DLA Discovery guide

Discovery guide 15.1 Discovery guide

Discovery 15.1.

Begin with a set of $xy$ -axes. Draw the vector $\uvec{x}_0 = (3,0)$ with its tail at the origin, and then draw the vector $\uvec{p} = (2,1)$ with its tail at the head of $\uvec{x}_0\text{.}$

🔗

(a)

Consider the expression $\uvec{x} = \uvec{x}_0 + t \uvec{p}$ in the parameter t. Think of $\uvec{x}$ as a variable vector: using different values of $t\text{,}$ $\uvec{x}$ evaluates to different vectors. Draw the vector $\uvec{x}$ for $t=1$ on your diagram with its tail at the origin and using a dashed line for the shaft of the arrow. Then do the same for $t=2\text{,}$ $t=-1\text{,}$ $t=1/2\text{,}$ $t=-3\text{.}$

🔗

(b)

Suppose you continued sketching in the different possible $\uvec{x}$ vectors forever, using every possible value for the parameter $t\text{.}$ What shape would be traced out by all of the points at the heads of the different versions of $\uvec{x}\text{?}$

🔗

Discovery 15.2.

The equation $x - 2 y = 3$ defines a line $\ell$ in $\R^2\text{.}$ We can also view this equation as a system of linear equations. Its solution requires one parameter.

🔗

(a)

Set $y = t$ and then compute the parametric equation for $x\text{.}$ Set $\uvec{x}$ to be the variable vector $\uvec{x} = \left[\begin{smallmatrix}x\\y\end{smallmatrix}\right]\text{.}$ Fill in the vectors at below. Then compare with Discovery 15.1.

$\begin{equation*} \uvec{x} \;\;=\;\;\begin{bmatrix}x\\y\end{bmatrix} \;\;=\;\;\begin{bmatrix}\hphantom{3+2t}\\t\end{bmatrix} \;\;=\;\;\begin{bmatrix}\hphantom{3}\\\hphantom{0}\end{bmatrix} \;\;+\;\; t\begin{bmatrix}\hphantom{2}\\\hphantom{1}\end{bmatrix} \end{equation*}$

🔗

(b)

Use the line equation $x - 2 y = 3$ to verify that the point $(4,1/2)$ lies on $\ell\text{.}$ Then determine the value of the parameter $t$ so that $\uvec{x} = (4,1/2)\text{.}$

🔗

Discovery 15.3.

Consider the two planes

$\begin{align*} \Pi_1\amp\colon\;\; 2 x - y + 5 z = -5, \amp \Pi_2\amp\colon\;\; x + 2 y - 5 z = 10 \end{align*}$

in $\R^3\text{.}$

🔗

(a)

Verify that $\Pi_1$ and $\Pi_2$ are not parallel.

Hint

Compare their normal vectors.

🔗

(b)

Two nonparallel planes must intersect in a line. Describe the line of intersection of $\Pi_1$ and $\Pi_2$ in the form $\uvec{x} = \uvec{x}_0 + t \uvec{p}\text{.}$

Hint

Any point in the intersection must lie on both planes at once. That is, any point in the intersection must be a solution to the system of equations formed by the two plane equations.

🔗

Discovery 15.4.

The equation $x - y + 5 z = -5$ defines a plane in $\R^3\text{.}$ We can also view this equation as a system of linear equations.

🔗

(a)

Similarly to Discovery 15.2, determine vectors $\uvec{x}_0\text{,}$ $\uvec{p}_1\text{,}$ and $\uvec{p}_2$ so that

$\begin{equation*} \uvec{x} = \left[\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right] = \uvec{x}_0 + s\uvec{p}_1 + t\uvec{p}_2 \end{equation*}$

describes all solutions to the equation (and hence all points on the plane).

🔗

(b)

Use the plane's equation $x - y + 5 z = -5$ to verify that the point $(1,1,-1)$ lies on the plane.

Then determine the values of the parameters $s$ and $t$ so that the formula

$\begin{equation*} \uvec{x} = \uvec{x}_0 + s\uvec{p}_1 + t\uvec{p}_2 \end{equation*}$

results in this point $\uvec{x} = (1,1,-1)\text{.}$

🔗

Discovery 15.5.

Draw a grid over the $xy$ -plane, with a vertical line at each integer value of $x$ and a horizontal line at each integer value of $y\text{.}$ Then draw $\uvec{e}_1$ and $\uvec{e}_2$ on your diagram.

What does the decomposition $(3,2) = 3\uvec{e}_1 + 2\uvec{e_2}$ look like on your grid?

How about $(-1,2) = (-1)\uvec{e}_1 + 2\uvec{e}_2\text{?}$

How about $(3/2,-2) = (3/2)\uvec{e}_1 + (-2)\uvec{e}_2\text{?}$

🔗

Discovery 15.6.

Draw a “grid” over the $xy$ -plane as follows: at each integer value along the $x$ -axis, draw both a vertical line and a slant line parallel to the line $y=x\text{.}$ Then draw $\uvec{u} = (1,1)$ and $\uvec{e}_2$ on your diagram.

What does the decomposition $(3,2) = 3\uvec{u} + (-1)\uvec{e}_2$ look like on your grid?

How about $(-1,2) = (-1)\uvec{u} + 3\uvec{e}_2\text{?}$

How about $(3/2,-2) = (3/2)\uvec{u} + (-7/2)\uvec{e}_2\text{?}$

🔗

Discovery 15.7.

The set of all solutions to the homogeneous equation $x - 2 y + 3 z = 0$ forms a plane in $\R^3\text{.}$ We can solve this equation by assigning parameters $y=s$ and $z=t\text{,}$ so that all solutions can be described parametrically by

$\begin{equation*} (x,y,z) = s(2,1,0) + t(-3,0,1) \text{.} \end{equation*}$

Discuss how the vectors $\uvec{p}_1 = (2,1,0)$ and $\uvec{p}_2 = (-3,0,1)$ create a “grid” on the plane defined by $x - 2 y + 3 z = 0\text{,}$ similarly to the grids you worked with in Discovery 15.5 and Discovery 15.6.

🔗

Discovery 15.8.

Determine the point of intersection of the line $\ell\text{,}$ described parametrically below left, and the plane $\Pi\text{,}$ described algebraically below right.

$\begin{align*} \ell\amp\colon\;\; \uvec{x} = (2,0,3) + t(-1,1,1) \amp \Pi\amp\colon\;\; 2 x + y - 3 z = 7 \end{align*}$

Hint

The point of intersection is simultaneously on the line and on the plane.

🔗

Discovery 15.9.

Set up a system of equations whose solution is the point of intersection of the line $\ell$ and the plane $\Pi\text{,}$ described parametrically below.

$\begin{align*} \ell\amp\colon\;\; \uvec{x} = (2,0,3) + t(-1,1,1) \amp \Pi\amp\colon\;\; \uvec{x} = (3,1,0) + r(1,1,1) + s(3,0,2) \end{align*}$

Hint

The point of intersection is simultaneously on the line and on the plane.