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 \(\uvec{x}\text{?}\)
Discovery14.2.
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 14.1.
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 \(\uvec{x} = (4,1/2)\text{.}\)
Discovery14.3.
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.
Discovery14.4.
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 14.2, determine vectors \(\uvec{x}_0\text{,}\)\(\uvec{p}_1\text{,}\) and \(\uvec{p}_2\) so that
results in this point \(\uvec{x} = (1,1,-1)\text{.}\)
Discovery14.5.
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 \((3/2,-2) = (3/2)\uvec{e}_1 + (-2)\uvec{e}_2\text{?}\)
Discovery14.6.
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 \((3/2,-2) = (3/2)\uvec{u} + (-7/2)\uvec{e}_2\text{?}\)
Discovery14.7.
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
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 14.5 and Discovery 14.6.
Discovery14.8.
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.
Discovery14.9.
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.
