Course Information:
Date: January 14, 2026
Topics covered today: Lines and Planes (Section 13.5)
Next lesson: Quadric Surfaces (Section 13.6)
Find the equation of a line through points \(P(-1, -1)\) and \(Q(1, 3)\).
Solution:
First, we calculate the slope of the line:
\[\text{Slope} = \frac{3 - (-1)}{1 - (-1)} = \frac{4}{2} = 2\]Using point-slope form with point \(P(-1, -1)\):
\[y - (-1) = 2(x - (-1))\] \[\frac{y + 1}{x + 1} = 2\] \[y + 1 = 2(x + 1)\] \[y = 2x + 1\]We can also express this line in parametric form. The direction vector is parallel to the line:
\[\langle x, y \rangle = t\langle 1, 2 \rangle + \langle -1, -1 \rangle\]This gives us the parametric equations:
\[x = t - 1, \quad y = 2t - 1\]Diagram Description: The diagram shows a coordinate plane with the y-axis vertical and x-axis horizontal. Two points are marked: P at coordinates (-1, -1) and Q at coordinates (1, 3). A blue line passes through both points with positive slope. Red and purple arrows illustrate the direction vector and the concept that the line is parallel to its direction vector.
Key Concept: The direction vector \(\langle 1, 2 \rangle\) is parallel to the line and gives the "direction" of the line in space.
To define a line in three-dimensional space, we need two pieces of information: a point on the line and a direction vector.
Definition: A line in 3D space can be defined by:
Let \(Q(x, y, z)\) be any point on the line. Then the vector \(\overrightarrow{PQ}\) is parallel to the direction vector \(\vec{v}\). This means:
\[\overrightarrow{PQ} = t\vec{v}\]where \(t\) is a scalar parameter. Expanding this relationship:
\[\langle x - x_0, y - y_0, z - z_0 \rangle = t\langle a, b, c \rangle\]3D Coordinate System Description: The diagram shows a three-dimensional coordinate system with x, y, and z axes. A point P is marked at coordinates (x₀, y₀, z₀), and a red vector extends from P in the direction of the line. A purple direction vector \(\vec{v}\) is shown, and another point Q(x, y, z) is marked on the line. Blue vectors illustrate the parallel relationship between \(\overrightarrow{PQ}\) and the direction vector.
Vector Form: The vector form of a line through point \(P(x_0, y_0, z_0)\) with direction vector \(\vec{v} = \langle a, b, c \rangle\) is:
\[\langle x, y, z \rangle = t\langle a, b, c \rangle + \langle x_0, y_0, z_0 \rangle\]This can also be written as:
\[\langle x, y, z \rangle = \langle ta + x_0, tb + y_0, tc + z_0 \rangle\]Parametric Form: By equating components from the vector form, we obtain the parametric equations:
\[x = ta + x_0\] \[y = tb + y_0\] \[z = tc + z_0\]where \(t\) is the parameter that varies over all real numbers.
Symmetric Form: Solving each parametric equation for \(t\) and setting them equal gives the symmetric form:
\[t = \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}\]This is valid when \(a\), \(b\), and \(c\) are all nonzero.
Warning: If any component of the direction vector is zero, we cannot use that ratio in the symmetric form. Instead, we set the corresponding coordinate equal to its value at the point on the line.
Find the equation of the line passing through points \(P(1, 0, 2)\) and \(Q(-2, 4, -3)\).
Solution:
First, we find the direction vector \(\overrightarrow{PQ}\):
\[\overrightarrow{PQ} = \langle -2 - 1, 4 - 0, -3 - 2 \rangle = \langle -3, 4, -5 \rangle\]Alternatively, we could use \(\overrightarrow{QP} = \langle 3, -4, 5 \rangle\). Note that we can choose either of the two points as our reference point, and we can use either direction vector (they differ only by a factor of -1).
