Problem 41 - University Calculus: Early Transcendentals

Question

find the distance from the point to the plane. $$(0,1,1), \quad 4 y+3 z=-12$$

Step-by-Step Solution

Answer

3.8 units.

Step 1: Identify the Given

We are given a point \((0,1,1)\) and the equation of the plane \(4y + 3z = -12\). From the plane equation, we identify the normal vector for the plane, \(\mathbf{n} = (0, 4, 3)\).

Step 2: Write Down the Distance Formula

The distance \(D\) from a point \((x_0, y_0, z_0)\) to a plane \(Ax + By + Cz + D = 0\) is given by: \[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

Step 3: Substitute the Values

Substitute \((x_0, y_0, z_0) = (0, 1, 1)\) and the plane \(4y + 3z = -12\) rewritten as \(0x + 4y + 3z + 12 = 0\), into the distance formula:\[ D = \frac{|0 \cdot 0 + 4 \cdot 1 + 3 \cdot 1 + 12|}{\sqrt{0^2 + 4^2 + 3^2}} \]

Step 4: Calculate Numerator

Calculate the numerator:\[ |0 + 4 + 3 + 12| = |19| = 19 \]

Step 5: Calculate Denominator

Calculate the denominator:\[ \sqrt{0^2 + 4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Step 6: Compute the Distance

Divide the numerator by the denominator:\[ D = \frac{19}{5} = 3.8 \]

Step 7: Conclude the Solution

The distance from the point \((0, 1, 1)\) to the plane \(4y + 3z = -12\) is 3.8 units.

Key Concepts

Normal Vector

The normal vector is a key concept when working with planes in three-dimensional geometry. It tells us the direction that is perpendicular to the plane. In a plane equation of the form \( Ax + By + Cz + D = 0 \), the coefficients \( A, B, \) and \( C \) define the normal vector. For example, given the plane equation \( 4y + 3z = -12 \), it can be rewritten in the standard format as \( 0x + 4y + 3z + 12 = 0 \). From this equation, the normal vector \( \mathbf{n} \) is \( (0, 4, 3) \), because the coefficients of \( x, y, \) and \( z \) are used as the components of the vector. This vector points directly outwards from the plane surface, and is crucial in calculating distances and angles related to the plane.
Understanding the normal vector is essential as it simplifies many geometric calculations, such as finding the shortest distance from a point to the plane. This vector helps in determining projections, reflections, and even rotations in three-dimensional space.

Distance Formula

The distance formula for finding the shortest perpendicular distance from a point to a plane is derived from the equation of the plane itself. The generic formula is given by:

• \( D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \)

Here, \((x_0, y_0, z_0)\) represents the coordinates of the point from which we are measuring the distance, and \(A, B, C, D\) are constants from the plane equation \(Ax + By + Cz + D = 0\). This formula efficiently uses the plane's normal vector components \((A, B, C)\) to calculate how far the point is along this perpendicular direction.
By plugging any specific coordinates of a point and plane equation into this formula, you can find the exact distance with ease. Keeping this formula in mind allows you to tackle various problems involving distances to planes, providing both a mathematical and geometric perspective to solve them.

Plane Equation

A plane equation in three-dimensional space is a linear equation that represents a flat, two-dimensional surface. It is often written in the form \( Ax + By + Cz + D = 0 \). Each term in the equation contributes to defining the position and orientation of the plane. The constants \( A, B, \) and \( C \) are particularly important as they define the normal vector of the plane, while \( D \) is a constant that adjusts the plane's position along the normal vector without altering its orientation.
For instance, the plane \( 4y + 3z = -12 \) can be rewritten as \( 0x + 4y + 3z + 12 = 0 \). Here, the coefficients \( A = 0 \), \( B = 4 \), and \( C = 3 \) help identify the normal vector, \( \mathbf{n} = (0, 4, 3) \). Solving plane equations is crucial in practical applications like computer graphics, physics simulations, and geometric modeling. Understanding how to manipulate and interpret these equations aids in visualizing spatial relationships and solving problems involving spatial reasoning.