GRAPHS

• Graph 1 — The line \(y = x/3 + 2\) with direction vector and fixed-point position vector.
• Graph 2 — The plane \(3x + 4y + 6z = 12\), with normal vector and fixed point.

EXERCISE 1 — Vector Equations of Lines

Background & Notes

General

Lines in the \(xy\)-plane and planes in 3-space can be described by linear equations:

\[\begin{align*}
&\text{lines:} & ax + by &= c\\
&\text{planes:} & ax + by + cz &= d\\
\end{align*}\]

where \(a, b, c, d\) are constants.

Geometrically, lines (in any dimension) and planes (in 3-space) can be described by a direction (vector) and a position (fixed point).

Lines

Lines in the \(xy\)-plane are described by the equation:

\[ax + by = c\]

where \(a, b, c\) are constant.

Lines in any dimension can be described by a direction vector \(\vec{v}\), and a fixed point on the line \(P_0\):

\[\vec{r}(t) = t\vec{v} + \vec{r}_0\]

• \(\vec{v}\) is a direction vector parallel to the line.
• \(P_0\) is a fixed point on the line with position vector \(\vec{r}_0\).
• \(t\) is the parameter (variable).

Parametric equations of a vector equation come from the coordinate functions (or component functions) of \(\vec{r}(t)\).

For a line in 3-space with direction \(\vec{v} = \big< a, b, c \big>\) and fixed point \(P_0 = ( x_0, y_0, z_0 )\):

\[\vec{r}(t) = t \vec{v} + \vec{r}_0 = \big< at + x_0, bt + y_0, ct + z_0 \big>\]

And the parametric equations are:

\[\begin{align*}
x(t) &= at + x_0\\
y(t) &= bt + y_0\\
z(t) &= ct + z_0
\end{align*}\]

Assumptions

Let \(L_1\) be the line in the \(xy\)-plane:

\[y = x/3 + 2\]

Instructions

0. Table

Copy this table into your notebook — make it wide:

1. Construct Equation & Graph (example)

Find a vector equation:

\[\vec{r}_1(t) = t \vec{v} + \vec{r}_0\]

for line \(L_1\) — the line \(y = x/3 + 2\).

Using scalar multiplication and vector addition, combine the right-hand side of the vector equation into a single vector, and identify the parametric equations \(x(t), y(t), z(t)\).

Complete the first row of the table, then go to graph 1 and:

On the graph, identify the line, the direction vector \(\vec{v} = \big< 3, 1 \big>\), and the position vector \(\vec{r}_0\) for the fixed point \(P_0(0,2)\).

In the control window, identify the parametric equations \(x(t), y(t), z(t)\).

2. Construct Examples

Find a vector equation \(\vec{r}_2(t) = t \vec{v} + \vec{r}_0\) for \(L_2\) — the line passing through the origin, and parallel to \(L_1\) (\(y = x/3 + 2\)) in the \(xy\)-plane.

Find a vector equation \(\vec{r}_3(t) = t \vec{v} + \vec{r}_0\) for \(L_3\) — the line passing through points \(P(2, 0, 1)\) and \(Q(-4, -2, 1)\).

Complete the table, then return graph 1 and add your two examples \(\vec{r}_2(t)\) and \(\vec{r}_3(t)\):

• Go the “Add to graph: Select …” pull-down menu, and select “Space Curve: r(t)”.
• In the control window, add the parametric equations for \(L_2\) to the new curve, and change the \(t\)-values to \(-5 \leq t \leq 5\).
• Repeat for \(L_3\). Use the `2D/3D' button to toggle between views.

3. Compare Direction Vectors

Compare the three lines \(\vec{r}_1(t), \vec{r}_2(t), \vec{r}_3(t)\) in graph 1, and the three direction vectors from the three vector equations.

• What’s the relationship between the three lines?
• What’s the relationship between the direction vectors of the three lines?
• How can you determine whether two lines are parallel based on their direction vectors?

4. Challenge

Write the line \(\vec{r}(t) = t \big< 2, 6 \big> + \big< -4, 0 \big>\) in slope-intercept form, \(y = mx + b\).

EXERCISE 2 — Planes & Normal Vectors

Background & Notes

General

Lines in the \(xy\)-plane and planes in 3-space can be described by linear equations:

\[\begin{align*}
&\text{lines:} & ax + by &= c\\
&\text{planes:} & ax + by + cz &= d\\
\end{align*}\]

where \(a, b, c, d\) are constants.

Geometrically, lines (in any dimension) and planes (in 3-space) can be described by a direction (vector) and a position (fixed point).

Planes

Planes in 3-space are described by the equation:

\[ax + by + cz = d\]

where \(a, b, c, d\) are constant.

