If a plane with normal vector $\mathbf{n}=\left( \begin{matrix} {{n}_{1}} \\ {{n}_{2}} \\ {{n}_{3}} \\ \end{matrix} \right)$ passes through the point $A$, with position vector $\mathbf{a}=\left( \begin{matrix} {{a}_{1}} \\ {{a}_{2}} \\ {{a}_{3}} \\ \end{matrix} \right)$. Vector equation (Parametric form) $\mathbf{r}=\mathbf{a}+\lambda {{\mathbf{m}}_{1}}+\mu {{\mathbf{m}}_{2}}$ where $\lambda ,\mathrm{ }\mu \in \mathbb{R}$, where ${{\mathbf{m}}_{1}}$ and ${{\mathbf{m}}_{2}}$ are vectors parallel to the plane. Note: ${{\mathbf{m}}_{1}}\times {{\mathbf{m}}_{2}}=\mathbf{n}$ Scalar Product Form: $\mathbf{r}\cdot \mathbf{n}=p\Rightarrow \mathbf{r}\cdot \left( \begin{matrix} {{n}_{1}} \\ {{n}_{2}} \\ {{n}_{3}} \\ \end{matrix} \right)=p$ Cartesian equation: ${{n}_{1}}x+{{n}_{2}}y+{{n}_{3}}z=p$

Equations and Coordinate Planes

A plane can be described by two independent in-plane directions, or by one nonzero normal vector.

Parametric plane

Normal form

Parametric and scalar-product forms

\[\mathbf r=\mathbf a+\lambda\mathbf p+\mu\mathbf q\quad(\mathbf p\not\parallel\mathbf q),\qquad (\mathbf r-\mathbf a)\cdot\mathbf n=0.\]

Equivalent plane equations

\[\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n=d,\qquad ax+by+cz=d,\quad \mathbf n=\begin{pmatrix}a\\b\\c\end{pmatrix}.\]

If $\widehat{\mathbf n}$ is a unit normal, $\mathbf r\cdot\widehat{\mathbf n}=d$ is signed: the distance of the plane from the origin is $|d|$.

Coordinate geometry in three dimensions

Object	Equation / direction
x-axis	$\mathbf r=\lambda\mathbf i$, so $y=z=0$
y-axis	$\mathbf r=\lambda\mathbf j$, so $x=z=0$
z-axis	$\mathbf r=\lambda\mathbf k$, so $x=y=0$
xy-plane	$z=0$, normal $\mathbf k$
xz-plane	$y=0$, normal $\mathbf j$
yz-plane	$x=0$, normal $\mathbf i$

The translated coordinate planes are $x=a_1$ (parallel to the $yz$-plane), $y=a_2$ (parallel to the $xz$-plane), and $z=a_3$ (parallel to the $xy$-plane).

Video Solution:

Video solution locked

A free account saves your progress; this video needs a course plan.

Create free account Sign in Get the H2 Math course · from S$39.98/mo

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs a course plan.

Create free account Sign in Get the H2 Math course · from S$39.98/mo H2 Math tuition

Create a free account to practise similar questions.

Create free account Sign in Get the H2 Math course · from S$39.98/mo H2 Math tuition

Need help? Join our JC Math tuition classes.

Learn more

Object	Equation / direction
x-axis	\(\mathbf r=\lambda\mathbf i\), so \(y=z=0\)
y-axis	\(\mathbf r=\lambda\mathbf j\), so \(x=z=0\)
z-axis	\(\mathbf r=\lambda\mathbf k\), so \(x=y=0\)
xy-plane	\(z=0\), normal \(\mathbf k\)
xz-plane	\(y=0\), normal \(\mathbf j\)
yz-plane	\(x=0\), normal \(\mathbf i\)