Vector Equations of a Plane
Parametric Form: r = a + λb + μc
The vector equation of a plane in its parametric form is given by:
$r = a + λb + μc$
Where:
- $r$ is the position vector of any point on the plane
- $a$ is the position vector of a fixed point on the plane
- $b$ and $c$ are non-parallel vectors lying within the plane
- $λ$ and $μ$ are scalar parameters
This equation represents all points on the plane as a linear combination of two non-parallel vectors ($b$ and $c$) added to a fixed point ($a$).
The vectors $b$ and $c$ must be non-parallel to ensure they span the entire plane.
Interpretation
- The vector $a$ anchors the plane at a specific point in space.
- Vectors $b$ and $c$ define the "directions" in which the plane extends.
- By varying $λ$ and $μ$, we can reach any point on the plane.
Let \(\mathbf{a} = (1,2,3) \), \(\mathbf{b} = (1,0,1) \), and \(\mathbf{c} = (0,1,-1) \). The equation of the plane is: \[\mathbf{r} = (1,2,3) + \lambda (1,0,1) + \mu (0,1,-1) \] This can be written in component form as: \[\begin{cases} x = 1 + \lambda\\ y = 2 + \mu\\ z = 3 + \lambda - \mu\end{cases} \]
Understanding Vector Equations: Lines vs. Planes
- A vector line equation extends a point one-dimensionally using a single direction vector, allowing movement along that direction.
- In contrast, a vector plane equation extends a point two-dimensionally using two direction vectors, defining a surface rather than a single path.
Normal Form: (r-a) • n = 0
Another way to represent a plane is using its normal vector:
$(r-a) • n = 0$
This can be restructured to be:
$r • n = a • n$
Where:
- $r$ is the position vector of any point on the plane
- $n$ is a vector normal (perpendicular) to the plane
- $a$ is the position vector of any point on the plane
This equation states that the dot product of the normal vector with any vector from $a$ to a point on the plane ($r - a$) is zero, which is the definition of perpendicularity.
To find a normal vector to a plane given in the form $r = a + λb + μc$, you can use the cross product: $n = b × c$
Interpretation
- The normal vector $n$ is perpendicular to every vector lying in the plane.
- The equation ensures that the projection of $(r - a)$ onto $n$ is always zero.
Given a plane with normal vector $n = (2, -1, 3)$ passing through the point $(1, 4, -2)$:
$r • (2, -1, 3) = (1, 4, -2) • (2, -1, 3)$
Simplifying the right side: $r • (2, -1, 3) = 2(1) + (-1)(4) + 3(-2) = -8$
So the equation of the plane is: $2x - y + 3z = -8$
Cartesian Equation: ax + by + cz = d
The Cartesian equation of a plane is:
$ax + by + cz = d$
Where:
- $(a, b, c)$ are components of the normal vector to the plane
- $d$ is a constant
This form is directly related to the normal form, as $(a, b, c)$ corresponds to the normal vector $n$.
Interpretation
Students often confuse the coefficients $(a, b, c)$ with a point on the plane. Remember, these form the normal vector, not a point!
Converting Between Forms
To convert from the normal vector form to the cartesian form:
- Expand $(r - a)• n = 0$ to $r • n = a • n$
- Write $r$ in its component form of $(x, y, z)$
- Simplify
Converting $r • (2, -1, 3) = -4$ to Cartesian form:
$(x, y, z) • (2, -1, 3) = -4$ $2x - y + 3z = -4$
This is already in the Cartesian form $ax + by + cz = d$
Applications and Problem Solving
Understanding these different forms allows for solving various geometric problems involving planes:
- Finding the intersection of planes
- Determining if a point lies on a plane
- Calculating the distance from a point to a plane
- Finding the angle between two planes
To find if a point $P(2, 1, 3)$ lies on the plane $2x - y + 3z = 5$:
Substitute the coordinates into the equation: $2(2) - 1 + 3(3) = 5$ $4 - 1 + 9 = 12$ $12 ≠ 5$
Therefore, the point does not lie on the plane.
The ability to switch between different representations of a plane is crucial for solving complex problems efficiently.
