Write vector, parametric and symmetric equations of the line through $(2, -1, 4)$ with direction $[3, 0, -2]$.

$\vec{r} = [2, -1, 4] + t[3, 0, -2]$; $x = 2 + 3t$, $y = -1$, $z = 4 - 2t$; $\frac{x - 2}{3} = \frac{z - 4}{-2}$, $y = -1$

Find vector and symmetric equations of the line through $A(1, 3, -2)$ and $B(4, 1, 0)$.

$\vec{r} = [1, 3, -2] + t[3, -2, 2]$ and $\frac{x - 1}{3} = \frac{y - 3}{-2} = \frac{z + 2}{2}$

Is $P(7, -1, 2)$ on the line $\vec{r} = [1, 3, -2] + t[3, -2, 2]$? Is $Q(4, 1, 1)$?

$P$ is on the line (at $t = 2$); $Q$ is not

Write parametric equations for the line $\frac{x + 1}{2} = \frac{y - 4}{-3} = \frac{z}{5}$, and find where it crosses the $xz$-plane.

$x = -1 + 2t$, $y = 4 - 3t$, $z = 5t$; it crosses at $\left(\frac{5}{3}, 0, \frac{20}{3}\right)$

Find a vector equation of the line through $A(2, 0, -1)$ parallel to $L: \frac{x - 3}{4} = \frac{y + 1}{-2} = \frac{z - 5}{2}$. Is it the same line as $L$? Where does it cross the $xy$-plane?

$\vec{r} = [2, 0, -1] + t[2, -1, 1]$, a distinct line parallel to $L$; it crosses the $xy$-plane at $(4, -1, 0)$

Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Equations of Lines in 3D

In 2D, $y = mx + b$ describes a line. In 3D there is no single slope, so a line is described the way you would give directions: start here, then walk this way. That needs a point and a direction vector.

How It Works

Vector form. $\vec{r} = \vec{r_0} + t\vec{d}$ , where $\vec{r_0}$ is the position vector of a point on the line and $\vec{d}$ is a direction vector. Each value of the parameter $t$ gives one point of the line; $t = 0$ is the starting point.

Parametric form. Split the vector equation into components. For $\vec{r_0} = [x_0, y_0, z_0]$ and $\vec{d} = [a, b, c]$ : $x = x_0 + at$ , $y = y_0 + bt$ , $z = z_0 + ct$ .

Symmetric form. Solve each parametric equation for $t$ and set them equal: $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ . If a component of $\vec{d}$ is $0$ , that coordinate never changes, so write it separately, such as $y = -1$ , and keep the other two fractions.

Reading an equation. The numbers added to $t$ (or subtracted from $x$ , $y$ , $z$ in symmetric form) give a point; the multipliers of $t$ (or the denominators) give the direction. Watch the signs: $\frac{x + 1}{2}$ means $x_0 = -1$ .

Through two points. Subtract to get a direction vector, then use either point.

The same line, many equations. Any point on the line and any nonzero multiple of $\vec{d}$ work. That is why two equations can look different and still describe one line.

Is a point on the line? Find $t$ from one coordinate, then check that same $t$ gives the other two. Parallel lines have collinear direction vectors. They are the same line if one line's point satisfies the other's equation, and distinct otherwise.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Write vector, parametric and symmetric equations of the line through $(2, -1, 4)$ with direction $[3, 0, -2]$ .
Show solution
1. Vector: $\vec{r} = [2, -1, 4] + t[3, 0, -2]$ .
2. Parametric: $x = 2 + 3t$ , $y = -1$ , $z = 4 - 2t$ .
3. The $y$ component of the direction is $0$ , so $y$ is fixed and stays out of the fractions.
Answer: $\vec{r} = [2, -1, 4] + t[3, 0, -2]$ ; $x = 2 + 3t$ , $y = -1$ , $z = 4 - 2t$ ; $\frac{x - 2}{3} = \frac{z - 4}{-2}$ , $y = -1$
Easy
Find vector and symmetric equations of the line through $A(1, 3, -2)$ and $B(4, 1, 0)$ .
Show solution
1. Direction: $\vec{AB} = [3, -2, 2]$ .
2. Use $A$ as the point.
Answer: $\vec{r} = [1, 3, -2] + t[3, -2, 2]$ and $\frac{x - 1}{3} = \frac{y - 3}{-2} = \frac{z + 2}{2}$
Medium
Is $P(7, -1, 2)$ on the line $\vec{r} = [1, 3, -2] + t[3, -2, 2]$ ? Is $Q(4, 1, 1)$ ?
Show solution
1. For $P$ : $1 + 3t = 7$ gives $t = 2$ . Then $y = 3 - 4 = -1$ and $z = -2 + 4 = 2$ . Both match.
2. For $Q$ : $1 + 3t = 4$ gives $t = 1$ . Then $y = 1$ , which matches, but $z = 0 \ne 1$ .
Answer: $P$ is on the line (at $t = 2$ ); $Q$ is not
Medium
Write parametric equations for the line $\frac{x + 1}{2} = \frac{y - 4}{-3} = \frac{z}{5}$ , and find where it crosses the $xz$ -plane.
Show solution
1. Point $(-1, 4, 0)$ and direction $[2, -3, 5]$ : $x = -1 + 2t$ , $y = 4 - 3t$ , $z = 5t$ .
2. The $xz$ -plane is $y = 0$ : $4 - 3t = 0$ , so $t = \frac{4}{3}$ .
3. Then $x = -1 + \frac{8}{3} = \frac{5}{3}$ and $z = \frac{20}{3}$ .
Answer: $x = -1 + 2t$ , $y = 4 - 3t$ , $z = 5t$ ; it crosses at $\left(\frac{5}{3}, 0, \frac{20}{3}\right)$
Test level
Find a vector equation of the line through $A(2, 0, -1)$ parallel to $L: \frac{x - 3}{4} = \frac{y + 1}{-2} = \frac{z - 5}{2}$ . Is it the same line as $L$ ? Where does it cross the $xy$ -plane?
Show solution
1. $L$ has direction $[4, -2, 2]$ ; divide by $2$ to use $[2, -1, 1]$ . So $\vec{r} = [2, 0, -1] + t[2, -1, 1]$ .
2. Same line only if $A$ is on $L$ : $\frac{2 - 3}{4} = -\frac{1}{4}$ but $\frac{0 + 1}{-2} = -\frac{1}{2}$ . Not equal, so $A$ is not on $L$ .
3. The $xy$ -plane is $z = 0$ : $-1 + t = 0$ , so $t = 1$ , giving $(4, -1, 0)$ .
Answer: $\vec{r} = [2, 0, -1] + t[2, -1, 1]$ , a distinct line parallel to $L$ ; it crosses the $xy$ -plane at $(4, -1, 0)$

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Martian Lab teaches this in the Equations of Lines in 3D lesson of its MCV4U Lines and Planes unit. Watch it, then take the Lines and Planes practice tests to see if it stuck.

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MCV4U: Equations of Lines in 3D

How It Works

5 Worked Examples

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