91Ó°ÊÓ

Parametric equations offer a unique way to represent lines, curves, and surfaces in coordinate systems. Unlike linear equations that directly relate coordinates like x and y, parametric equations express them in terms of a third parameter, often denoted as 't'. This method is particularly useful for capturing motion and changing quantities.

In our given problem, the line is described by a parametric equation: \( \mathbf{r}(t) = \langle 2t+1, -t+4, t-6 \rangle \). Here:

• The equation for \(x\) is \(x = 2t + 1\).
• The equation for \(y\) is \(y = -t + 4\).
• The equation for \(z\) is \(z = t - 6\).

By changing the value of \(t\), we can determine different points along this line. For any given \(t\), these equations yield a unique combination of \(x, y,\) and \(z\). This makes parametric forms hugely beneficial in establishing precise positions, particularly when calculating intersections as seen in the exercise.

Finding the intersection point between a plane and a line is often possible by substituting the parametric equations of the line into the equation of the plane. The goal is to find a specific value of the parameter, typically called \(t\), where the line intersects the plane.

In the problem we've tackled, the plane is given by \(y = -2\). We substitute the line's parametric equation for \(y\), which is \(-t + 4\), into the plane equation. This gives us the equation:

\(-t + 4 = -2\).

By solving this, we find the value for \(t\) that actually places the line exactly on the plane. This determination of the parameter \(t\) is crucial, as it will subsequently allow us to find the exact intersection point in 3D space. This process ensures the line touches the plane at a single distinct point, facilitating a deeper understanding of intersection calculations.

Solving linear equations is a fundamental skill necessary when finding intersection points. In our example, we encounter a simple linear equation from the substitution step: \(-t + 4 = -2\). Solving it step-by-step involves:

• First, subtracting 4 from both sides to isolate terms including \(t\). This gives us \(-t = -6\).
• Next, multiplying both sides by \(-1\) to solve for \(t\), resulting in \(t = 6\).

Linear problems like these are about one variable and require operations that preserve equality, carefully isolating the variable on one side.

Once \(t\) is known, it's straightforward to substitute it back into the original parametric equations. This lets us calculate the precise intersection coordinates: \(x = 13, y = -2, z = 0\). This step-by-step approach highlights the connection between linear equation solving and finding intersection points, showing how values calculated here feed back into providing the comprehensive answer.