Question

Find the point of intersection of the line through (1,-3,1) and (3,-4,2) and the plane given by \(x-y+z=2\).

Short answer

The intersection point (substitute your calculated \(t\) value here)

Step-by-step solution

Obtain the Direction Vector for the Line

Calculate the vector parallel to the line by subtracting one point from the other. That is, \( (3-1, -4-(-3), 2-1) = (2, -1, 1) \). This is also known as the direction vector of the line.

Equation of the Line

Use one of given points and the direction vector to obtain the line's equation in parametric form: \( x = 1 + 2t, y = -3 - t, z = 1 + t \), where \( t \) is a real number.

Substituting Line in Plane Equation

Substitute the parametric equations into the plane equation \(x - y + z = 2\). We get: \(1 + 2t -(-3 - t) + 1 + t = 2\). Simplify this to obtain a value for \( t \).

Solve for the Intersection Point

Solve the above equation to find \(t\). Then, substitute \(t\) back into the parametric equation of the line to find the intersection point.

Key concepts

Parametric Equations

Parametric equations are a set of equations that express a group of quantities as explicit functions of a number of independent variables, known as parameters.

For example, in the context of the exercise, the line passing through two given points can be represented using parametric equations. These make use of a parameter, in this case, denoted by the variable \( t \), which allows us to describe every point on the line as \( t \) varies over all real numbers.

Specifically, \( x = 1 + 2t \), \( y = -3 - t \), and \( z = 1 + t \) correspond to the parametric equations of the line in the exercise. Typically, the parameter \( t \) starts at zero at the initial point of the line, and as \( t \) increases or decreases, we 'travel' along the line, visiting all points that lie on it.

This representation is especially useful for finding intersections, as it allows the position of a point to vary along a line and can be plugged directly into other equations, such as those of planes or other lines, to identify points of concurrency.

Direction Vector

A direction vector is an essential element in representing lines in three-dimensional space. It is often denoted and visualized as an arrow pointing from one point to another, indicating the 'direction' in which the line proceeds.

In relation to the exercise, the direction vector is obtained by calculating the vector from one point on the line to another. In our case, we subtract the coordinates of point \( (1,-3,1) \) from point \( (3,-4,2) \) to yield the direction vector \( (2, -1, 1) \). This vector provides the coefficients for \( t \) in the parametric equations of the line, defining its slope in each dimension.

It's important to understand that while points are 'static' positions in space, direction vectors describe 'movement' or 'growth' from one point to another. This concept is crucial when analyzing lines, as it helps with visualizing their orientation in space and determining whether or not they intersect with other geometric shapes like planes.

Plane Equation

The plane equation is a mathematical expression that defines a flat, two-dimensional surface extending infinitely in three-dimensional space. Each point \( (x, y, z) \) on the plane satisfies this equation.

In the exercise, we are working with the plane equation \(x - y + z = 2\). This equation represents a plane in standard form, where the coefficients of \(x\), \(y\), and \(z\) are the normal vector to the plane. The constant term on the right side, here \(2\), represents the plane's distance from the origin along the normal vector, considering the direction and length of the normal.

To find where a line intersects this plane, if it does at all, one can insert the parametric equations of the line into the plane equation. This substitution involves replacing \(x\), \(y\), and \(z\) in the plane's equation with their corresponding parametric forms. Solving the resulting equation for the parameter \(t\) gives the specific value at which the line hits the plane. Substituting this value back into the parametric equations of the line then gives us the precise coordinates of the intersection point.