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

${\begin{aligned}{\vec {u}}&={\overrightarrow {B-A}}=\left\langle 3,\,3,\,3\right\rangle \\{\vec {v}}&={\overrightarrow {C-A}}=\left\langle -6,\,-7,\,-8\right\rangle \end{aligned}}$

Take their cross product: ${\begin{aligned}{\vec {n}}={\vec {u}}\times {\vec {v}}&=3\left({\hat {\imath }}\left(-8+7\right)-{\hat {\jmath }}\left(-8+6\right)+{\hat {k}}\left(-7+6\right)\right)\\&=3\left\langle -1,\,2,\,-1\right\rangle \end{aligned}}$

Therefore using normal form:

${\begin{aligned}-\left(x-1\right)+2\left(y-2\right)-\left(z-3\right)&=0\\-x+2y-z&=0\end{aligned}}$

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 ${\vec {n}}=\left\langle 1,2,-3\right\rangle$, 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 $\left|\mathrm {comp} _{\vec {n}}\left({\overrightarrow {P-Q}}\right)\right|$

${\begin{aligned}\mathrm {comp} _{\vec {n}}\left({\overrightarrow {P-Q}}\right)&={\frac {\left({\overrightarrow {P-Q}}\right)\cdot {\vec {n}}}{\left\|{\vec {n}}\right\|}}\\&={\frac {3.5}{\sqrt {14}}}\end{aligned}}$

This guy has a lot of problems ;)

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

"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 ${\vec {n}}=\left\langle 4,0,2\right\langle$ to the line's directional vector ${\vec {v}}=\left\langle 1,4,-2\right\rangle$.

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 $\left|\mathrm {comp} _{\vec {n}}\left({\overrightarrow {P-Q}}\right)\right|$

${\begin{aligned}\mathrm {comp} _{\vec {n}}\left({\overrightarrow {P-Q}}\right)&=\left|{\frac {\left({\overrightarrow {P-Q}}\right)\cdot {\vec {n}}}{\left\|{\vec {n}}\right\|}}\right|\\&={\frac {1}{2{\sqrt {5}}}}\end{aligned}}$

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"