Problem 15 - 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 $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} $$ $$ (1,5,-4), 3 x-y+2 z=6 $$

Step-by-Step Solution

Answer

The distance between the given point and the plane is approximately 2.94.

Step 1: Understand the provided information

Observe the given data and confirm what have been provided: the position of the point and the equation of a plane. The point coordinates are \(1, 5, -4\) and the equation of the plane is \(3x - y + 2z = 6\).

Step 2: Rewrite the equation of the plane in standard form

The standard form of a plane is \(ax + by + cz + d = 0\). We can rearrange the provided equation \(3x - y + 2z = 6\) to align with this by shifting '6' to the other side of the equation. Doing this gives us \(3x - y + 2z - 6 = 0\). Now, the values of a, b, c, and d are 3, -1, 2 and -6 respectively.

Step 3: Substitute all the given values into the formula for distance

The next step is to take the point's coordinates and the plane's coefficients (from step 2) and substitute them into the distance formula. This gives us; \[D=\frac{\left|3*1 + (-1)*5 + 2 * (-4) - 6\right|}{\sqrt{3^2 + (-1)^2 + 2^2}}\]

Step 4: Simplify the equation

On simplifying the above equation, \[D= \frac{\left|-11\right|}{\sqrt{14}}\]After that, the denominator will be simplified.\[D= \frac{11}{\sqrt{14}}\] Now, the distance \(D\) will be found to be approximately 2.94, after rounding to 2 decimal places.

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that focuses on vector fields and operations on these fields. In the context of the given exercise, it helps us understand concepts such as gradients, divergences, and curls, which are used to describe multiple dimensional spaces.

When finding the distance from a point to a plane, vector calculus provides the necessary tools to handle the spatial relationships involved. One major concept in vector calculus is the dot product, which is integral in determining perpendicularity and projections, both required in distance calculations.

By examining these vector interactions, we can find how far a point is from a given plane, which is essentially a measure of projection along the plane's normal vector.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, allows us to describe geometric shapes using algebraic equations. It provides us with a framework to work with points, lines, and planes in a coordinate system, which is essential for solving the distance problem.

In the problem, we have a point with given coordinates and a plane defined by its equation. Coordinate geometry allows us to manipulate these points and equations easily. The plane's equation is rearranged into its standard form, aligning it for further calculations.

Through understanding coordinate geometry, one can visualize the geometric configurations, making it easier to approach and solve problems involving distances or other spatial relationships.

Distance Formula

The distance formula is key to calculating how far an entity is in space from another reference point or surface. For a point to a plane, the formula used requires specific inputs: the point coordinates and coefficients from the plane's equation.

The standard distance formula is given by:

• \[D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}\]

Key figures like \(a, b, c\) are coefficients of the plane's normal vector and \(d\) is the constant in the plane equation.

Substituting the values efficiently and simplifying, as shown in the solution, achieves the shortest perpendicular distance between the point and the plane.

Problem Solving

Problem solving in mathematics involves breaking down a problem into manageable steps, as evidenced by the solution to the given exercise. It involves understanding the problem, translating it into mathematical language, and systematically working through the solution.

In this exercise:

• First, comprehend the problem statement and gather all information, such as point coordinates and the plane equation.
• Next, rearrange and rewrite the given equation into a useful form.
• Substitute all gathered data into a mathematical formula for solving the problem.
• Finally, simplify and compute the results, ensuring accuracy in the final answer.

Approaching problems in this structured manner helps tackle complex calculations and achieves accurate results.