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Section 15.4 Examples

Subsection 15.4.1 Describing lines and planes parametrically

First we will work out some of the activities from Discovery guide 15.1 that involve describing lines and planes parametrically.

Example 15.4.1. Parametrically describing a line in \(\R^2\).

In Discovery 15.2, we considered the equation \(x-2y=3\) for a line in the plane. Setting parameter \(y=t\) and isolating \(x\) in this equation leads to general solution

\begin{equation*} \uvec{x} = \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix} 3+2t \\ t \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \end{bmatrix} + t\begin{bmatrix}2\\1\end{bmatrix}\text{.} \end{equation*}

Geometrically, the vector \(\uvec{x}_0 = (3,0)\) represents an “base” point on the line, and algebraically represents the particular solution to the system obtained from parameter value \(t=0\text{.}\) The vector \(\uvec{p} = (2,1)\) represents a vector parallel to the line, and every other point on the line (i.e. every other solution to the system) can be obtained by adding a scalar multiple of \(\uvec{p}\) to \(\uvec{x}_0\text{.}\) For example, the point \(\uvec{x} = (4,1/2)\) lies on the line, as we can verify by checking

\begin{equation*} 4 - 2 \cdot \frac{1}{2} = 4 - 1 = 3 \text{,} \end{equation*}

so that the coordinates of the point satisfy the line equation \(x - 2 y = 3\text{.}\) We can solve for \(t\) in the vector equation

\begin{equation*} \begin{bmatrix}4\\\frac{1}{2}\end{bmatrix} = \begin{bmatrix}3\\0\end{bmatrix} + t \begin{bmatrix}2\\1\end{bmatrix} \end{equation*}

to see exactly how this point is a multiple of \(\uvec{p}\) away from \(\uvec{x}_0\text{:}\)

\begin{gather} \begin{bmatrix}4\\\frac{1}{2}\end{bmatrix} - \begin{bmatrix}3\\0\end{bmatrix} = \begin{bmatrix}1\\\frac{1}{2}\end{bmatrix} = \begin{bmatrix}2t\\t\end{bmatrix} \qquad\qquad\implies\qquad\qquad t=\frac{1}{2}\text{.}\label{equation-system-geom-examples-line-in-R2-point-to-parameter}\tag{\(\star\)} \end{gather}

Finally, we have

\begin{equation*} \begin{bmatrix}4\\\frac{1}{2}\end{bmatrix} = \uvec{x} + \frac{1}{2}\uvec{p} \text{.} \end{equation*}
Warning 15.4.2.

In the previous example, we determined the value of the parameter \(t\) that corresponds to the point \((4,1/2)\) on the line by solving vector equation (\(\star\)). When you are solving vector equations for parameters like this, make sure you check that your solution works in every coordinate! For example, in (\(\star\)) we see that \(t=1/2\) immediately from comparing the second coordinate on left and right. But it is important to check that this parameter value also works in the first coordinate (which it does).

Example 15.4.3. Parametrically describing the intersection of two planes in \(\R^3\).

In Discovery 15.3, we considered a system consisting of two equations in three variables,

\begin{equation*} \left\{\begin{array}{rcrcrcr} 2 x \amp - \amp y \amp + \amp 5 z \amp = \amp -5 \text{,} \\ x \amp + \amp 2 y \amp - \amp 5 z \amp = \amp 10 \text{.} \end{array}\right. \end{equation*}

