Question

Let \(P_{0}\) be a point with vector \(\mathbf{p}_{0},\) and let \(a x+b y+c z=d\) be the equation of a plane with normal \(\mathbf{n}=\left[\begin{array}{l}a \\\ b \\ c\end{array}\right]\) a. Show that the point on the plane closest to \(P_{0}\) has vector \(\mathbf{p}\) given by $$ \mathbf{p}=\mathbf{p}_{0}+\frac{d-\left(\mathbf{p}_{0} \cdot \mathbf{n}\right)}{\|\mathbf{n}\|^{2}} \mathbf{n} $$ \(\left[\right.\) Hint \(: \mathbf{p}=\mathbf{p}_{0}+t \mathbf{n}\) for some \(t,\) and \(\left.\mathbf{p} \cdot \mathbf{n}=\mathbf{d} .\right]\) b. Show that the shortest distance from \(P_{0}\) to the plane is \(\frac{\left|d-\left(\mathbf{p}_{0} \cdot \mathbf{n}\right)\right|}{\|\mathbf{n}\|}\). c. Let \(P_{0}^{\prime}\) denote the reflection of \(P_{0}\) in the planethat is, the point on the opposite side of the plane such that the line through \(P_{0}\) and \(P_{0}^{\prime}\) is perpendicular to the plane. Show that \(\mathbf{p}_{0}+2 \frac{d-\left(\mathbf{p}_{0} \cdot \mathbf{n}\right)}{\|\mathbf{n}\|^{2}} \mathbf{n}\) is the vector of \(P_{0}^{\prime}\)

Short answer

The closest point on the plane is \( \mathbf{p} = \mathbf{p}_0 + \frac{d - (\mathbf{p}_0 \cdot \mathbf{n})}{\|\mathbf{n}\|^2} \mathbf{n} \), shortest distance is \( \frac{|d - (\mathbf{p}_0 \cdot \mathbf{n})|}{\|\mathbf{n}\|} \), and reflection vector is \( \mathbf{p}_0 + 2 \frac{d - (\mathbf{p}_0 \cdot \mathbf{n})}{\|\mathbf{n}\|^2} \mathbf{n} \).

Step-by-step solution

Solve for the vector of the point on the plane (part a)

We begin by assuming the point on the plane closest to \( P_0 \) is of the form \( \mathbf{p} = \mathbf{p}_0 + t \mathbf{n} \), where \( t \) is a scalar. To lie on the plane, \( \mathbf{p} \cdot \mathbf{n} = d \) must be true. Substituting for \( \mathbf{p} \), we get: \[ (\mathbf{p}_0 + t \mathbf{n}) \cdot \mathbf{n} = d \] This simplifies to: \[ \mathbf{p}_0 \cdot \mathbf{n} + t(\mathbf{n} \cdot \mathbf{n}) = d \] Rearranging for \( t \), we get: \[ t = \frac{d - \mathbf{p}_0 \cdot \mathbf{n}}{\| \mathbf{n} \|^2} \] Thus, the vector \( \mathbf{p} \) is given by: \[ \mathbf{p} = \mathbf{p}_0 + \frac{d - \mathbf{p}_0 \cdot \mathbf{n}}{\| \mathbf{n} \|^2} \mathbf{n} \]

Determine the shortest distance to the plane (part b)

The shortest distance from \( P_0 \) to the plane is along the line perpendicular to the plane, which is parallel to the normal \( \mathbf{n} \). The magnitude of this distance is given by:\[ \text{Distance} = \left\| \mathbf{p} - \mathbf{p}_0 \right\| = \left\| \mathbf{p}_0 + \frac{d - \mathbf{p}_0 \cdot \mathbf{n}}{\| \mathbf{n} \|^2} \mathbf{n} - \mathbf{p}_0 \right\| = \left\| \frac{d - \mathbf{p}_0 \cdot \mathbf{n}}{\| \mathbf{n} \|^2} \mathbf{n} \right\| = \frac{|d - \mathbf{p}_0 \cdot \mathbf{n}|}{\| \mathbf{n} \|^2} \cdot \| \mathbf{n} \| = \frac{|d - \mathbf{p}_0 \cdot \mathbf{n}|}{\| \mathbf{n} \|} \]

