Animated Solution for Mathematics - Three Dimensional Geometry: The line passing through the points $(5,1,a)$ and $(3,b,1)$ crosses the $yz$-plane at the point $(0,\frac{17}{2},-\frac{13}{2})$. Then

Imagine standing in a vast, three-dimensional room. You have two fixed points, $P(5,1,a)$ and $Q(3,b,1)$, floating in the air. A straight, invisible wire connects them, stretching out into infinity.

This wire is our line. We are told that this wire pierces through a specific 'wall' in our room—the $yz$-plane—at a precise location, $R(0,\frac{17}{2},-\frac{13}{2})$.

Our mission is to uncover the hidden values of $a$ and $b$ that define the positions of our starting points. It sounds like a puzzle, doesn't it? Let's solve it.

The $yz$-Plane

Before we dive into the algebra, let's look at the geometry. The $yz$-plane is a fundamental boundary.

In our 3D coordinate system, any point that sits on this plane has a very specific trait: its $x$-coordinate is absolutely, undeniably $0$. This is our master key.

The problem gives us the intersection point $R(0,\frac{17}{2},-\frac{13}{2})$. The fact that the first coordinate is $0$ confirms it lies on the $yz$-plane. We will use this fact to unlock the entire problem.

To connect the points $P$ and $Q$ to the intersection point $R$, we need the equation of the line. We use the two-point form of a line in 3D space:

Substituting our points $P(5,1,a)$ and $Q(3,b,1)$, we get:

Simplifying the first denominator, we have $\frac{x-5}{-2}=\frac{y-1}{b-1}=\frac{z-a}{1-a}$. Now, to make this manageable, we introduce a parameter, $\lambda$. By setting each fraction equal to $\lambda$, we can express any point $(x,y,z)$ on this line as a function of $\lambda$:

This is the 'parametric bridge.' It allows us to travel along the line just by changing the value of $\lambda$.

We know that at the intersection point $R$, the $x$-coordinate is $0$. So, we set our expression for $x$ to $0$:

With $\lambda =\frac{5}{2}$ in our pocket, we can now find $b$ and $a$ by looking at the $y$ and $z$ coordinates of $R$. For $y$:

And for $z$:

And there we have it! Through the elegance of parametric geometry, we have found that $a=6$ and $b=4$. The line is defined, the points are anchored, and the mystery is solved.