Three-dimensional vector geometry

Welcome to the World of 3D Vectors!

In our previous look at vectors, we mostly dealt with 2D space or basic operations. Now, we are stepping into Three-Dimensional Vector Geometry. Think of this as the "GPS of Mathematics." Whether you are designing a skyscraper, programming a 3D video game, or navigating a flight path, you need to understand how lines and planes interact in space.

Don’t worry if this seems a bit abstract at first! We will break down every concept into simple steps. By the end of these notes, you’ll be able to visualize and calculate exactly how objects sit in 3D space. Let's dive in!

1. Lines in 3D Space

To define a straight line in 3D, we need two things: a starting point (position vector) and a direction to travel in.

A. Vector Equation of a Line

Imagine you are at point \(A\) and you want to walk in the direction of vector \(\mathbf{d}\). Your position at any time depends on how far you’ve walked. The equation is:
\(\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}\)
Where:
• \(\mathbf{r}\) is the position vector of any point on the line.
• \(\mathbf{a}\) is the position vector of a known point on the line.
• \(\mathbf{d}\) is the direction vector (the "compass" of the line).
• \(\lambda\) is a scalar parameter (a number that tells you how far along the line you are).

B. Cartesian Equation of a Line

If we write out the components of the vector equation for \(x, y,\) and \(z\), and then solve for \(\lambda\), we get the Cartesian form:
\(\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3}\)
Example: If a line passes through (1, 2, 3) with direction (4, 5, 6), the equation is \(\frac{x-1}{4} = \frac{y-2}{5} = \frac{z-3}{6}\).

Quick Tip:

If one of the direction components is zero (e.g., \(d_3 = 0\)), the Cartesian form looks like this: \(\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2}, z = a_3\). This just means the line is "flat" relative to the z-axis!

Key Takeaway: A line is defined by a point and a direction. If two lines have the same direction vector (or multiples of each other), they are parallel.

2. Planes in 3D Space

A plane is a flat, 2D surface that extends forever in 3D space—like a giant sheet of paper.

A. Scalar Product Form (The "Normal" Form)

This is the most common way to represent a plane in H2 Math. To define a plane, we need a point on the plane and a Normal Vector (\(\mathbf{n}\)), which is a vector that sticks straight out of the plane at a 90-degree angle.
\(\mathbf{r} \cdot \mathbf{n} = d\)
Where \(d = \mathbf{a} \cdot \mathbf{n}\) (the dot product of a point on the plane and the normal).

B. Cartesian Equation of a Plane

If \(\mathbf{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}\), the equation is simply:
\(ax + by + cz = d\)
Example: If the normal vector is \(\begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix}\) and it passes through a point such that \(d=10\), the plane is \(2x - 3y + 5z = 10\).

Did you know?

You can instantly see the "tilt" of a plane just by looking at the coefficients of \(x, y,\) and \(z\). Those numbers are the components of the normal vector!

Key Takeaway: The normal vector \(\mathbf{n}\) is the "boss" of the plane. It tells you everything about the plane's orientation.

3. Angles between Lines and Planes

When calculating angles, we almost always use the Dot Product. Remember: \(\cos \theta = \frac{|\mathbf{u} \cdot \mathbf{v}|}{|\mathbf{u}| |\mathbf{v}|}\).

A. Angle between Two Lines

Use the direction vectors of the two lines (\(\mathbf{d}_1\) and \(\mathbf{d}_2\)).
\(\cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|}\)

B. Angle between Two Planes

The angle between two planes is the same as the angle between their normal vectors (\(\mathbf{n}_1\) and \(\mathbf{n}_2\)).
\(\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}\)

C. Angle between a Line and a Plane

Watch out! This one is a bit different. The angle \(\theta\) between the line (direction \(\mathbf{d}\)) and the plane (normal \(\mathbf{n}\)) uses sine instead of cosine:
\(\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}\)
Why? Because the dot product gives the angle between the line and the normal. Since the normal is 90° to the plane, we use the complementary angle (90 - \(\phi\)), which turns cosine into sine!

Key Takeaway: Line-Line and Plane-Plane use \(\cos \theta\). Line-Plane uses \(\sin \theta\).

4. Intersections and Relationships

A. Two Lines

In 3D, two lines can be:
1. Parallel: Directions are multiples.
2. Intersecting: They meet at one point.
3. Skew: They are not parallel but never meet (like two planes flying at different altitudes in different directions).
4. Coincident: They are actually the same line.

B. Line and Plane

To find where a line \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}\) hits a plane \(\mathbf{r} \cdot \mathbf{n} = d\):
1. Substitute the line's expression into the plane's equation.
2. Solve for \(\lambda\).
3. Plug \(\lambda\) back into the line equation to find the point of intersection.

C. Two Planes

If two planes are not parallel, they intersect to form a straight line. You can find the direction of this line by taking the vector product (cross product) of the two normal vectors: \(\mathbf{d}_{line} = \mathbf{n}_1 \times \mathbf{n}_2\).

Key Takeaway: If the dot product of a line's direction and a plane's normal is zero (\(\mathbf{d} \cdot \mathbf{n} = 0\)), the line is parallel to the plane.

5. Perpendiculars and Distances

A. Foot of the Perpendicular (Point to Line)

To find the "shadow" of a point \(P\) on a line:
1. Let \(F\) be the foot of the perpendicular on the line. Express \(F\) using the line's parameter \(\lambda\).
2. Create vector \(\vec{PF}\).
3. Since \(\vec{PF}\) is perpendicular to the line, set \(\vec{PF} \cdot \mathbf{d} = 0\).
4. Solve for \(\lambda\) and find the coordinates of \(F\).

B. Foot of the Perpendicular (Point to Plane)

To find the foot of the perpendicular \(F\) from point \(P\) to a plane:
1. Create a "mini-line" that starts at \(P\) and goes in the direction of the plane's normal \(\mathbf{n}\).
2. Find where this mini-line intersects the plane. That intersection is \(F\)!

C. Distance from a Point to a Plane

If you have a point \(P(x_1, y_1, z_1)\) and a plane \(ax + by + cz = d\), the shortest distance is:
Length = \(\frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}\)
This is just the length of the vector \(\vec{PF}\)!

Common Mistake: Forgetting the absolute value signs in the distance formula. Distance can never be negative!

Key Takeaway: Whenever you hear "shortest distance" or "perpendicular," think Dot Product = 0 or use the specific formulas provided.

Summary Checklist

• Can I convert between Vector and Cartesian forms for lines and planes?
• Do I remember to use \(\sin \theta\) for Line-Plane angles?
• Can I find the intersection point of a line and a plane by substituting \(\lambda\)?
• Do I know that skew lines are non-parallel lines that never meet?
• Am I comfortable finding the foot of a perpendicular using the dot product?

Great job! 3D Geometry is all about visualization. If you get stuck, try drawing a quick sketch of the line and the normal vector to help you see the relationship. You've got this!