Find the reflection of the point across the plane (part c)

The reflection of \( P_0 \) across the plane, \( P_0' \), is achieved by moving twice the distance along the normal beyond \( P \). This is because the reflection must be symmetric about the plane:\[ \mathbf{p}_0' = \mathbf{p}_0 + 2\left(\frac{d - \mathbf{p}_0 \cdot \mathbf{n}}{\| \mathbf{n} \|^2} \mathbf{n}\right) \]

Key concepts

Vector Equation of Plane

Understanding the vector equation of a plane is crucial in geometry and calculus. A plane in a three-dimensional space can be defined by a point and a normal vector. The vector equation of the plane can be expressed as: \[ a x + b y + c z = d \] where:

• \( a,\ b,\ c \) are coefficients that dictate the orientation of the plane.
• \( x,\ y,\ z \) are coordinates of any point on the plane.
• \( d \) is a constant that sets the plane's position relative to the origin.

The normal vector \( \mathbf{n} = \begin{bmatrix} a \ b \ c \end{bmatrix} \) is perpendicular to the plane, and plays a key role in defining the plane's characteristics. This vector allows us to describe not just where the plane is, but how it is oriented in space. With this vector, we can solve numerous problems regarding distances from points to planes, reflections over planes, and more. The vector gives the plane its distinct spatial "tilt".

Normal Vector

The normal vector is central to understanding planes. It provides a straightforward way to determine distances and angles with respect to the plane. A normal vector \( \mathbf{n} \) is a nonzero vector that is perpendicular to a given plane, and for every plane described by \( a x + b y + c z = d \), the normal vector is \( \mathbf{n} = \begin{bmatrix} a \ b \ c \end{bmatrix} \).
This vector enhances calculations, such as finding distances or reflecting points, because:

• The formula for the shortest distance from a point \( P_0 \) to the plane uses the normal to easily find a perpendicular direction: \( \text{Distance} = \frac{|d - (\mathbf{p}_0 \cdot \mathbf{n})|}{\| \mathbf{n} \|} \).
• To locate specific points, like reflections, we use the normal vector in our calculations: the vector indicating the closest point on the plane from a point \( P_0 \) is adjusted by the normal, thus reflecting: \( \mathbf{p} = \mathbf{p}_0 + t \mathbf{n} \).

The magnitude, \( \| \mathbf{n} \| \), often appears in formulas, revealing just how integral the normal vector is to understanding planes.

Reflection Across Plane

Reflecting a point across a plane implies finding the mirrored point on the opposite side. With a point \( P_0 \) and its reflection \( P_0' \), the line connecting these two points is perpendicular to the plane.
The reflection formula makes use of the normal vector to ensure symmetry across the plane. Reflection can be calculated as:

• First, calculate the vector that points to the closest point on the plane: \( \mathbf{p} = \mathbf{p}_0 + \frac{d - (\mathbf{p}_0 \cdot \mathbf{n})}{\| \mathbf{n} \|^2} \mathbf{n} \).
• To find the reflection \( P_0' \), you effectively "step" past the closest point on the plane by using double this vector distance: \( \mathbf{p}_0' = \mathbf{p}_0 + 2 \left( \frac{d - (\mathbf{p}_0 \cdot \mathbf{n})}{\| \mathbf{n} \|^2} \mathbf{n} \right) \).

By utilizing the perpendicular nature of the normal vector, the symmetry consistent in reflections is maintained and accurately calculated. This aspect of linear algebra is not only fascinating but reveals the depth behind geometric transformations.