Equations of planes (HL)

A plane, $\Pi$, is a flat surface, like an infinitely large square. It is the 2-D equivalent of a line. Two dimensional also means it requires two direction vectors to specify a plane.

Contents

• Using two directional vectors
• Normal vector
• Vector equation of a plane
• Scalar (Cartesian) equation of a plane
• Wait, these are planes?

Using two directional vectors

where $\bm{d_1}$ and $\bm{d_2}$ are direction (displacement) vectors on the plane and are not scalar multiples of each other, and $(x_0, y_0, z_0)$ is a point on the plane.

This roughly means that any displacement vector on the plane is a linear combination of $\bm{d_1}$ and $\bm{d_2}$, meaning there exists a unique $(\lambda, \mu)$.

Normal vector

A normal vector $\bm n$, or $(a, b, c)$, is one that is perpendicular to all displacement vectors on the plane. In particular, it can be found using the cross product of two direction vectors of the plane, meaning

where $k$ means the normal vector is any non-zero scalar multiple of the cross product of the two direction vectors.

Each displacement vector with respect to some point on the plane $(x_0, y_0, z_0)$ can be represented by $(x - x_0,\,y - y_0,\, z - z_0)$

Perpendicular vectors dot to $0$, so

Vector equation of a plane

Rearranging the equation, we get

Scalar (Cartesian) equation of a plane

Simplifying the dot product on the right hand side, we get

To calculate distance between a plane and a point, it is useful to keep in mind $d$ is a dot product.

Note that unlike other forms, the Cartesian equation of a plane does not provide a point.

To find a point $P(x_0, y_0, z_0)$ on the plane, guess the component whose corresponding $a, b, c$ is not zero yet closest to zero. This ensures smallest denominator of fractions we have to work with. Then solve a system of two equations.

Because of the likelihood of working with fractions, you are very likely to show that a point is on the plane, or otherwise be given a point.

Wait, these are planes?

The $xy$-plane is represented by $z = 0$. Similarly $xz$-plane is $y = 0$, and $yz$-plane is $x = 0$.

So in 2-D, $x = 0$ is a line, but in 3-D it is a plane.

Normally, the plane $ax + by + cz = d$ passes through the points ${\left(\frac da, 0, 0\right)}$, ${\left(0, \frac db, 0, \right)}$, and ${\left(0, 0, \frac dc\right)}$ on the respective axes. However for the plane $y = 7$, both $a$ and $c$ are $0$, meaning $y = 7$ intersects neither the $x$-axis nor the $z$-axis. Hence $y = 7$ is parallel to the $xz$-plane.