MCV4U Planes: Normals and Equations: Notes and Practice

A three-legged stool never wobbles, even on a bumpy floor, but a four-legged chair often does. The reason is geometry: any three points that are not in a line lie on exactly one flat surface, so three feet always rest on a plane, while a fourth foot might not touch it. Planes are everywhere in design: a solar panel on a roof, a ramp, the faces of a building. To describe one, you need either a point and the direction that sticks straight out of it (the normal), or a point and two directions that lie flat in it. Those two ideas give the two main kinds of plane equation.

Two weathered wooden stools, each standing on three legs, photographed against a plain grey background. — PhotoTwo wooden three-legged stools from the collection of the Museo Egizio in Turin. Any three feet that are not in a line touch exactly one plane, which is why a three-legged stool does not rock the way a four-legged chair can.Unknown author, CC0, via Wikimedia Commons

One linear equation in 3-space is a plane

The same equation can describe different shapes depending on the space it lives in. In 2-space, $x = 0$ is the $y$ -axis, a line. In 3-space, $x = 0$ allows any $y$ and any $z$ , so its solutions fill the whole $yz$ -plane. In general, the solutions $(x, y, z)$ of one linear equation $A x + B y + C z + D = 0$ form a plane. A variable that is missing from the equation is free, so the plane runs parallel to that axis.

Equation	Solutions in 2-space	Solutions in 3-space
$x = 0$	The $y$ -axis	The $yz$ -plane
$y = 0$	The $x$ -axis	The $x z$ -plane
$y = x$	The line through the origin with slope $1$	The vertical plane through the $z$ -axis and the line $y = x$
$z = 5$	(no $z$ in 2-space)	The horizontal plane $5$ units above the $x y$ -plane
$y - z = 3$	(no $z$ in 2-space)	A plane parallel to the $x$ -axis
$x + z = 1$	(no $z$ in 2-space)	A plane parallel to the $y$ -axis

A missing variable is free, so the plane is parallel to that variable's axis.

The normal vector and the scalar equation

Fig 1A plane with normal vector n = (A, B, C) perpendicular to it, a point P₀ in the plane and a vector P₀P lying in the plane. Because n · P₀P = 0 for every point P of the plane, the plane has equation Ax + By + Cz + D = 0.

A NORMAL to a plane is a non-zero vector perpendicular to it. Let $n = (A, B, C)$ be a normal and $P_{0} (x_{0}, y_{0}, z_{0})$ a point of the plane. A point $P (x, y, z)$ is in the plane exactly when $P_{0} P$ lies flat in the plane, that is, when $n \cdot P_{0} P = 0$ . Expanding gives $A (x - x_{0}) + B (y - y_{0}) + C (z - z_{0}) = 0,$ which tidies up to the SCALAR equation $A x + B y + C z + D = 0$ with $D = - (A x_{0} + B y_{0} + C z_{0})$ . So the coefficients of $x$ , $y$ and $z$ are a normal: one normal to $3 x + 5 y - 2 z = 6$ is $(3, 5, - 2)$ .

A x + B y + C z + D = 0, n = (A, B, C), D = - n \cdot O P_{0}

The direction of a normal never changes across the plane, and every non-zero scalar multiple of a normal is also a normal: $(3, 5, - 2)$ and $(- 6, - 10, 4)$ are normals to the same plane.
Three points that are not collinear determine exactly one plane (the stool). Two intersecting lines, or a line and a point off it, also determine one.
If two vectors lie in a plane, so does their sum, and so does any combination $s a + t b$ . That fact is what makes the vector equation work.
Two planes are parallel (or coincident) when their normals are parallel, and perpendicular when their normals are perpendicular: $n_{1} \cdot n_{2} = 0$ .

VideoMr. Cioffi's MCV4U lesson on the scalar (Cartesian) equation of a plane and how it comes from a normal vector and a point.Watch on YouTube

The vector and parametric equations

Fig 2A plane through P₀ spanned by two non-parallel direction vectors a and b. Every point of the plane is reached as r = r₀ + s a + t b.

r = r_{0} + s a + t b x = x_{0} + s a_{1} + t b_{1}, y = y_{0} + s a_{2} + t b_{2}, z = z_{0} + s a_{3} + t b_{3}

s and t are real numbers; a and b are non-parallel vectors that lie in the plane.

A plane has two free directions, so its vector equation needs two parameters. The direction vectors $a$ and $b$ must not be parallel; if they were, $s a + t b$ would only sweep out a line. To convert from vector to scalar form, find a normal with the cross product, $n = a \times b$ , which is perpendicular to both, then substitute the known point to get $D$ . To convert from scalar to vector form, find three non-collinear points (the intercepts are often easiest) and use two of the vectors between them as $a$ and $b$ , checking that each has a zero dot product with $n$ .

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Rotate a 3-D view of planes drawn from their equations, each with its normal vector (untick planes 2 and 3 to study one plane). Type new coefficients to see A, B and C tilt the plane and D shift it.

Checking a point against each form

With the scalar equation, substitute the point: it is in the plane when the left side equals

0

. With the vector equation, you must find parameters: solve two of the three component equations for

s

and

t

, then check that the third also holds. Converting to scalar form first is often quicker.

Planes: Normals and Equations

Key things to remember

One linear equation in 3-space is a plane

The normal vector and the scalar equation

The vector and parametric equations

Worked examples

Where marks get lost

Flashcards 8

Practice