This equation can be derived using a direction vector \(\vec{n}\), and a fixed point on the plane \(P_0\):

\[\vec{n} \cdot \vec{P_0P} = 0\]

• \(\vec{n} = \big< a, b, c \big>\) is a normal vector perpendicular to the plane.
• \(P_0(x_0, y_0, z_0)\) is a fixed point on the plane — \(x_0, y_0, z_0\) are constant.
• \(P(x, y, z)\) is a second point — \(x, y, z\) are variables.

Since \(P_0\) is a point in the plane, the point \(P\) is also in the plane when \(\vec{P_0P}\) is orthogonal to \(\vec{n}\).

Assumptions

Let \(\text{P}\) be the plane with normal vector \(\vec{n}\), passing through the point \(P_0\):

\[\begin{align*}
\vec{n} &= \big< 3, 4, 6 \big>\\
P_0 &= ( 2, 0, 1 )
\end{align*}\]

Instructions

1. Graph & Compute (example)

Go to graph 2 and identify:

• The normal vector \(\vec{n} = \big< 3, 4, 6 \big>\) (red)
• The fixed point \(P_0( 2, 0, 1 )\) (blue)
• The second point \(P( x, y, z )\) (green)
• The vector \(\vec{P_0P} = \big< 2 - x, 0 - y_0, 1 - z_0 \big>\) from \(P_0\) to \(P\) (green)

Then, set up and evaluate the vector equation:

\[\vec{n} \cdot \vec{P_0P} = 0\]

To show that the equation for the plane \(\text{P}\) is:

\[3x + 4y + 6z = 12\]

What’s the relationship between this equation of the plane, and the normal vector \(\vec{n}\) used in the equation \(\vec{n} \cdot \vec{P_0P} = 0\)?

2. Construct Examples

Using the equation:

\[\vec{n} \cdot \vec{P_0P} = 0\]

• Find the equation (\(ax + by + cz = d\)) of the plane that passes through the point \(P_0(1, 2, 3)\), and is parallel to the plane \(4x - 2y + 7z = 42\).
• Find the equation (\(ax + by + cz = d\)) of the plane that passes through the origin, and is perpendicular to the line \(\vec{r}(t) = \big< 2t + 1, t - 3, 5t + 4 \big>\).

A line perpendicular to a plane is called a normal line.

3. Challenge

Using the equation:

\[\vec{n} \cdot \vec{P_0P} = 0\]

find the equation (\(ax + by + cz = d\)) of the plane that contains the origin \(O(0,0,0)\) and the points \(P(1,2,-1)\) and \(Q(3,0,5)\).

BIG IDEAS

Vector Equations of Lines (exercise 1)

Lines in any dimension can be described by a direction vector \(\vec{v}\), and a fixed point on the line \(P_0\):

\[\vec{r}(t) = t\vec{v} + \vec{r}_0\]

• \(\vec{v}\) is a direction vector parallel to the line.
• \(P_0\) is a fixed point on the line with position vector \(\vec{r}_0\).
• \(t\) is the parameter (variable).

Parametric equations of a vector equation come from the coordinate functions (or component functions) of \(\vec{r}(t)\).

For a line in 3-space with direction \(\vec{v} = \big< a, b, c \big>\) and fixed point \(P_0 = ( x_0, y_0, z_0 )\):

\[\vec{r}(t) = t \vec{v} + \vec{r}_0 = \big< at + x_0, bt + y_0, ct + z_0 \big>\]

And the parametric equations are:

\[\begin{align*}
x(t) &= at + x_0\\
y(t) &= bt + y_0\\
z(t) &= ct + z_0
\end{align*}\]

Parallel lines have parallel direction vectors.

More generally: The angle between two lines is the angle between their direction vectors.

Planes & Normal Vectors (exercise 2)

Planes in 3-space are described by the equation:

\[ax + by + cz = d\]

where \(a, b, c, d\) are constant.

This equation can be derived using a direction vector \(\vec{n}\), and a fixed point on the plane \(P_0\):

\[\vec{n} \cdot \vec{P_0P} = 0\]

• \(\vec{n} = \big< a, b, c \big>\) is a normal vector perpendicular to the plane.
• \(P_0(x_0, y_0, z_0)\) is a fixed point on the plane — \(x_0, y_0, z_0\) are constant.
• \(P(x, y, z)\) is a second point — \(x, y, z\) are variables.

The coefficients in the equation of a plane correspond to the components of a normal vector to the plane:

\[ax + by + cz = d \quad \iff \quad \vec{n} = \big< a, b, c \big>\]

Parallel planes have parallel normal vectors.

More generally: The angle between two planes is the angle between their normal vectors.

The angle between a plane and a line is the angle between the normal vector of the plane and the direction vector of the line.

A line perpendicular to a plane is called a normal line.