Problem 17 - Calculus an Applied Approach

Question

Find the distance between the point and the plane (see figure). The distance \(D\) between a point \(\left(x_{0}, y_{0}, z_{0}\right)\) and the plane \(a x+b y+c z+d=0\) is $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} $$ $$ (1,0,-1), 2 x-4 y+3 z=12 $$

Step-by-Step Solution

Answer

The distance between the point and the plane is \(\frac{11 \sqrt{29}}{29}\).

Step 1 - Identify required parameters

From the equation of the plane, identify the coefficients of \(x\), \(y\), and \(z\) as \(a\), \(b\), and \(c\) respectively. Also, represent \(d\) as the rest of the equation. Here, our plane equation is \(2 x - 4 y + 3 z = 12\). So, \(a = 2\), \(b = -4\), \(c = 3\), and \(d = -12\). For the point \((x_0, y_0, z_0)\), we have been given \(x_0 = 1\), \(y_0 = 0\), and \(z_0 = -1\).

Step 2 - Substitute into the formula

Now, substitute these values into the formula : \(D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\). We get \(D=\frac{\left|2*1+(-4)*0 + 3*(-1) -12\right|}{\sqrt{2^{2}+(-4)^{2}+3^{2}}}\).

Step 3 - Compute the answer

Compute the numerator and the denominator separately. For numerator we have: \(2*1+(-4)*0 + 3*(-1) - 12 = -11\). Remember absolute value is distance, so we take the absolute value and get \(11\). For denominator: \(\sqrt{2^{2}+(-4)^{2}+3^{2}} = \sqrt{4+16+9} = \sqrt{29}\). Now, divide the absolute of numerator by the denominator \(11 / \sqrt{29}\).

Step 4 - Simplify the answer

To simplify \(\frac{11}{\sqrt{29}}\), multiply by \(\frac{\sqrt{29}}{\sqrt{29}}\) to get \(\frac{11 \sqrt{29}}{29}\).

Key Concepts

Plane Equation

Understanding the equation of a plane is essential for tackling problems in coordinate geometry. A plane in three-dimensional space can be defined by a linear equation of the form \( ax+by+cz+d=0 \). This equation represents a flat, two-dimensional surface that extends infinitely in three dimensions. Here, \(a\), \(b\), and \(c\) are the coefficients that determine the orientation of the plane, while \(d\) is the constant term that affects the plane's position relative to the origin.
When we're given a specific plane equation, such as \(2x-4y+3z=12\), we can interpret the coefficients \(2\), \( -4\), and \(3\) as directing the plane's normal vector, which is perpendicular to the surface. This vector is critical when calculating distances from points to the plane, as it helps us understand the plane's orientation in space.

Absolute Value

The concept of absolute value represents the magnitude of a number regardless of its sign. In a geometrical context, it signifies the distance of a number from zero on the number line. \( |x| \) denotes the absolute value of \(x\), which is \(x\) if \(x \geq 0\) and \( -x\) if \(x < 0\).
In the context of finding the distance between a point and a plane, absolute value serves a critical role by ensuring that the distance, which can never be negative, is measured correctly. The absolute value strips away the directionality associated with the signed distance, leaving us with the pure magnitude of separation, which is what we require when measuring distances.

Distance Formula

The distance formula is a derivative of the Pythagorean theorem and is used to calculate the distance between two points in space. To find the distance \(D\) between a point and a plane, we use the formula \[ D=\frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}} \], where \(\left(x_{0}, y_{0}, z_{0}\right)\) are the coordinates of the point in question, and \(a\), \(b\), and \(c\) are the coefficients of the plane equation, deriving from the normal vector to the plane.
In this calculation, the numerator involves the absolute value of the point's coordinates plugged into the plane equation, which effectively measures how far the point is from satisfying the plane's equation - this is the signed distance. The denominator normalizes this distance by the length of the plane's normal vector, giving us the perpendicular distance to the plane.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, unifies algebra and geometry through the use of a coordinate system. It helps in describing geometric figures and properties algebraically. By assigning coordinates to points in space, we can apply algebraic equations to solve geometric problems, such as finding the distance between a point and a plane.
In the process of solving for distances in three-dimensional space, we visualize three mutually perpendicular lines, typically labeled as the \(x\)-axis, \(y\)-axis, and \(z\)-axis, to form a three-dimensional grid. Each point can then be represented by an ordered triple of numbers known as its coordinates, which indicate its position relative to these axes. By using the principles of coordinate geometry, we can determine geometric quantities that would be more complex to assess using traditional Euclidean methods alone.