3D Geometry: Finding Points On A Line

Hey guys! Today, we're diving into a super interesting problem involving points and lines in 3D space. We've got a point P with coordinates (p, 1, q) that lies on the line passing through two other points, A(3, -1, 4) and B(4, -3, -1). Our mission, should we choose to accept it, is to find the values of p and q. Sounds like fun, right? Let's break it down!

The Problem: Finding p and q

The core of this problem lies in understanding the relationship between points on a line in three-dimensional space. Understanding this relationship is fundamental not just for math problems, but also for various applications in computer graphics, physics, and engineering. The main concept we'll be using is that if a point P lies on a line defined by two points A and B, then the vector AP is a scalar multiple of the vector AB. This means that the direction from A to P is the same as the direction from A to B, just scaled by some factor. We can express this mathematically, but before we jump into the equations, let's visualize what's happening. Imagine a straight line slicing through 3D space. We have two fixed points, A and B, defining this line. Now, picture point P somewhere along this line. The position of P is determined by how far along the line it is, relative to A and B. This 'how far' is what our scalar multiple represents. It's a factor that stretches or shrinks the vector AB to match the vector AP. So, the key here is not just to crunch numbers, but to really grasp this geometric relationship. Think about it this way: if the scalar multiple is 0, P is at A. If it's 1, P is at B. If it's 0.5, P is halfway between A and B. And so on. This intuitive understanding will not only help in solving this particular problem but will also build a solid foundation for tackling more complex concepts in linear algebra and geometry. This foundational understanding allows us to tackle a variety of problems involving spatial relationships, and it is also key in fields like robotics, where understanding spatial movement and planning is crucial.

Step 1: Finding the Direction Vector AB

First things first, we need to figure out the direction vector of the line. To find the direction vector AB, we subtract the coordinates of point A from the coordinates of point B. This gives us the vector that points from A to B. So, let's do the math:

AB = B - A = (4 - 3, -3 - (-1), -1 - 4) = (1, -2, -5)

What we've just calculated, the vector (1, -2, -5), is more than just a set of numbers; it's a directional arrow in 3D space. Think of it as the 'blueprint' for the line's orientation. It tells us how the line stretches out along the x, y, and z axes. The numbers 1, -2, and -5 are the components of this vector, and each one is super important. The '1' tells us that for every unit we move along the line in the x-direction, we also move -2 units in the y-direction and -5 units in the z-direction. These components are the line's 'slope' in each dimension, and they're the key to understanding how the line is positioned in space. Visualizing this vector can be super helpful. Imagine starting at point A and then moving 1 unit along the x-axis, then -2 units along the y-axis, and finally -5 units along the z-axis. You'll end up exactly at point B. This vector AB is the direct path, the straightest route, from A to B. This simple subtraction has unlocked a wealth of information about the line's direction. This vector is going to be our main tool for finding other points on the line, like our mysterious point P. Remember, this vector isn't just specific to points A and B; it represents the direction of the entire line. Any segment of this line will be parallel to this vector, and this is a crucial concept that we'll use to find the coordinates of point P.

Step 2: Expressing AP as a Scalar Multiple of AB

Now comes the cool part. Since P lies on the line AB, the vector AP must be a scalar multiple of AB. Expressing AP as a scalar multiple means there's a number, let's call it 't', such that AP = t * AB. This 't' is our scaling factor, telling us how much to stretch or shrink AB to get to AP. First, let's find the vector AP:

AP = P - A = (p - 3, 1 - (-1), q - 4) = (p - 3, 2, q - 4)

Now, we set AP equal to t * AB:

(p - 3, 2, q - 4) = t * (1, -2, -5)

This equation is the heart of our solution. It’s saying that the vector from A to P is just a scaled version of the vector from A to B. The scalar 't' is like a dial that we can turn to move along the line. If t = 0, then P is at A. If t = 1, then P is at B. If t = 0.5, then P is halfway between A and B. And so on. This concept of scalar multiplication is so powerful because it allows us to describe an infinite number of points along the line using just one equation and one variable, 't'. To truly grasp this, imagine the vector AB as a rubber band. Multiplying it by 't' is like stretching or compressing that rubber band. The direction stays the same, but the length changes. If 't' is negative, we're stretching the rubber band in the opposite direction. So, 't' is our magic number that connects point P to the line defined by A and B. But how do we find 't'? Well, we've turned our geometric problem into an algebraic one. We now have an equation with three components, each representing a direction in space. And, as we'll see in the next step, this gives us a system of equations that we can solve. This is a common technique in 3D geometry: transforming spatial relationships into algebraic equations. It's a powerful way to solve problems, and it's a skill that will come in handy in many different contexts.

Step 3: Solving for t, p, and q

Now we have a vector equation, which we can break down into three separate equations by equating the corresponding components:

p - 3 = t

2 = -2t

q - 4 = -5t

Look at that! We've transformed our single vector equation into a system of three simple algebraic equations. Solving this system of equations is now the key to unlocking the values of p, q, and t. The second equation, 2 = -2t, is our starting point because it only involves 't'. We can easily solve for 't' by dividing both sides by -2:

t = -1

Fantastic! We've found our scaling factor. This 't' = -1 has a geometric interpretation: it tells us that point P is located on the line, but in the opposite direction from B, relative to A. It's like going 'backwards' along the line. Now that we know 't', we can plug it into the other two equations to find p and q. Let's start with the first equation:

p - 3 = t p - 3 = -1 p = 2

So, we've found p! It's equal to 2. This means that the x-coordinate of point P is 2. Now, let's use the third equation to find q:

q - 4 = -5t q - 4 = -5(-1) q - 4 = 5 q = 9

And there we have it! We've found q, which is equal to 9. The z-coordinate of point P is 9. This step-by-step process of breaking down a vector equation into scalar equations is a fundamental technique in linear algebra. It allows us to solve for multiple unknowns by systematically isolating them. And it's not just useful in geometry; it's also a key skill in physics, engineering, and computer science, where systems of equations are used to model and solve a wide variety of problems. The beauty of this method is its clarity and precision. By carefully tracking each step, we can be confident in our results. And now that we've found p and q, we're ready to state our solution.

Step 4: Stating the Solution

So, after all that awesome math, we've found that p = 2 and q = 9. Stating the solution clearly is important, because it shows that we've understood the problem and its requirements. This corresponds to option C, which is (2, 9).

Therefore, the values of p and q are 2 and 9, respectively. We did it!

Final Thoughts

This problem was a great exercise in understanding 3D geometry and vector operations. Remember, the key is to break down complex problems into smaller, manageable steps. By visualizing the problem and understanding the underlying concepts, you can tackle even the trickiest questions. Keep practicing, and you'll become a 3D geometry master in no time!