frac{5-2}{3}=1, 2+46=1\frac{2+4}{6}=1. The first fraction gives 22 while the other two give 11 — the three values are not equal, so (5,5,2)(5,5,2) does not lie on the line (only a point giving the same λ\lambda in all three fractions would qualify).

Competitive stretch

For the two-point case, the line through (−2,4,−5)(-2,4,-5) and (1,2,3)(1,2,3) has direction ratios a=1−(−2)=3a=1-(-2)=3, b=2−4=−2b=2-4=-2, c=3−(−5)=8c=3-(-5)=8, giving x+23=y−4−2=z+58\frac{x+2}{3}=\frac{y-4}{-2}=\frac{z+5}{8}. A fast integrity check without re-deriving anything: substitute the other point, (1,2,3)(1,2,3), into this equation. 1+23=1\frac{1+2}{3}=1, 2−4−2=1\frac{2-4}{-2}=1, 3+58=1\frac{3+5}{8}=1 — all three equal 11, confirming the equation is correct, since a genuine point on the line must produce one common λ\lambda value across all three fractions.

Common traps

  • Sign errors when the anchor point has negative coordinates. x−(−2)x-(-2) must be simplified to x+2x+2, not left as x−−2x--2 or mis-simplified to x−2x-2.
  • Thinking b⃗\vec b must be a unit vector. Any nonzero vector parallel to the line works in r⃗=a⃗+λb⃗\vec r=\vec a+\lambda\vec b; there is no need to divide by its magnitude first.
  • Forgetting λ\lambda ranges over all reals, not just positive values. The line extends infinitely in both directions, and negative λ\lambda reaches points on the other side of a⃗\vec a.
  • Checking a point by eye instead of computing all three fractions. A point lies on the line only if all three fractions equal the same number — as the worked example shows, getting one fraction to match while the others don't means the point is off the line.

Ask the tutor: Want to practise converting between vector and Cartesian form, or work through another two-point example step by step? Ask the AI tutor.

Why can two Cartesian equations that look completely different still represent the same line?

Because neither the anchor point nor the direction ratios in the symmetric form are unique. You can use any point that lies on the line as (x1,y1,z1)(x_1,y_1,z_1), and any nonzero scalar multiple of a valid direction triple (a,b,c)(a,b,c) still points along the same line. So x−12=y−24=z+16\frac{x-1}{2}=\frac{y-2}{4}=\frac{z+1}{6} and an equation anchored at a different point with direction ratios (1,2,3)(1,2,3) can both be correct descriptions of one line. To confirm two equations match, check that their direction ratios are proportional and that a point from one equation satisfies the other.

Do I always need a direction vector to write a line's equation, or can two points be enough?

Two points are enough on their own — you do not need a direction vector handed to you separately. Given P(x1,y1,z1)P(x_1,y_1,z_1) and Q(x2,y2,z2)Q(x_2,y_2,z_2), compute the direction ratios as (x2−x1, y2−y1, z2−z1)(x_2-x_1,\ y_2-y_1,\ z_2-z_1) and then use either point as the anchor in the usual symmetric form. This is exactly the point-and-direction method in disguise, with the direction vector built from the two given points instead of stated outright.

What does it actually mean for a point to satisfy the Cartesian equation of a line?

It means substituting that point's coordinates into all three fractions gives the same numerical value, which is the λ\lambda that would reach that point starting from the anchor point along the direction vector. If even one of the three fractions comes out different from the other two, as in the worked example's check of (5,5,2)(5,5,2), the point is simply not on the line — there is no λ\lambda that lands on it in all three coordinates at once.

Before this

Where this leads