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Wize University Linear Algebra Textbook > Equations of Lines and Planes

Hyperplanes

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Hyperplanes

Hyperplanes are objects of dimension n−1n-1 in Rn\reals^n that are able to split the space into two half-spaces.
  • Lines (1D objects) are hyperplanes in R2\reals^2 Example: x=1x=1

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  • Planes (2D objects) are hyperplanes in R3\reals^3 Example: x=1x=1


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Point-Normal Form

A general point on the hyperplane x⃗=⟨x1,x2,…,xn⟩\vec x = \lang x_1, x_2, \dots, x_n \rang appears along with a known point on the hyperplane, PP, and a normal vector 𝑛⃗=\vec{𝑛}= ⟨a1,a2,…,an⟩\lang a_1, a_2, \dots, a_n \rang:
n⃗ ⋅ (x⃗−P⃗)=0\boxed{\quad \vec{n}\ \cdot\ \left(\vec{x}-\vec{P}\right)=0 \quad}

Standard Form

As before, we use a normal vector 𝑛⃗=\vec{𝑛}= ⟨a1,a2,…,an⟩\lang a_1, a_2, \dots, a_n \rang and a point on the hyperplane PP to compute d=n⃗⋅P⃗d = \vec n \cdot \vec{P} :
a1x1+a2x2+⋯+anxn=d\boxed{\quad a_1 x_1 + a_2 x_2 + ⋯ + a_n x_n = d \quad}


Wize Tip
An equation in standard form always represents a hyperplane in Rn\reals^n.

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Example: Hyperplanes

Find the standard form equation for the hyperplane in R5\reals^5 with normal vector ⟨1,−1,0,3,2⟩\lang 1,−1,0,3,2\rang and containing the point P(4,−2,1,0,−1).P(4,−2,1,0,−1).

We have a normal vector n⃗=⟨1,−1,0,3,2⟩\vec{n} = \lang 1, −1,0,3,2 \rangand a known point on the hyperplane P⃗=⟨4,−2,1,0,−1⟩ \vec{P} = \lang 4, −2,1,0, −1 \rang.

So, the standard form equation of the hyperplane in R5ℝ^5 is:
1x1+(−1)x2+0x3+3x4+2x5=[1−1032]⋅[4−210−1]x1−x2+3x4+2x5=4+2+0+0−2\begin{array}{rcl} 1 x_1 + (− 1) x_2 + 0 x_3 + 3 x_4 + 2 x_5 &=&\begin{bmatrix} 1\\ -1\\ 0\\ 3\\ 2 \end{bmatrix} \cdot \begin{bmatrix} 4\\ -2\\ 1\\ 0\\ -1 \end{bmatrix}\\[3em] x_1 − x_2 + 3 x_4 + 2 x_5 &=& 4+2+0+0-2\\[0.5em] \end{array}
  ⟹  x1−x2+3x4+2x5=4\implies \boxed{x_1 − x_2 + 3 x_4 + 2 x_5 = 4}
Note: we can also written the point-normal form:
[1−1032]⋅([x1x2x3x4x5]−[4−210−1])=0\begin{aligned} \begin{bmatrix} 1\\ -1\\ 0\\ 3\\ 2 \end{bmatrix} \cdot \left( \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\\ x_5 \end{bmatrix} - \begin{bmatrix} 4\\ -2\\ 1\\ 0\\ -1 \end{bmatrix} \right) =0\\[3em] \end{aligned}

Practice: Hyperplanes

Determine if the following hyperplanes are orthogonal, parallel, or neither. [Answer with a single word]

Part A)

Π1: x1−x2+x3−4x4=3\Pi_1:\ x_1-x_2+x_3-4x_4=3 and Π2: 5x1−5x2+5x3−20x4=−6\Pi_2:\ 5x_1−5x_2+5x_3-20x_4=−6

Part B)

Π1: 2x1+5x2+x3−x4=−2\Pi_1:\ 2x_1+5x_2+x_3-x_4=−2 and Π2: ⟨−2,1,1,2⟩⋅(⟨x1,x2,x3,x4⟩−⟨1,0,2,−1⟩)=0\Pi_2 :\ \lang −2,1,1,2 \rang\cdot\big( \lang x_1,x_2,x_3,x_4 \rang−\lang 1,0,2,-1\rang \big)=0


Extra Practice
133 - FML 3 - 18.1W e.g. 36
For two lines having directional vectors v1\mathbf{v}_1 and v2\mathbf{v}_2, respectively, the shortest distance between them can be found by creating a vector that connects the two lines (by subtracting a point on one line from a point on the other), i.e. (r2−r1)\left(\mathbf{r}_2 - \mathbf{r}_1\right), and projecting it onto the unit normal vector connecting the two lines. The unit normal is easily found by crossing their respective directionals and dividing by the magnitude of the resulting vector, as illustrated in Fig. 5. i.e.
s=∣projn^(r2−r1)∣=∣(r2−r1)⋅v1×v2∣v1×v2∣∣ s = \left| \text{proj}_{\mathbf{\hat{n}}} \left(\mathbf{r}_2 - \mathbf{r}_1\right)\right| = \left| \left(\mathbf{r}_2 - \mathbf{r}_1\right) \cdot \frac{\mathbf{v}_1 \times \mathbf{v}_2}{\left|\mathbf{v}_1 \times \mathbf{v}_2\right|}\right|
Find the equations of the line passing through the point (2,3,−4)\bcb{(2, 3, -4)} parallel to the vector i−4k\bcb{\mathbf{i} - 4\mathbf{k}} ; express it in vector-parametric, scalar-parametric and standard form.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.2
Given the points P1=(0,0,1)\bcb{P_1 = (0,0,1)} and P2=(1,−2,3)\bcb{P_2 = (1,-2,3)} and the vector v⃗=<2,−3,1>\bcb{\vec{v} = \left< 2, -3, 1 \right>}
Find the minimum distance between the point P2\bcb{P_2} and the plane Π: 2x−3y+z=1\bcb{\Pi:\ 2x-3y + z = 1}.