Geometrically, each of the equations represents a plane in space, and solutions to the system represent points that are in common to both planes (that is, points in the intersection of the two planes). From the coefficients of the equations we may take \(\uvec{n}_1 = (2,-1,5)\) and \(\uvec{n}_2 = (1,2,-5)\) as normal vectors for the two planes, respectively. Since these normal vectors are not parallel, neither are the planes, and so they must intersect. Algebraically, this means that the coefficient parts of the two equations are not multiples of each other, so when we row reduce we will find two leading ones, representing the two independent equations with which we started. And so we will only require one parameter to express the general solution, which will then take the form of a line. The free variable in this system is \(z\text{,}\) so setting \(z\) to parameter \(t\) and solving we get

\begin{equation*} \uvec{x} = \begin{bmatrix}x \\[0.5ex] y \\[0.5ex] z \end{bmatrix} = \begin{bmatrix} -t \\ 5 + 3 t \\ t \end{bmatrix} = \begin{bmatrix} 0 \\ 5 \\ 0 \end{bmatrix} + t\left[\begin{array}{r} -1 \\ 3 \\ 1 \end{array}\right]. \end{equation*}

Here, the “base” point corresponding to \(t=0\) is \(\uvec{x}_0 = (0,5,0)\text{,}\) and the vector \(\uvec{p} = (-1,3,1)\) is parallel to the line.

Example 15.4.4. Parametrically describing a plane in \(\R^3\).

When we have just a single plane in space, as in Discovery 15.4, we can view its equation as a system of equations, just as we did with the line in Discovery 15.2. In that activity, we worked with equation \(x - y + 5 z = -5\text{.}\) For this equation, we will need two parameters to express the general solution, and each parameter will provide us with a vector parallel to the plane. Setting \(y = s\) and \(z = t\text{,}\) we can then use the plane equation to express \(x\) in terms of these parameters. This leads to general solution, in vector form:

\begin{equation*} \uvec{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -5 + s - 5t \\ s \\ t \end{bmatrix} = \left[\begin{array}{r} -5 \\ 0 \\0 \end{array}\right] + s \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + t \left[\begin{array}{r} -5 \\ 0 \\ 1 \end{array}\right]\text{.} \end{equation*}

Here, the “base” point on the plane is \(\uvec{x}_0 = (-5,0,0)\text{,}\) which corresponds to parameter values \(s = t = 0\text{.}\) Every other point on the plane corresponds to other choices of parameter values. For example, as in the discovery activity, the point \((1,1,-1)\) is on the plane. We can verify this by checking

\begin{equation*} 1 - 1 + 5 (-1) = -5 \text{,} \end{equation*}

so that the coordinates of the point satisfy the plane equation \(x - y + 5 z = -5\text{.}\) We can also describe this point using the vector equation

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

as follows:

\begin{equation*} \left[\begin{array}{r} 1 \\ 1 \\ -1 \end{array}\right] = \left[\begin{array}{r} -5 \\ 0 \\ 0 \end{array}\right] + s \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + t \left[\begin{array}{r} -5 \\ 0 \\ 1 \end{array}\right]\text{.} \end{equation*}

Solving this vector equation for \(s\) and \(t\) leads to a … system of linear equations!

\begin{align*} \left[\begin{array}{r} 1 \\ 1 \\ -1 \end{array}\right] - \left[\begin{array}{r} -5 \\ 0 \\ 0 \end{array}\right] \amp = \begin{bmatrix} s \\ s \\ 0 \end{bmatrix} + \begin{bmatrix} -5 t \\ 0 \\ t \end{bmatrix} \\ \left[\begin{array}{r} 6 \\ 1 \\ -1 \end{array}\right] \amp= \begin{bmatrix} s - 5 t \\ s \\ t \end{bmatrix}. \end{align*}

From the second and third coordinates we immediately see \(s = 1\) and \(t = -1\text{.}\) However, it's important to also verify that \(s - 5 t = 6\) for this choice of parameter values, to satisfy the equality of the two first coordinates on right and left.

Subsection 15.4.2 Determining points of intersection

When lines and/or planes are described using algebraic equations, determining points of intersection only requires solving the linear systems that those equations form together. Here we will demonstrate determining points of intersection when some or all of the lines and/or planes involved are described parametrically by working out some of the activities from Discovery guide 15.1.

Example 15.4.5. Intersection of a parametrically-described line and an algebraically-described plane in \(\R^3\).