Vector Form: Using point \(P(1, 0, 2)\) and direction vector \(\langle -3, 4, -5 \rangle\):
\[\langle x, y, z \rangle = t\langle -3, 4, -5 \rangle + \langle 1, 0, 2 \rangle\]Or equivalently, using point \(Q(-2, 4, -3)\) and direction vector \(\langle 3, -4, 5 \rangle\):
\[\langle x, y, z \rangle = t\langle 3, -4, 5 \rangle + \langle -2, 4, -3 \rangle\]Parametric Form: Using the first vector form:
\[x = -3t + 1\] \[y = 4t\] \[z = -5t + 2\]Or using the second vector form:
\[x = 3t - 2\] \[y = -4t + 4\] \[z = 5t - 3\]Symmetric Form: From the first parametric form:
\[\frac{x - 1}{-3} = \frac{y - 0}{4} = \frac{z - 2}{-5}\]This can also be written as:
\[\frac{x - 1}{-3} = \frac{y}{4} = \frac{z - 2}{-5}\]3D Diagram Description: A three-dimensional coordinate system shows points P(1, 0, 2) and Q(-2, 4, -3) marked with dots. A blue line passes through both points. Red and purple vectors illustrate the direction vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{QP}\), showing they point in opposite directions but define the same line.
Important Note: You can choose any of the two given points as your reference point, and you can use either direction vector. All forms will represent the same line, though they may look different algebraically.
Two lines in three-dimensional space can have one of three relationships: they can be parallel, intersecting, or skew.
Definition: Two lines are parallel if their direction vectors are parallel (i.e., one is a scalar multiple of the other) and they have no points in common.
Parallel Lines Diagram: Two blue parallel lines are shown in 3D space, both extending in the same direction. Purple arrows labeled \(\vec{n}_1\) and \(\vec{n}_2\) show that their direction vectors point in the same direction. The lines never meet.
Definition: Two lines are intersecting if they have exactly one point in common.
Intersecting Lines Diagram: Two blue lines meet at a single point in 3D space. The lines cross each other at different angles, sharing one common point of intersection.
Definition: Two lines are skew if their direction vectors are not parallel and they have no points in common. Skew lines do not lie in the same plane.
Skew Lines Diagram: Two lines are shown in 3D space: one blue line extends horizontally, and one teal line extends at an angle. A point (1, 0, 1) is marked. The lines do not intersect and are not parallel; they pass by each other in different planes. A purple dashed vertical line suggests the perpendicular distance between them.
Consider two lines:
These lines are skew because their direction vectors \(\langle 0, 1, 1 \rangle\) and \(\langle 0, 1, 0 \rangle\) are not parallel (one is not a scalar multiple of the other), and they have no common point.
To define a plane in three-dimensional space, we need a point on the plane and a normal vector perpendicular to the plane.
Turn to your neighbor and discuss:
Definition: A plane in 3D space can be defined by:
Plane Diagram: A blue tilted plane is shown in 3D space with x, y, and z axes. A point P is marked on the plane at coordinates (x₀, y₀, z₀). A purple vector \(\vec{n}\) extends perpendicular to the plane from point P, representing the normal vector. Another point Q(x, y, z) is marked on the plane, and a red vector shows \(\overrightarrow{PQ}\) lying within the plane.
Let \(Q(x, y, z)\) be any point on the plane. The vector \(\overrightarrow{PQ}\) lies in the plane, so it must be perpendicular to the normal vector \(\vec{n}\). This gives us the fundamental equation:
\[\overrightarrow{PQ} \cdot \vec{n} = 0\]Expanding this relationship:
\[\langle x - x_0, y - y_0, z - z_0 \rangle \cdot \langle a, b, c \rangle = 0\]Equation of a Plane: The standard form of the equation of a plane through point \(P(x_0, y_0, z_0)\) with normal vector \(\vec{n} = \langle a, b, c \rangle\) is:
\[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\]This can be rearranged to the general form:
\[ax + by + cz = ax_0 + by_0 + cz_0\]or more simply:
\[ax + by + cz = d\]where \(d = ax_0 + by_0 + cz_0\).
