91Ó°ÊÓ

The y-intercept of a line is where the line crosses the y-axis. This is a vital part of understanding the behavior of linear equations, especially when visualizing them. Generally, for any function, the intercept can be found by determining the point where the non-y variables (here, represented by x) are set to zero.

When dealing with vector equations, such as \( \mathbf{r} = (3 + 2t) \mathbf{i} + 5t \mathbf{j} \), the process to find the y-intercept is straightforward:

• Set the parameter \( t = 0 \).
• Substitute \( t = 0 \) into the equation.
• Evaluate the resulting vector components.

For the equation in our example, plugging \( t = 0 \) directly gives \( \mathbf{r} = 3 \mathbf{i} + 0 \mathbf{j} \). However, this indicates a slight oversight, since real observation suggests we need to ensure that x truly hits zero, guiding to redo the calculation properly.

Finally, upon proper evaluation, we determine that setting \( t = \frac{-3}{2} \) gives a y-intercept at a different correct condition which actually renders y itself. This shows how critical cautious computation is in vector mathematics.

Finding where a line intersects a plane involves comparing their equations and determining the common point. In 3D space, a line can be given in terms of a parameter, as in the equation \( \mathbf{r} = t \mathbf{i} + (1 + 2t) \mathbf{j} - 3t \mathbf{k} \). Simultaneously, a plane's equation can be expressed in the form of \( 3x - y - z = 2 \).

To find the intersection, the task is to substitute the line's parametric expressions into the plane's equation. This may look complex, but it boils down to these steps:

• Use the parametric representation: \( x = t \), \( y = 1 + 2t \), \( z = -3t \).
• Substitute these into the plane's equation.
• Solve the resulting equation for the parameter \( t \).

Upon substitution, the equation evolves into \( 4t - 1 = 2 \). Solving this gives \( t = \frac{3}{4} \).

Thus, to locate the exact intersection point, insert \( t = \frac{3}{4} \) back into each parametric component. The coordinates of the intersection are hence \( \left( \frac{3}{4}, 2.5, -2.25 \right) \). This method systematically reveals the precise meeting point of the given line and plane.

Parametric equations enable us to express geometric entities, like lines and curves, by assigning a set of parameters. Such representations can swiftly solve complex systems, especially in vector spaces.

In 3-dimensional space, parametric equations are very concise in defining lines. A parameter, generally denoted by \( t \), allows the equations:

• \( x = x_0 + at \)
• \( y = y_0 + bt \)
• \( z = z_0 + ct \)

where \( (x_0, y_0, z_0) \) is a point on the line, and \( (a, b, c) \) is the direction the line points towards.

This is especially useful when solving intersections or simulating trajectories since each specific value of \( t \) corresponds to a unique point on the line or curve. In our exercise's context, we saw \( \mathbf{r} = t \mathbf{i} + (1 + 2t) \mathbf{j} - 3t \mathbf{k} \) was broken down to understand and compute intersections with a plane, showcasing the practical benefits of this form.