Question

In each case, find all points of intersection of the given plane and the line \(\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}1 \\\ -2 \\ 3\end{array}\right]+t\left[\begin{array}{r}2 \\ 5 \\\ -1\end{array}\right]\) a. \(x-3 y+2 z=4\) b. \(2 x-y-z=5\) c. \(3 x-y+z=8\) d. \(-x-4 y-3 z=6\)

Short answer

a. \((\frac{1}{3}, \frac{13}{3}, \frac{10}{3})\), b. \((11, 23, -2)\), c. No intersection, d. \((\frac{11}{19}, \frac{22}{19}, \frac{61}{19})\).

Step-by-step solution

Understand the Intersection Problem

To find the intersection of a plane and a line, we substitute the parametric equations of the line into the equation of the plane. If possible, we solve for the parameter \( t \) which gives us the specific point or points of intersection.

Line Parametric Equations Setup

The given line is defined by the equations \( x = 1 + 2t \), \( y = -2 + 5t \), \( z = 3 - t \). These expressions will be substituted into each plane's equation.

Step 3a: Plane (a) Calculation

For the plane equation \( x - 3y + 2z = 4 \), substitute \( x = 1 + 2t \), \( y = -2 + 5t \), and \( z = 3 - t \) into the equation: \[(1 + 2t) - 3(-2 + 5t) + 2(3 - t) = 4\]. Simplify and solve for \( t \): \[1 + 2t + 6 - 15t + 6 - 2t = 4\], which simplifies to \[-15t - 1 = 4\], giving \( t = -\frac{1}{3} \). Substituting \( t \) back gives the point of intersection: \( x = \frac{1}{3} \), \( y = \frac{13}{3} \), \( z = \frac{10}{3} \).

Step 3b: Plane (b) Calculation

For the plane equation \( 2x - y - z = 5 \), substitute: \[2(1 + 2t) - (-2 + 5t) - (3 - t) = 5\]. Simplifying gives \[2 + 4t + 2 - 5t - 3 + t = 5\]. This results in \[t = 5\]. Substituting back gives the point: \( x = 11 \), \( y = 23 \), \( z = -2 \).

Step 3c: Plane (c) Calculation

For the plane equation \( 3x - y + z = 8 \), substitute: \[3(1 + 2t) - (-2 + 5t) + (3 - t) = 8\]. Simplify: \[3 + 6t + 2 - 5t + 3 - t = 8\], leading to \[8 = 8\]. No specific \( t \) solves this (dependent equation), meaning the line is parallel to the plane with no intersection.

Step 3d: Plane (d) Calculation

For the plane equation \(-x - 4y - 3z = 6 \), substitute: \[-(1 + 2t) - 4(-2 + 5t) - 3(3 - t) = 6\]. Simplifying results in \[-1 - 2t + 8 - 20t - 9 + 3t = 6\], which simplifies to \[-19t - 2 = 6\], giving \( t = -\frac{8}{19} \). Substituting \( t \) back gives the point of intersection: \( x = \frac{11}{19} \), \( y = \frac{22}{19} \), \( z = \frac{61}{19} \).

Key concepts

Parametric Equations

Parametric equations provide us with an elegant way to describe a line in three-dimensional space. Instead of just using standard Cartesian coordinates, parametric equations express each of the coordinates, such as \(x\), \(y\), and \(z\), in terms of a third variable, typically known as the parameter \(t\).
Using this parameter, each coordinate is represented as a function of \(t\). In the current exercise, the line is expressed by:

• \(x = 1 + 2t\)
• \(y = -2 + 5t\)
• \(z = 3 - t\)

These functions collectively define a line by illustrating how the point on the line moves as \(t\) changes its value. This representation is crucial since it makes it straightforward to find the intersection of the line with other geometric structures, such as planes.
By substituting these parametric equations into a plane's equation, we can solve for specific values of \(t\) that satisfy the intersection conditions.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of mathematics that uses algebraic equations to represent geometric figures. It allows us to analyze geometric problems in terms of coordinates and equations.
In the realm of three-dimensional geometry, the equation of a plane can be given in a standard form: \(ax + by + cz = d\). This form represents all the points \((x, y, z)\) that lie on a plane.
In the exercise, we have multiple planes represented by such equations, like \(x - 3y + 2z = 4\). By substituting the parametric line equations into these, we transform the spatial problem into one of algebra, enabling us to identify intersections.
This method simplifies what could be a complex geometric relationship into manageable algebraic equations, thereby making solving for intersection points more accessible.

System of Equations

The concept of a system of equations is foundational when determining intersections in coordinate geometry. A system arises when multiple equations are used simultaneously to find common solutions. In our exercise, each plane equation combined with the parametric line equations creates a system.
For instance, substituting the parametric equations into a plane equation like \(x - 3y + 2z = 4\) forms an algebraic equation in terms of \(t\). Solving this equation gives us the specific \(t\) values where intersection occurs. If \(t\) is an invalid result (like a contradiction), it indicates no intersection occurs, possibly due to parallelism, like in case (c).
The resulting system from substituting is often a single equation when dealing with line-plane intersections, and solving it provides precise intersection points. Such solutions emphasize how the power of algebra helps in navigating complex geometric queries with efficiency.