1.4 Equations of Lines and Planes

Vector and Parametric Equations of Lines

Lines and planes are the fundamental linear objects in 3D space. To describe a line, you need two pieces of information: a point on the line and a direction it travels. To describe a plane, you need a point and a vector that's perpendicular to the surface. These representations show up constantly in multivariable calculus, so getting comfortable with them now pays off throughout the course.

Vector Equations of 3D Lines

The vector equation of a line is:

$\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$

Each piece has a specific role:

• $\mathbf{r}_0$ is the position vector of a known point on the line. It anchors the line in space.
• $\mathbf{v}$ is the direction vector. It tells you which way the line points.
• $t$ is a scalar parameter. As $t$ varies over all real numbers, $\mathbf{r}(t)$ traces out every point on the line.

Think of it this way: you start at $\mathbf{r}_0$, and the term $t\mathbf{v}$ slides you forward or backward along the direction $\mathbf{v}$. At $t = 0$ you're at the initial point; at $t = 1$ you've moved exactly one copy of $\mathbf{v}$ away from it.

Example: The line through $(5, 2, -1)$ with direction vector $\langle 3, -1, 4 \rangle$ has vector equation:

$\mathbf{r}(t) = \langle 5, 2, -1 \rangle + t\langle 3, -1, 4 \rangle$

Parametric Equations of 3D Lines

Parametric equations are just the vector equation split into its three components:

$x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct$

Here $(x_0, y_0, z_0)$ is the point on the line and $\langle a, b, c \rangle$ is the direction vector.

Example: For the line through $(2, -1, 3)$ with direction $\langle 1, 2, -3 \rangle$:

$x = 2 + t, \quad y = -1 + 2t, \quad z = 3 - 3t$

Plugging in any value of $t$ gives you a specific point on the line. For instance, $t = 2$ gives $(4, 3, -3)$.

Symmetric Equations of 3D Lines

You can eliminate the parameter $t$ from the parametric equations by solving each one for $t$ and setting them equal:

$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$

This is the symmetric form. It's compact and useful for quickly reading off the point and direction.

Example: $\frac{x - 2}{3} = \frac{y + 1}{-2} = \frac{z - 4}{5}$ describes a line through $(2, -1, 4)$ with direction $\langle 3, -2, 5 \rangle$.

One thing to watch: if any direction component is zero (say $a = 0$), you can't divide by it. Instead, you write the corresponding coordinate as a constant. For example, if the line passes through $(2, -1, 4)$ with direction $\langle 0, -2, 5 \rangle$, you'd write:

$x = 2, \quad \frac{y + 1}{-2} = \frac{z - 4}{5}$

Relationships Between Lines in 3D

Three-dimensional space allows for a relationship between lines that doesn't exist in 2D:

• Parallel lines have direction vectors that are scalar multiples of each other. For example, $\langle 2, 4, -6 \rangle$ and $\langle -1, -2, 3 \rangle$ are parallel because $\langle 2, 4, -6 \rangle = -2\langle -1, -2, 3 \rangle$.
• Intersecting lines have non-parallel direction vectors and share exactly one point.
• Skew lines have non-parallel direction vectors but never intersect. This only happens in 3D or higher. Skew lines lie in different parallel planes.

To determine which case you're in, first check whether the direction vectors are parallel. If they aren't, try to find an intersection point (see below). If no intersection exists, the lines are skew.

Finding Intersection Points of Lines

Given two lines in parametric form, here's how to find their intersection:

1. Write out the parametric equations for each line, using different parameter names (say $t$ for the first line and $s$ for the second).
2. Set the corresponding coordinates equal: $x_1(t) = x_2(s)$, $y_1(t) = y_2(s)$, $z_1(t) = z_2(s)$. This gives you a system of three equations in two unknowns.
3. Solve any two of the three equations for $t$ and $s$.
4. Check the third equation. Plug your values of $t$ and $s$ into the remaining equation. If it holds, the lines intersect. If not, they're skew.
5. Substitute the valid $t$ (or $s$) back into the corresponding line's equations to get the intersection point.

That check in step 4 is critical. In 2D, two non-parallel lines always intersect. In 3D, they might miss each other entirely.

Equations of Planes

Normal Form of Plane Equations

A plane is determined by a normal vector $\mathbf{n}$ (perpendicular to the plane) and a point $\mathbf{r}_0$ on the plane. Any other point $\mathbf{r}$ lies on the plane exactly when the vector from $\mathbf{r}_0$ to $\mathbf{r}$ is perpendicular to $\mathbf{n}$:

$\mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0$

The dot product equals zero precisely when two vectors are perpendicular, so this equation captures every point in the plane and nothing else.

Example: The plane through $(1, 2, -3)$ with normal vector $\langle 3, -2, 1 \rangle$ satisfies:

$\langle 3, -2, 1 \rangle \cdot \langle x - 1,\; y - 2,\; z + 3 \rangle = 0$

Expanding: $3(x - 1) - 2(y - 2) + 1(z + 3) = 0$, which simplifies to $3x - 2y + z + 4 = 0$.

General Form of Plane Equations

The general form is:

$ax + by + cz + d = 0$

The coefficients $(a, b, c)$ form the normal vector to the plane, and $d$ determines how far the plane is shifted from the origin. You can always expand the normal form to get the general form (as shown in the example above).

Reading off the normal vector is straightforward: for $2x - 3y + 4z - 5 = 0$, the normal vector is $\langle 2, -3, 4 \rangle$.

Two planes are parallel if their normal vectors are scalar multiples of each other. They're perpendicular if the dot product of their normal vectors is zero.

Distance from a Point to a Plane

The shortest distance from a point $(x_0, y_0, z_0)$ to the plane $ax + by + cz + d = 0$ is:

$D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$

The numerator plugs the point into the left side of the plane equation and takes the absolute value (distance is always non-negative). The denominator is the magnitude of the normal vector, which scales the result to give a true geometric distance.

Example: Find the distance from $(2, -1, 3)$ to the plane $x + 2y - 3z + 4 = 0$.

1. Plug the point into the plane equation: $|1(2) + 2(-1) + (-3)(3) + 4| = |2 - 2 - 9 + 4| = |-5| = 5$

2. Compute the magnitude of the normal vector: $\sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$

3. Divide: $D = \frac{5}{\sqrt{14}} = \frac{5\sqrt{14}}{14}$