\( \newcommand{\R}{\mathbb{R}} \newcommand{\norm}[1]{\left\lVert #1 \right\rVert} \newcommand{\la}{\left\langle} \newcommand{\ra}{\right\rangle} \newcommand{\bbm}{\begin{bmatrix}} \newcommand{\ebm}{\end{bmatrix}} \newcommand{\orr}{\overrightarrow} \newcommand{\i}{\hat{\imath}} \newcommand{\j}{\hat{\jmath}} \newcommand{\k}{\hat{k}} \newcommand{\dotp}{\boldsymbol{\cdot}} \newcommand{\uu}{\vec{u}} \newcommand{\vv}{\vec{v}} \newcommand{\ww}{\vec{w}} \newcommand{\amp}{&} \newcommand{\proj}{\operatorname{proj}} \)

Math 1410, Spring 2020

Lines and Planes

January 21, 2020

Sean Fitzpatrick
University of Lethbridge

Recap

Warm-up

  1. Compute the cross product of \(\vec{v}=\la 3, -1, 4\ra\) and \(\vec{w} = \la 2,0,-1\ra\text{.}\)

  2. Find a vector equation for the line through \(P=(1,2,3)\) and \(Q=(3,-1,4)\)

Vector and parametric equations of lines

Points on a line in space are given in terms of a parameter (usually \(t\) — we can think of motion in a straight line, with \(t\) as time).

  • The vector equation of a line through \(P_0=(x_0,y_0,z_0)\) in the direction of \(\vv = \la a,b,c\ra\) is

    \begin{equation*} \la x,y,z\ra = \la x_0,y_0,z_0\ra + t \la a,b,c\ra\text{,} \end{equation*}
    or \(\vec{x}=\vec{x}_0+t\vv\text{,}\) for short.

  • Sometimes see \(\vec{r}\) or \(\vec{r}(t)\) instead of \(\vec{x}\text{.}\)

  • Equating coefficients in the vector equation gives the parametric equations:

    \begin{align*} x \amp = x_0+at\\ y \amp = y_0+bt\\ z \amp = z_0+ct\text{.} \end{align*}

Examples with lines

Finding equations

Find the vector equations of the lines:

  1. Through the point \(P_0 = (2,-5,1)\) in the direction of \(\vv = \la 6,-7,3\ra\)

  2. Through the points \(P=(4,5,-3)\) and \(Q=(-1,8,2)\text{.}\)

  3. Through the point \(P_0=(4,5,-9)\) and parallel to the line

    \begin{equation*} \vec{r}(t) = \la 3-4t, -6+8t, 2-7t\ra\text{.} \end{equation*}

Comparing lines: intersecting, parallel, skew

  • Two lines in \(\R^3\) are parallel if their direction vectors are parallel.

  • Not all parallel lines intersect: some are skew.

  • Checking for intersection leads to a system of equations. (Use a different parameter for each line.)

Example:

Determine if the following pairs of lines are parallel, skew, or if they intersect.

  1. \begin{align*} \vec{r}_1(s) \amp = \la 3, -2, 4\ra + s\la 4,-2,6\ra\\ \vec{r}_2(t) \amp = \la 2,2,2\ra + t\la -6, 3, -9\ra \end{align*}

  2. \begin{align*} \vec{r}_1(s) \amp = \la 3, 1, 1\ra + s\la 2, -1, 3\ra\\ \vec{r}_2(t) \amp = \la 1,0,-1\ra + t\la 1, 2, 1\ra \end{align*}

  3. \begin{align*} \vec{r}_1(s) \amp = \la 0, 1, 2\ra + s\la 4, -2, 1\ra\\ \vec{r}_2(t) \amp = \la -2, -3, 7 \ra + t\la 3, 1, -2\ra \end{align*}

Projection and distance

Orthogonal projection

Given vectors \(\uu\) (not equal to \(\vec{0}\)) and \(\vv\text{,}\) often useful to write \(\vv\) as the sum of a vector parallel to \(\uu\text{,}\) and a vector orthogonal to \(\uu\text{.}\)

Vector \(\vec{a}\) called the projection of \(\vv\) onto \(\uu\text{.}\) Notation and formula:

\begin{equation*} \proj_{\uu}\vv = \left(\frac{\uu\dotp \vv}{\norm{\uu}^2}\right)\uu\text{.} \end{equation*}

Orthogonal decomposition

Given \(\vv = \la 3,-1\ra\) and \(\ww = \la 2,5\ra\text{,}\) find vectors \(\vec{a}\) and \(\vec{b}\) such that:

  1. \(\vec{a}\) is parallel to \(\vec{v}\)

  2. \(\vec{b}\) is orthogonal to \(\vec{v}\)

  3. \(\vec{a}+\vec{b} = \vec{w}\)

Distance

Example:

  1. Find the distance from the point \(P=(3,1,2)\) to the line

    \begin{equation*} \la x,y,z\ra = \la 0, 2, -1\ra + t \la 2, -1, 2\ra\text{.} \end{equation*}

  2. Find the distance between the parallel lines

    \begin{equation*} \vec{r}_1(t) = \la 2, -1, 3\ra + t\la 1, 2, 3\ra, \text{ and } \vec{r}_2(t) = \la -4, 1, 3\ra + t\la 2, 4, 6\ra\text{.} \end{equation*}

Planes in \(\R^3\)

Equations of planes

Two ways to describe a plane:

  1. A point, and two vectors parallel to the plane.

  2. A point, and one vector perpendicular to the plane.

Second option is simpler. Suppose \(\vec n = \la a,b,c\ra\) is perpendicular to the plane. Suppose also \(P_0 = (x_0,y_0,z_0)\) and \(P=(x,y,z)\) are in the plane. Then \(\orr{P_0P}\) is parallel to the plane, so \(\vec n \dotp \orr{P_0P}=0\text{.}\)

Finding equations

Find the equation of the plane:

  1. Through \(P_0=(2,-1,3)\) and perpendicular to \(\vec n = \la 5, -3, 4\ra\)

  2. Perpendicular to the line \(\vec{r}(t) = \la 3+2t, -5+3t, -2-9t\ra\text{,}\) and containing the origin.

  3. Containing the points \(P=(1,0,4), Q=(2,-1,3), R=(5,2,4)\)

  4. Containing the lines

    \begin{align*} \vec{r}_1(s) \amp = \la 0, 1, 2\ra + s\la 4, -2, 1\ra\\ \vec{r}_2(t) \amp = \la -2, -3, 7 \ra + t\la 3, 1, -2\ra \end{align*}
    (We found that these intersect at \(P_0=(4,-1,3)\text{.}\))