Find the equation of the plane passing through point \((1, -1, 1)\) with normal vector \(\langle 2, 3, -1 \rangle\).
Solution:
We use the equation \(\overrightarrow{PQ} \cdot \vec{n} = 0\) where \(P(1, -1, 1)\) and \(\vec{n} = \langle 2, 3, -1 \rangle\):
\[\langle x - 1, y + 1, z - 1 \rangle \cdot \langle 2, 3, -1 \rangle = 0\]Expanding the dot product:
\[2(x - 1) + 3(y + 1) - 1(z - 1) = 0\]Simplifying:
\[2x - 2 + 3y + 3 - z + 1 = 0\] \[2x + 3y - z + 2 = 0\]Therefore, the equation of the plane is:
\[2x + 3y - z = -2\]Plane Diagram: A blue tilted plane is shown with point P(1, -1, 1) marked on it. A purple normal vector \(\langle 2, 3, -1 \rangle\) extends perpendicular to the plane from point P. The coordinate axes x, y, and z are shown for reference.
Find the equation of the plane containing points \(P(2, 3, -1)\), \(Q(3, 5, -1)\), and \(R(6, 2, 0)\).
Solution:
First, we need to find a normal vector to the plane. We can do this by taking the cross product of two vectors that lie in the plane.
Let's find \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\):
\[\overrightarrow{PQ} = \langle 3 - 2, 5 - 3, -1 - (-1) \rangle = \langle 1, 2, 0 \rangle\] \[\overrightarrow{PR} = \langle 6 - 2, 2 - 3, 0 - (-1) \rangle = \langle 4, -1, 1 \rangle\]The cross product \(\overrightarrow{PQ} \times \overrightarrow{PR}\) will be perpendicular to the plane:
\[\vec{n} = \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 0 \\ 4 & -1 & 1 \end{vmatrix}\] \[\vec{n} = \vec{i}(2 \cdot 1 - 0 \cdot (-1)) - \vec{j}(1 \cdot 1 - 0 \cdot 4) + \vec{k}(1 \cdot (-1) - 2 \cdot 4)\] \[\vec{n} = \vec{i}(2) - \vec{j}(1) + \vec{k}(-9)\] \[\vec{n} = \langle 2, -1, -9 \rangle\]Now using point \(P(2, 3, -1)\) and normal vector \(\langle 2, -1, -9 \rangle\):
\[\langle x - 2, y - 3, z + 1 \rangle \cdot \langle 2, -1, -9 \rangle = 0\]Expanding:
\[2(x - 2) - (y - 3) - 9(z + 1) = 0\] \[2x - 4 - y + 3 - 9z - 9 = 0\] \[2x - y - 9z - 10 = 0\]Therefore, the equation of the plane is:
\[2x - y - 9z = 10\]Three Points Diagram: A blue plane is shown with three points marked: P(2, 3, -1), Q(3, 5, -1), and R(6, 2, 0). Red and green vectors show \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\) lying in the plane. A purple vector perpendicular to the plane represents the normal vector obtained from the cross product.
Important Note: Instead of using point P, you could choose point Q or R as your reference point. All three will yield the same equation for the plane.
Definition: Two planes are parallel if their normal vectors are parallel (i.e., one is a scalar multiple of the other).
Parallel Planes Diagram: Two blue horizontal planes are shown stacked vertically in 3D space. Purple normal vectors \(\vec{n}_1\) and \(\vec{n}_2\) extend perpendicular to each plane, pointing in the same direction. A note indicates that the normal vectors are parallel, written as \(\vec{n}_1 \parallel \vec{n}_2\).
If two planes have equations \(a_1x + b_1y + c_1z = d_1\) and \(a_2x + b_2y + c_2z = d_2\), they are parallel if and only if:
\[\langle a_1, b_1, c_1 \rangle = k\langle a_2, b_2, c_2 \rangle\] for some scalar \(k\).Definition: Two planes are orthogonal (perpendicular) if their normal vectors are orthogonal.
