1e871b…: “Subsequent to the breezy lake”

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The following exercise explores how geometric information about lines can be obtained from vector-valued functions that describe them.

First, recall that once we start discussing lines in $\R ^3$ , the notion of “slope” needs to be revisited. A key observation is that lines with different “slopes” in the $xy$ -plane actually point in different directions, so it is a natural idea to replace “slope” with a vector parallel to the line.

Given a vector $\vec {v}$ and a point $P_0$ , a parameterization of the line that is parallel to $\vec {v}$ that passes through $P_0$ is

$\vec {l}(t) = \vec {v}t+\vec {P}_0, t \in \R .$

Find a parametrization of the line that is parallel to $\vec {v} = \vector {3,-2,5}$ that passes through $(2,9,4)$ .

$\vec {l}(t) = \vector {\answer {3},\answer {-2},\answer {5}}t+\vector {\answer {2},\answer {9},\answer {4}} = \vector {\answer {3t+2},\answer {-2t+9},\answer {5t+4}}$

Now, note that the line exists independently of the parameterization that we use to describe it; different choices of the parameter will trace out the line in different fashions, but they trace out the same line.

Consider the line in $\R ^2$ given by $y=4x+5$ .

If we require that $x(t)=t$ , then $y(t) = \answer {4t+5}$ , and a parameterization of the line is
$\vec {l}_1(t) = \vector {x(t),y(t)} = \vector {\answer {t},\answer {4t+5}}, t \in \R .$
If we require that $y(t)=t$ , then $x(t) = \answer {\frac {1}{4}t-\frac {5}{4}}$ , and a parameterization of the line is
$\vec {l}_2(t) = \vector {x(t),y(t)} = \vector {\answer {\frac {1}{4}t-\frac {5}{4}},\answer {t}}, t \in \R .$

Note that the line passes through the point $(x,y) = (2,13)$ . The vector-valued function $\vec {l}_1(t)$ will trace through this point at $t=\answer {2}$ , while the vector-valued function $\vec {l}_2(t)$ will trace through this point at $t=\answer {13}$ .

Given a parametric description of a line, we can extract a vector parallel to the line without too much difficulty.

To find a vector parallel to the line $\vec {l}(t) = \vector {2t+3,4-5t,2}$ , note that we can write the line in the form

$\vec {l}(t) = \vec {v}t+\vec {P}_0 = \vector {\answer {2},\answer {-5},\answer {0}}t+\vector {\answer {3},\answer {4},\answer {2}}$

A vector parallel to the line is thus $\vector {\answer {2},\answer {-5},\answer {0}}$ .

To explore this a little more deeply, note that we can find a vector parallel to the line from two points on the line. Note that $\vec {l}(0) = \vector {3,4,2}$ and $\vec {l}(1) = \vector {\answer {5},\answer {-1},\answer {2}}$ .

A vector $\vec {v}$ parallel to the line can be found from the corresponding points

$\vec {v} = \vector {5-3,-1-4,2-2} = \vector {\answer {2},\answer {-5},\answer {0}}.$

Now that we have a quick way to find a vector parallel to a given line, we can use these to determine if two lines described by vector-valued functions are parallel.

Let $\vecl _{1}(t)=\vector {2t+4,t+5,4t+8}$ and $\vecl _{2}(t)=\vector {4t+1,2t,8t-3}$ .

Then $\vecl _{1}$ and $\vecl _{2}$ are

parallel. intersecting. non-intersecting. the same line.