How to find where a line intersects a plane?
Understand the Problem
The question is asking for the method to determine the point where a line crosses or meets a plane in three-dimensional space. This involves using equations of the line and the plane to find the intersection point mathematically.
Answer
The intersection point is given by: $$ \mathbf{P} = \begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix} + \left(-\frac{A x_0 + B y_0 + C z_0 + D}{A a + B b + C c}\right) \begin{pmatrix} a \\ b \\ c \end{pmatrix} $$
Answer for screen readers
The intersection point of the line and the plane can be represented as:
$$ \mathbf{P} = \begin{pmatrix} x_0 \ y_0 \ z_0 \end{pmatrix} + \left(-\frac{A x_0 + B y_0 + C z_0 + D}{A a + B b + C c}\right) \begin{pmatrix} a \ b \ c \end{pmatrix} $$
Steps to Solve
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Identify the equations of the line and the plane
Let the equation of the line be represented parametrically as: $$ \mathbf{L}(t) = \begin{pmatrix} x_0 \ y_0 \ z_0 \end{pmatrix} + t \begin{pmatrix} a \ b \ c \end{pmatrix} $$
where $(x_0, y_0, z_0)$ is a point on the line and $\begin{pmatrix} a \ b \ c \end{pmatrix}$ is the direction vector of the line.
The equation of the plane can be written as:
$$ Ax + By + Cz + D = 0 $$
where $A$, $B$, and $C$ are the coefficients of the plane's normal vector and $D$ is a constant. -
Substitute the line's equation into the plane's equation
Substitute the parametric equations from the line into the plane's equation. This involves figuring out what $x$, $y$, and $z$ represent in terms of the parameter $t$.
For example:
$$ A(x_0 + at) + B(y_0 + bt) + C(z_0 + ct) + D = 0 $$ -
Solve for the parameter $t$
Rearranging the equation gives: $$ A x_0 + B y_0 + C z_0 + D + (A a + B b + C c)t = 0 $$
Carrying out the algebra to isolate $t$, we get: $$ t = -\frac{A x_0 + B y_0 + C z_0 + D}{A a + B b + C c} $$ -
Find the intersection point
Use the value of $t$ you found in the previous step and substitute it back into the parametric equation of the line to find the coordinates of the intersection point: $$ \mathbf{P} = \begin{pmatrix} x_0 \ y_0 \ z_0 \end{pmatrix} + t \begin{pmatrix} a \ b \ c \end{pmatrix} $$ -
Evaluate the coordinates
This will give you the coordinates at which the line intersects the plane.
The intersection point of the line and the plane can be represented as:
$$ \mathbf{P} = \begin{pmatrix} x_0 \ y_0 \ z_0 \end{pmatrix} + \left(-\frac{A x_0 + B y_0 + C z_0 + D}{A a + B b + C c}\right) \begin{pmatrix} a \ b \ c \end{pmatrix} $$
More Information
Finding the intersection between a line and a plane in three-dimensional space is crucial in various fields such as physics, engineering, and computer graphics. The method described is commonly used in 3D modeling and simulations.
Tips
- Forgetting to substitute correctly: Ensure that each variable from the line is correctly substituted into the plane's equation.
- Not checking for parallelism: If the denominator $A a + B b + C c$ equals zero, the line is either parallel to the plane or lies in it. This should be checked before finding $t$.
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