In Discovery 15.8, we have a line described by a vector equation and a plane described algebraically, and would like to determine their point of intersection (if there is one). Any such point of intersection must be on the line, and so its coordinates can be described in terms of the parameter \(t\text{:}\)

\begin{equation*} \uvec{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 3 \end{bmatrix} + t \left[\begin{array}{r} -1 \\ 1 \\ 1 \end{array}\right] = \begin{bmatrix} 2-t \\ t \\ 3+t \end{bmatrix}\text{.} \end{equation*}

If this point is also on the plane, its coordinates must satisfy the equation for the plane:

\begin{align*} 2x+y-3z \amp= 7 \\ 2(2-t) + t - 3(3+t) \amp= 7 \\ -5 - 4t \amp= 7 \\ t \amp= -3\text{.} \end{align*}

Substituting this parameter value in our expression for \(\uvec{x}\) gives us the point of intersection:

\begin{equation*} \left[\begin{array}{r} 5 \\ -3 \\ 0 \end{array}\right] \text{.} \end{equation*}
Example 15.4.6. Intersection of parametrically-described line and plane in \(\R^3\).

In Discovery 15.9, we again want to determine the point of intersection (if any) of a line and a plane, but this time both line and plane are described by vector equations. If a point lies on both line and plane, its coordinates must have a simultaneous description by both vector equations in terms of parameters:

\begin{align*} \uvec{x} \amp = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 3 \end{bmatrix} + t \left[\begin{array}{r} -1 \\ 1 \\ 1 \end{array}\right] = \begin{bmatrix} 2-t \\ t \\ 3+t \end{bmatrix},\\ \\ \uvec{x} \amp = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix} + r\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + s\begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 3+r+3s \\ 1+r \\ r+2s \end{bmatrix}. \end{align*}

Now, this point can only have one set of coordinates, so these two descriptions must actually be the same. This lets us set up a … system of linear equations!

\begin{equation*} \begin{array}{rcl} x\colon \amp \qquad \amp 2-t = 3+r+3s\text{,} \\ y\colon \amp \qquad \amp t = 1+r\text{,} \\ z\colon \amp \qquad \amp 3+t=r+2s\text{,} \end{array} \qquad\implies\qquad \left\{\begin{array}{rcrcrcr} r \amp + \amp 3 s \amp + \amp t \amp = \amp -1\text{,} \\ r \amp \amp \amp - \amp t \amp = \amp -1\text{,} \\ r \amp + \amp 2s \amp - \amp t \amp = \amp 3\text{.} \end{array}\right. \end{equation*}

We could put this system in a matrix and row reduce, but we only really care about the value of parameter \(t\) in the solution, because knowing \(t\) allows us to determine \(\uvec{x}\) from the vector description for the line. So we can use Cramer's rule instead. Write \(A\) for the coefficient matrix of this system, and \(A_3\) for the matrix obtained by replacing the third column of \(A\) by the vector of constants. Then,

\begin{align*} \det A_3 \amp= \left\lvert\begin{array}{rrr} 1 \amp 3 \amp -1 \\ 1 \amp 0 \amp -1 \\ 1 \amp 2 \amp 3 \end{array}\right\rvert = -12,\\ \\ \det A \amp= \left\lvert\begin{array}{rrr} 1 \amp 3 \amp 1 \\ 1 \amp 0 \amp -1 \\ 1 \amp 2 \amp -1 \end{array}\right\rvert = 4\\ \\ \implies \quad t \amp= \frac{\det A_3}{\det A} = -3. \end{align*}

Now that we have \(t=-3\text{,}\) we can determine the point of intersection:

\begin{equation*} \uvec{x} = \begin{bmatrix} 2-t \\ t \\ 3+t \end{bmatrix} = \left[\begin{array}{r} 5 \\ -3 \\ 0 \end{array}\right]\text{.} \end{equation*}