Problem 102 - Calculus Early Transcendentals

Question

a. Show that the point in the plane \(a x+b y+c z=d\) nearest the origin is \(P\left(a d / D^{2}, b d / D^{2}, c d / D^{2}\right),\) where \(D^{2}=a^{2}+b^{2}+c^{2} .\) Conclude that the least distance from the plane to the origin is \(|d| / D\). (Hint: The least distance is along a normal to the plane.) b. Show that the least distance from the point \(P_{0}\left(x_{0}, y_{0}, z_{0}\right)\) to the plane \(a x+b y+c z=d\) is \(\left|a x_{0}+b y_{0}+c z_{0}-d\right| / D\).

Step-by-Step Solution

Answer

#Results# a. Nearest point to the origin in the given plane: $$ P = \left(\frac{ad}{D^2}, \frac{bd}{D^2}, \frac{cd}{D^2}\right) $$ Least distance from the plane to the origin: $$ d_\text{min} = \frac{|d|}{D} $$ b. Least distance from a point \(P_0(x_0, y_0, z_0)\) to the plane: $$ \text{Distance} = \frac{|a x_0 + b y_0 + c z_0 - d|}{D} $$

Step 1: Finding the normal vector to the plane

The normal vector to the plane \(ax + by + cz = d\) is given by the coefficients of \(x\), \(y\), and \(z\). Thus, the normal vector \(\vec{n}\) is: $$\vec{n} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}.$$ 2. Project the origin onto the plane

Step 2: Project the origin onto the plane

The nearest point to the origin on the plane is the orthogonal projection of the origin along the normal vector. Let's call this point P. To find its coordinates, we can first find the scaling factor \(s\), and then multiply the normal vector by this scaling factor: $$ s = \frac{d}{D^2} $$ Where \(D^2 = a^2 + b^2 + c^2\), as given in the exercise. The coordinates of point P are then: $$ P = s \vec{n} = \left(\frac{ad}{D^2}, \frac{bd}{D^2}, \frac{cd}{D^2}\right) $$ 3. Find the least distance from the plane to the origin

Step 3: Find the least distance from the plane to the origin

The least distance from the plane to the origin is the length of the vector from the origin to point P (the distance along the normal): $$ d_\text{min} = \frac{|d|}{D} $$ 4. Find the least distance from a point \(P_0\) to the plane

Step 4: Find the least distance from a point \(P_0\) to the plane

For this, we first find the dot product between the vector \(P_0 - P\) (where P is the nearest point to the origin in the plane) and the normal vector: $$ \text{Distance} = \frac{\left| (P_0 - P) \cdot \vec{n} \right|}{D} = \frac{\left|\left( \begin{bmatrix}x_0 - \frac{ad}{D^2} \\ y_0 - \frac{bd}{D^2} \\ z_0 - \frac{cd}{D^2}\end{bmatrix}\right) \cdot \begin{bmatrix} a \\ b \\ c \end{bmatrix}\right|}{D} = \frac{|a x_0 + b y_0 + c z_0 - d|}{D} $$

Key Concepts

Normal Vector to a Plane

In geometry, understanding the orientation of a plane in a three-dimensional space is fundamental. A normal vector helps us describe this orientation succinctly. It's a vector that is perpendicular to every line lying on the plane.

• If you have the equation of a plane written as \(ax + by + cz = d\), the normal vector \(\vec{n}\) is directly derived from the coefficients of \(x\), \(y\), and \(z\).
• The normal vector is \(\vec{n} = \begin{bmatrix} a \ b \ c \end{bmatrix}\).
• This vector is crucial for many calculations involving the plane, such as finding distances from points to the plane.

This normal vector allows us to calculate projections and distances efficiently by giving a direction along which these tasks are simplest. In essence, the normal vector acts as a "directional tool" toward the shortest path away from the plane.

Orthogonal Projection

The concept of orthogonal projection is about dropping a perpendicular from a point to a plane, effectively finding the shortest path down. This involves using the normal vector to determine exactly what "straight down" means in mathematical terms.

• For instance, to find the closest point on the plane \(ax + by + cz = d\) to the origin \((0, 0, 0)\), we project the origin onto the plane.
• The orthogonal projection is achieved by using the scaling factor \(s = \frac{d}{D^2}\), where \(D^2 = a^2 + b^2 + c^2\).
• The point \(P\) on the plane is found with \(s \times \vec{n}\), resulting in coordinates \(\left(\frac{ad}{D^2}, \frac{bd}{D^2}, \frac{cd}{D^2}\right)\).

This approach is not only efficient but ensures the least distance, or shortest path, is determined accurately, due to the unique properties of orthogonality.

Least Distance Calculation

Calculating the least distance from a point to a plane involves a series of logical steps that leverage the properties of vectors and projections.

• The minimum distance from the origin to the plane \(ax + by + cz = d\) is given by the magnitude of the vector to the point \(P\), which is \(\frac{|d|}{D}\).
• Likewise, the least distance from another point \(P_0(x_0, y_0, z_0)\) to the plane is calculated using: \(\frac{|a x_0 + b y_0 + c z_0 - d|}{D}\).
• This calculation involves finding the dot product of the vector \(P_0 - P\) with the normal vector \(\vec{n}\), guiding us directly to the shortest path vertically down to the plane.

The process highlights the efficiency of using linear algebra concepts like dot products and normalization to simplify otherwise complex problems, ensuring distances are calculated in the most straightforward manner. By running these steps, one can see the intertwining roles of both projections and geometrical vectors in clear, tangible ways.