Welcome back, everyone. So far, we've been building our vector toolkit. We've mastered the dot product, which tells us about alignment and angles, and the cross product, which gives us perpendicular vectors and areas. Today, we're going to use these tools to do something fundamental: describe lines and planes in 3D space.
In a 2D plane, you defined a line with a point and a slope. But how would you define the 'slope' of a line floating in 3D space? The concept doesn't quite work. We need a more robust way to define orientation, and that's where vectors come in.
It turns out there are two key recipes we need to remember:
This is where our dot and cross products become the essential tools for the job. They are the exact building blocks we need to construct these geometric descriptions.
Let's start with lines. Imagine a line floating in space. To describe it, we first need to anchor it with a point. Let's call that point $P_0$. But that's not enough; there are infinitely many lines that go through $P_0$. We also need to specify its direction. We can do that with a vector, $\mathbf{v}$, that is parallel to our line.
Imagine you're at the origin and want to get to any point $P$ on a specific line. Your journey has two parts:
Adding these two parts gives us the position vector $\mathbf{r}$ of any point on the line:
$$\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$$The variable $t$ is called the parameter. As $t$ changes, the vector $\mathbf{r}(t)$ traces out the entire line.
The vector equation is great for theory, but for computation, we usually break it into its components. This gives us the parametric equations.
If we let our known point be $P_0(x_0, y_0, z_0)$ and our direction vector be $\mathbf{v} = \langle a, b, c \rangle$, then the vector equation $\langle x, y, z \rangle = \langle x_0, y_0, z_0 \rangle + t\langle a, b, c \rangle$ becomes:
If we solve each of these for the parameter $t$ (assuming $a,b,c$ are not zero) and set them equal, we get the symmetric equations:
The symmetric form is a nice, clean way to represent a line without a parameter. Its main value is that it gives us a quick way to check if a point lies on the line. If you plug the point's $(x,y,z)$ coordinates into the symmetric equations and all three fractions are equal, then the point is on the line. It's also useful for seeing the direction numbers $\langle a,b,c \rangle$ at a glance.
Often, you won't be given a direction vector directly, but two points, say $A(a_1, a_2, a_3)$ and $B(b_1, b_2, b_3)$. Remember that the vector that connects them is their displacement vector, which is found by subtracting the coordinates of the initial point from the terminal point.
$$\vec{AB} = \langle b_1-a_1, b_2-a_2, b_3-a_3 \rangle$$This displacement vector gives us the perfect direction vector for the line passing through A and B.
Find the vector, parametric, and symmetric equations of the line that passes through the points A(1, 2, -1) and B(3, 0, 4).
Step 1: Find a point on the line. We can use either A or B. Let's choose A(1, 2, -1). So, $\mathbf{r}_0 = \langle 1, 2, -1 \rangle$.
Step 2: Find a direction vector. The vector that goes from A to B, $\vec{AB}$, will be parallel to the line.
$$\mathbf{v} = \vec{AB} = \langle 3-1, 0-2, 4-(-1) \rangle = \langle 2, -2, 5 \rangle$$
Step 3: Write the equations.
Problem: Find the parametric equations for the line through the point (-2, 0, 4) and parallel to the vector $\mathbf{v} = \langle 2, 4, -2 \rangle$. At what point does this line intersect the xy-plane?
Now, let's think about planes. A point isn't enough, and a direction vector isn't enough either. The key ingredient for a plane is a vector that is perpendicular to it. We call this the normal vector, $\mathbf{n}$.
Let's say we know a point $P_0$ is in our plane and we have a normal vector $\mathbf{n}$ that is perpendicular to the plane. For any other point $P$ in that plane, the vector connecting them, $\vec{P_0P}$, must lie flat in the plane.
What does that mean for its relationship with $\mathbf{n}$? They must be orthogonal! And our test for orthogonality is the dot product. The dot product of the normal vector and the vector in the plane must be zero.
If $\mathbf{r}_0$ is the position vector of $P_0$ and $\mathbf{r}$ is the position vector of $P$, then $\vec{P_0P} = \mathbf{r} - \mathbf{r}_0$. This gives us the vector equation:
$$\mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0$$If we let the normal vector be $\mathbf{n} = \langle a, b, c \rangle$ and expand the dot product, we get the more common scalar equation of the plane:
If we multiply this out and group the constants, we get the general form: $ax + by + cz + d = 0$. This is a crucial takeaway: whenever you see a linear equation of a plane, you can immediately identify its normal vector! It's simply the vector of the coefficients: $\mathbf{n} = \langle a, b, c \rangle$.
Find the equation of the plane passing through (1, 4, 5) with normal vector $\mathbf{n} = \langle 3, 2, -1 \rangle$.
We have our point $(x_0, y_0, z_0) = (1, 4, 5)$ and our normal vector $\langle a, b, c \rangle = \langle 3, 2, -1 \rangle$. We just plug them into the scalar equation:
$$3(x - 1) + 2(y - 4) - 1(z - 5) = 0$$
$$3x - 3 + 2y - 8 - z + 5 = 0$$
$$3x + 2y - z - 6 = 0$$
Find the equation of the plane that passes through the three points P(1, 3, 2), Q(3, -1, 6), and R(5, 2, 0).
