Equation of a line

Vector form. Fix a point $A$ on the line with position vector $\vec a$, and let $\vec b$ be any vector parallel to the line (its direction vector, not necessarily a unit vector). Every point $\vec r$ on the line is reached by walking from $A$ some multiple of $\vec b$:

where: $\vec r$ is the position vector of a general point on the line, $\vec a$ is the position vector of one known point on the line, $\vec b$ is a direction vector of the line, and $\lambda$ is a scalar parameter — as it ranges over all real numbers, $\vec r$ traces out the whole line, one point at a time.

Cartesian (symmetric) form. Write $\vec a$ as the point $(x_1,y_1,z_1)$ and $\vec b$ as direction ratios $(a,b,c)$. Substituting $\vec r = x\hat i+y\hat j+z\hat k$ into the vector equation and comparing $\hat i,\hat j,\hat k$ components in turn gives $x-x_1=\lambda a$, $y-y_1=\lambda b$, $z-z_1=\lambda c$. Eliminating $\lambda$ between the three ties the coordinates together:

where: $(x,y,z)$ is a general point on the line, $(x_1,y_1,z_1)$ is the known point, and $(a,b,c)$ are direction ratios of the line — they need not be normalised into direction cosines; any scalar multiple of a valid $(a,b,c)$ is an equally valid final answer.

Two points instead of point-and-direction. If the line passes through $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$, the direction ratios are simply the difference of coordinates, $x_2-x_1,\ y_2-y_1,\ z_2-z_1$, so

where the second form uses $\vec a,\vec b$ as the position vectors of $P,Q$ themselves — the parenthesis $(\vec b - \vec a)$ is the direction vector $PQ$.

Axes and lines parallel to axes. The $x$-axis is the line through the origin with direction ratios $(1,0,0)$, written $\dfrac{x}{1}=\dfrac{y}{0}=\dfrac{z}{0}$ and understood as the constraint pair $y=0,\ z=0$ with $x$ free; the $y$- and $z$-axes follow the same pattern. A line through $(x_1,y_1,z_1)$ parallel to the $x$-axis is simply $y=y_1,\ z=z_1$ with $x$ free, and similarly for lines parallel to the $y$- or $z$-axis — these are the degenerate cases where one or two of the "denominators" in the symmetric form are zero, and the fraction notation is only shorthand, not literal division.

Find the vector and Cartesian equations of the line passing through the point $(1,2,-3)$ and parallel to the vector $2\hat i-\hat j+4\hat k$. Then find the coordinates of the point on the line at $\lambda=2$, and verify it satisfies the Cartesian form.

Step 1. Identify $\vec a$ and $\vec b$. The given point gives $\vec a = \hat i+2\hat j-3\hat k$, and the given direction gives $\vec b = 2\hat i-\hat j+4\hat k$.

Step 2. Write the vector equation directly from the point-and-direction template: $\vec r = (\hat i+2\hat j-3\hat k)+\lambda(2\hat i-\hat j+4\hat k)$, $\lambda\in\mathbb R$.

Step 3. Convert to Cartesian form using $(x_1,y_1,z_1)=(1,2,-3)$ and direction ratios $(2,-1,4)$:

Step 4. Substitute $\lambda=2$ into the vector form: $\vec r = (1+4)\hat i+(2-2)\hat j+(-3+8)\hat k = 5\hat i+0\hat j+5\hat k$, i.e. the point $(5,0,5)$.

Step 5. Check against the Cartesian form: $\dfrac{5-1}{2}=2$, $\dfrac{0-2}{-1}=2$, $\dfrac{5+3}{4}=2$ — all three fractions equal $2$, confirming $(5,0,5)$ lies on the line. The vector equation is $\vec r = (\hat i+2\hat j-3\hat k)+\lambda(2\hat i-\hat j+4\hat k)$, the Cartesian form is $\dfrac{x-1}{2}=\dfrac{y-2}{-1}=\dfrac{z+3}{4}$, and the point at $\lambda=2$ is $(5,0,5)$.

Competitive stretch

For the line through $A(3,-2,1)$ and $B(1,4,-3)$, the direction ratios are $1-3=-2,\ 4-(-2)=6,\ -3-1=-4$, giving the Cartesian equation $\dfrac{x-3}{-2}=\dfrac{y+2}{6}=\dfrac{z-1}{-4}$; taking $\lambda=2$ in $\vec r=(3\hat i-2\hat j+\hat k)+\lambda(-2\hat i+6\hat j-4\hat k)$ lands on $(-1,10,-7)$, and back-substitution confirms $\dfrac{-1-3}{-2}=\dfrac{10+2}{6}=\dfrac{-7-1}{-4}=2$.

• Writing the symmetric form with the numerator flipped — it is $x-x_1$, not $x_1-x$; a sign slip here silently reverses which way $\lambda$ increases along the line.
• Treating $\lambda$ (or $t$) as a fixed unknown to "solve for," rather than a free parameter labelling a whole family of points on the line.
• Writing an illegal division by zero in the symmetric form when a direction ratio is $0$ (line parallel to a coordinate plane), instead of stating that coordinate directly as a constant, e.g. $y=y_1$.
• When a line is given by two points, mixing the base point from one point with direction ratios computed the other way round — either point works as the base, but it must stay paired consistently with the direction ratios computed from it.

Ask the tutor: if your equation of a line isn't matching the answer key, share whether you were given a point-and-direction or two points, and the tutor will walk through the setup with you.