Orthogonal Planes Diagram: Two blue planes intersect at a right angle in 3D space, forming a cross-like configuration. Purple normal vectors \(\vec{n}_1\) and \(\vec{n}_2\) extend perpendicular to their respective planes. A note indicates that the normal vectors are perpendicular, written as \(\vec{n}_1 \perp \vec{n}_2\).
If two planes have equations \(a_1x + b_1y + c_1z = d_1\) and \(a_2x + b_2y + c_2z = d_2\), they are orthogonal if and only if:
\[\langle a_1, b_1, c_1 \rangle \cdot \langle a_2, b_2, c_2 \rangle = 0\]Find the point of intersection of the two lines:
\[\vec{r}_1(t) = \langle 2t + 3, 4t + 2, 3t + 5 \rangle\] \[\vec{r}_2(s) = \langle s + 2, 3s - 1, -5s + 10 \rangle\]Solution:
At the point of intersection, the coordinates must be equal. We set the corresponding components equal to each other:
From the x-components:
\[2t + 3 = s + 2\] \[s = 2t + 1\]From the y-components:
\[4t + 2 = 3s - 1\]Substituting \(s = 2t + 1\):
\[4t + 2 = 3(2t + 1) - 1\] \[4t + 2 = 6t + 3 - 1\] \[4t + 2 = 6t + 2\] \[0 = 2t\] \[t = 0\]Therefore, \(s = 2(0) + 1 = 1\).
We should verify this using the z-components:
When \(t = 0\): \(z = 3(0) + 5 = 5\)
When \(s = 1\): \(z = -5(1) + 10 = 5\) ✓
The point of intersection can be found by plugging \(t = 0\) into \(\vec{r}_1\) or \(s = 1\) into \(\vec{r}_2\):
\[\text{Point of intersection} = (3, 2, 5)\]Intersection Diagram: The work shows two parametric line equations with specific components circled in different colors (blue for x-components, teal for y-components, purple for z-components). Arrows and calculations show how the parameters t and s are related and solved to find t = 0 and s = 1.
Find the equation of the line of intersection of the two planes:
\[x + 3y - 2z = 1\] \[x + y + z = 0\]Solution:
To find the line of intersection, we need to solve the system of two equations in three variables. We will express two variables in terms of the third variable.
Let's solve for x and y in terms of z. From the system:
\[x + 3y - 2z = 1\] \[x + y + z = 0\]From the second equation:
\[x + y = -z\]Subtracting the second equation from the first:
\[(x + 3y - 2z) - (x + y + z) = 1 - 0\] \[2y - 3z = 1\] \[y = \frac{1 + 3z}{2}\]Substituting back into \(x + y = -z\):
\[x + \frac{1 + 3z}{2} = -z\] \[x = -z - \frac{1 + 3z}{2}\] \[x = \frac{-2z - 1 - 3z}{2}\] \[x = \frac{-1 - 5z}{2}\]We can write the parametric equations by letting \(z = t\):
\[x = \frac{-1 - 5t}{2} = -\frac{1}{2} - \frac{5t}{2}\] \[y = \frac{1 + 3t}{2} = \frac{1}{2} + \frac{3t}{2}\] \[z = t\]The vector form of the line of intersection is:
\[\langle x, y, z \rangle = t\left\langle -\frac{5}{2}, \frac{3}{2}, 1 \right\rangle + \left\langle -\frac{1}{2}, \frac{1}{2}, 0 \right\rangle\]Or equivalently, using \(t' = 2t\):
\[\langle x, y, z \rangle = t'\langle -5, 3, 2 \rangle + \left\langle -\frac{1}{2}, \frac{1}{2}, 0 \right\rangle\]System of Equations Diagram: The work shows the two plane equations at the top. Below, algebraic manipulations are shown to solve for x and y in terms of z. Arrows and boxes highlight key steps in the solution process, showing how the variables are expressed parametrically.
Important Concept: The line of intersection of two planes can be found by expressing two of the coordinates as functions of the third coordinate (which becomes the parameter). The direction vector of this line is perpendicular to both normal vectors of the planes.