Step 1: Find the normal vector. We don't have a normal vector, but we can make one! The vectors $\vec{PQ}$ and $\vec{PR}$ both lie in the plane. Their cross product will be perpendicular to both, giving us our normal vector $\mathbf{n}$.
$$\vec{PQ} = \langle 3-1, -1-3, 6-2 \rangle = \langle 2, -4, 4 \rangle$$
$$\vec{PR} = \langle 5-1, 2-3, 0-2 \rangle = \langle 4, -1, -2 \rangle$$
$$\mathbf{n} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -4 & 4 \\ 4 & -1 & -2 \end{vmatrix} = \langle 12, 20, 14 \rangle$$
Step 2: Use a point and the normal vector. Now we have $\mathbf{n} = \langle 12, 20, 14 \rangle$ and we can use any of the three points. Let's use P(1, 3, 2).
$$12(x - 1) + 20(y - 3) + 14(z - 2) = 0$$
$$12x - 12 + 20y - 60 + 14z - 28 = 0$$
$$12x + 20y + 14z - 100 = 0$$
We can simplify by dividing by 2: $6x + 10y + 7z - 50 = 0$.
Problem: Find the equation of the plane that contains the point (1, -1, 2) and is parallel to the plane $3x - y + 5z = 2$.
Now let's put it all together to see how these objects interact.
Think of the parametric equations of a line, like $\langle x, y, z \rangle = \langle 1+t, 2-t, 4t \rangle$, as a recipe for generating points. The parameter $t$ is like a time variable. At $t=0$, you're at the starting point, $(1, 2, 0)$. At $t=1$, you've moved one full direction vector away to the point $(1+1, 2-1, 4(1))$, which is $(2, 1, 4)$. The line is the path of all points for all possible values of $t$.
A plane's equation, like $x+y+z=12$, is a rule. It says 'you're only on my plane if your coordinates add up to 12'. To find where the line intersects the plane, we need to find the one specific moment in time, the one value of $t$, where the line's point satisfies the plane's rule.
Find the point where the line $\langle x, y, z \rangle = \langle 1+t, 2-t, 4t \rangle$ intersects the plane $x+y+z=12$.
Step 1: Substitute the line's equations into the plane's equation. We replace $x, y,$ and $z$ in the plane's equation with their recipes in terms of $t$.
$$(1+t) + (2-t) + (4t) = 12$$
Step 2: Solve for the parameter $t$. This gives us the 'time' of collision.
$$1 + t + 2 - t + 4t = 12$$
$$3 + 4t = 12 \implies 4t = 9 \implies t = 9/4$$
Step 3: Find the point. Now that we know the exact moment of intersection, we plug this value of $t$ back into the line's parametric equations to find the coordinates.
$$x = 1 + \frac{9}{4} = \frac{13}{4}$$
$$y = 2 - \frac{9}{4} = \frac{8-9}{4} = -\frac{1}{4}$$
$$z = 4\left(\frac{9}{4}\right) = 9$$
The point of intersection is $(\frac{13}{4}, -\frac{1}{4}, 9)$.
Two distinct planes in 3D space are either parallel or they intersect in a straight line. To describe that line, we need our two ingredients: a direction vector and a point.
Direction Vector: The line of intersection lies in both planes. This means it must be perpendicular to both of the planes' normal vectors. What tool gives us a vector perpendicular to two others? The cross product! So, the direction vector of the line is $\mathbf{v} = \mathbf{n}_1 \times \mathbf{n}_2$.
A Point on the Line: This part is a little trickier. We need to find any single point that satisfies both plane equations simultaneously. A common strategy is to simplify the problem by setting one variable to zero (say, $z=0$) and then solving the resulting 2x2 system for $x$ and $y$.
Find the parametric equations for the line of intersection of the planes $P_1: x+y+z=1$ and $P_2: x-2y+3z=1$.
Step 1: Find the direction vector. First, we identify the normal vectors from the coefficients: $\mathbf{n}_1 = \langle 1, 1, 1 \rangle$ and $\mathbf{n}_2 = \langle 1, -2, 3 \rangle$. The direction vector $\mathbf{v}$ is their cross product.
$$\mathbf{v} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix} = \langle 5, -2, -3 \rangle$$
Step 2: Find a point on the line. Let's set $z=0$ in both equations to see where the line crosses the xy-plane.
$$x+y = 1$$
$$x-2y = 1$$
Subtracting the second equation from the first gives $3y=0$, so $y=0$. Plugging this back in gives $x=1$. So, a point on the line is $(1, 0, 0)$.
Step 3: Write the parametric equations. Using the point $(1,0,0)$ and direction $\langle 5, -2, -3 \rangle$:
$$x = 1 + 5t, \quad y = -2t, \quad z = -3t$$
Problem: Find the angle between the two planes from the previous example, $x+y+z=1$ and $x-2y+3z=1$. (Hint: The angle between the planes is the angle between their normal vectors!).
Today, we've seen how to describe fundamental geometric objects—lines and planes—using the language of vectors. A line is defined by a point and a parallel direction vector. A plane is defined by a point and a perpendicular normal vector. We've put our dot and cross products to work to find equations, intersections, and angles, which are the building blocks for so much of what's to come.
In our next lectures, we'll move on from these 'flat' objects to 'curved' ones, exploring vector-valued functions and the calculus of motion through space. See you then! 🚀