MATH 251 Lecture 6

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Problem

Given A(1, 2, 3), B(4, 5, 6), and C(-5, -5, -5), find the equation of the plane through these points.

Find 2 vectors among the points

$\begin{array}{rl}\overrightarrow{u}& =\overrightarrow{B-A}=⟨3,3,3⟩\\ \overrightarrow{v}& =\overrightarrow{C-A}=⟨-6,-7,-8⟩\end{array}$

Take their cross product: $\begin{array}{rl}\overrightarrow{n}=\overrightarrow{u}\times \overrightarrow{v}& =3(\hat{ı}(-8+7)-\hat{ȷ}(-8+6)+\hat{k}(-7+6))\\ & =3⟨-1,2,-1⟩\end{array}$

Therefore using normal form:

$\begin{array}{rl}-(x-1)+2(y-2)-(z-3)& =0\\ -x+2y-z& =0\end{array}$

Another Problem

Given the planes $x+2y-3z=1$ and $2x+3y-z=0$, find the distance between them.

The distance is 0 because they intersect: they have different normal vectors: (1, 2, -3) and (2, 3, -1)

Yet Another Problem

Given $\alpha =x+2y-3z=1$ and$\beta =2x+4y-6z=9$, find the distance between the planes.

Both planes have the same normal vectors $\overrightarrow{n}=⟨1,2,-3⟩$, so they are parallel.

Pick a point on plane $\alpha$: P(1,0,0). Pick one from plane $\beta$: Q(9/2,0,0).

The distance between the planes is $\left|{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}}_{\overrightarrow{n}}\left(\overrightarrow{P-Q}\right)\right|$

$\begin{array}{rl}{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}}_{\overrightarrow{n}}\left(\overrightarrow{P-Q}\right)& =\frac{\left(\overrightarrow{P-Q}\right)\cdot \overrightarrow{n}}{\Vert \overrightarrow{n}\Vert }\\ & =\frac{3.5}{\sqrt{14}}\end{array}$

This guy has a lot of problems ;)

"line": $\alpha \left(t\right)=\left(\begin{array}{c}0\\ -1\\ 2\end{array}\right)+t\left(\begin{array}{c}1\\ 4\\ -2\end{array}\right)$

"plane": $4x+2z=3$

Find the distance between these two things. We can determine whether they intersect by comparing the dot product of the plane's normal vector $\overrightarrow{n}=⟨4,0,2⟨$ to the line's directional vector $\overrightarrow{v}=⟨1,4,-2⟩$.

Their dot product is 0, so they are indeed parallel.

Find a point on the line: P(0,-1,2).
Find a point on the plane: Q(0,0,3/2).

The distance between the plane and the point is $\left|{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}}_{\overrightarrow{n}}\left(\overrightarrow{P-Q}\right)\right|$

$\begin{array}{rl}{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}}_{\overrightarrow{n}}\left(\overrightarrow{P-Q}\right)& =\left|\frac{\left(\overrightarrow{P-Q}\right)\cdot \overrightarrow{n}}{\Vert \overrightarrow{n}\Vert }\right|\\ & =\frac{1}{2\sqrt{5}}\end{array}$

Quadric Surfaces

Determined by a quadratic equation:

For example, $x^2-2xy+z^2+3z-1=0$ is a quadric surface.

For our purposes, we are not going to have terms like $xy$, $yz$, or $xz$ because adding those terms just rotates the surface. We will also omit linear terms of a single variable because that justs shifts the graph up and down.

$\frac{x^2}{a^2}\pm \frac{y^2}{b^2}\pm \frac{z^2}{c^2}=1$

Ellipsoid

$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

A 3D oval...

Paraboloid

$\frac{x^2}{a^2}+\frac{y^2}{b^2}-z=0$

Start with a parabola and rotate it around the x axis

Hyperboloid of One Sheet

$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$

Start with $x^2-y^2=1$ (a hyperbola) and rotate around the $x$ axis

Hyperboloid of Two Sheets

$-\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

Start with $x^2-y^2=1$ (a hyperbola) and rotate around the $y$ axis

Cylinder

hyperbolic, elliptic, parabolic

Omit $z$ variable, then $xy$ plot sweeps out a "cylinder"