Intersections of Lines and Planes
Line-Plane Intersection
- The intersection of a line and a plane in three-dimensional space is a fundamental concept in analytic geometry.
- This intersection can result in one of three outcomes:
- A single point of intersection
- No intersection (the line is parallel to the plane)
- The entire line (when the line lies entirely within the plane)
To find the point of intersection, we typically use a system of equations.
The geometric interpretation of this solution is crucial. It represents the unique point where the line pierces the plane in 3D space.
Derivation of the Angle Between a Line and a Plane
- The angle $\theta$ between a line and a plane is the complement of the angle between the direction vector of the line and the normal vector of the plane.
- To derive this formula, consider a line with direction vector $\mathbf{v}$ and a plane with normal vector $\mathbf{n}$.
- The angle $\phi$ between the line and the plane's normal vector can be calculated using the dot product formula:
$$ \phi = \frac{\mathbf{v} \cdot \mathbf{n}}{|\mathbf{v}||\mathbf{n}|}$$
- Since the angle $\theta$ between the line and the plane is the complement of $\phi$, we use:
$$\theta = 90^\circ - \phi$$
- Taking the sine of both sides and rearranging, we obtain:
$$\theta = \frac{|\mathbf{v} \cdot \mathbf{n}|}{|\mathbf{v}||\mathbf{n}|}$$
This formula shows that the shortest angle between a line and a plane depends only on their directional relationship and provides a straightforward way to compute it.
Consider a line given by the parametric equations:
$x = 1 + 2t$ $y = 3 - t$ $z = 2 + 3t$
And a plane given by the equation:
$2x + 3y - z = 4$
To find the intersection, we substitute the line equations into the plane equation:
$2(1 + 2t) + 3(3 - t) - (2 + 3t) = 4$
Simplifying:
$2 + 4t + 9 - 3t - 2 - 3t = 4$ $9 - 2t = 4$ $-2t = -5$ $t = \frac{5}{2}$
Substituting this value of $t$ back into the line equations gives us the point of intersection:
$x = 1 + 2(\frac{5}{2}) = 6$ $y = 3 - \frac{5}{2} = \frac{1}{2}$ $z = 2 + 3(\frac{5}{2}) = \frac{17}{2}$
Therefore, the point of intersection is $(6, \frac{1}{2}, \frac{17}{2})$.
Intersection of Two Planes
When two planes intersect, they typically form a line. However, there are two special cases:
- The planes are parallel and don't intersect
- The planes are identical and overlap completely
To find the line of intersection, we solve the equations of the two planes simultaneously.
Consider two planes:
Plane 1: $2x + y - z = 3$ Plane 2: $x - y + 2z = 1$
To find their intersection, we solve these equations together with a third equation $z = t$ (where $t$ is a parameter). This gives us:
$2x + y - t = 3$ $x - y + 2t = 1$ $z = t$
Solving this system (you can use substitution or elimination methods), we get:
$x = 1 + \frac{1}{3}t$ $y = -1 + \frac{4}{3}t$ $z = t$
This represents the line of intersection in parametric form.
Intersection of Three Planes
Three planes can intersect in several ways:
- At a single point
- Along a line (if two planes intersect and the third contains that line)
- Not at all (if at least two planes are parallel)
- In a plane (if all three planes are identical)
Finding the intersection point (if it exists) involves solving a system of three linear equations.
Consider three planes:
Plane 1: \( x + y + z = 1 \)
Plane 2: \( 2x - y + z = 3 \)
Plane 3: \( x + 3y - 2z = 4 \)
We can solve this system using matrix methods or elimination:
\[\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 3 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 4 \end{bmatrix} \]
The solution is \( x = 1, y = -1, z = 1 \), representing the point of intersection \( (1, -1, 1) \).
When solving systems of equations for intersections, always check your solution by substituting it back into the original equations.
Angles Between Lines and Planes
Angle Between a Line and a Plane
- The angle between a line and a plane is defined as the complement of the angle between the line and its projection onto the plane.
- It can be calculated using the dot product of the direction vector of the line and the normal vector of the plane.
- If $\vec{v}$ is the direction vector of the line and $\vec{n}$ is the normal vector of the plane, the angle $\theta$ is given by:
$$\sin \theta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}||\vec{n}|}$$
Consider a line with direction vector $\vec{v} = (1, 2, 3)$ and a plane with normal vector $\vec{n} = (2, -1, 2)$.
The angle between them is:
$$\sin \theta = \frac{|1(2) + 2(-1) + 3(2)|}{\sqrt{1^2 + 2^2 + 3^2}\sqrt{2^2 + (-1)^2 + 2^2}}$$
$$\sin \theta = \frac{|2 - 2 + 6|}{\sqrt{14}\sqrt{9}} = \frac{6}{\sqrt{126}} \approx 0.5345$$
Therefore, $\theta \approx 32.2°$
Angle Between Two Planes
- The angle between two planes is the same as the angle between their normal vectors. It can be calculated using the dot product of the normal vectors.
- If $\vec{n_1}$ and $\vec{n_2}$ are the normal vectors of the two planes, the angle $\theta$ between them is given by:
$$\cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}||\vec{n_2}|}$$
Consider two planes with normal vectors $\vec{n_1} = (1, 2, 3)$ and $\vec{n_2} = (2, -1, 2)$.
The angle between them is:
$$\cos \theta = \frac{1(2) + 2(-1) + 3(2)}{\sqrt{1^2 + 2^2 + 3^2}\sqrt{2^2 + (-1)^2 + 2^2}}$$
$$\cos \theta = \frac{6}{\sqrt{14}\sqrt{9}} = \frac{6}{\sqrt{126}} \approx 0.5345$$
Therefore, $\theta \approx 57.8°$
Students often confuse the formulas for the angle between a line and a plane (which uses sine) and the angle between two planes (which uses cosine). Remember that these are different scenarios and require different approaches.
