Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(P\) be a point on a plane with normal vector \(\mathbf{n}\) and \(Q\) be a point off the plane. Show that the result of Example 10 of Section \(11.3\), the distance \(d\) between the point \(Q\) and the plane, can be expressed as $$ d=\frac{|\overrightarrow{P Q} \cdot \mathbf{n}|}{\|\mathbf{n}\|} $$ and use this result to find the distance from \((4,-2,3)\) to the plane \(4 x-4 y+2 z=2\).

Step-by-Step Solution

Answer

The distance is \( \frac{14}{3} \).

Step 1: Understand the formula

The given formula \( d=\frac{|\overrightarrow{P Q} \cdot \mathbf{n}|}{\|\mathbf{n}\|} \) calculates the distance \( d \) from a point \( Q \) to a plane. Here, \( \overrightarrow{PQ} \) is the vector from point \( P \) on the plane to point \( Q \), and \( \mathbf{n} \) is the normal vector to the plane. \( \cdot \) represents the dot product, and \( \|\mathbf{n}\| \) is the magnitude of the normal vector.

Step 2: Determine vector PQ

Choose a point \( P \) on the plane. Since the plane equation is \( 4x - 4y + 2z = 2 \), a point on the plane can be \( P = (0, 0, 1) \) since it satisfies the equation: \( 4(0) - 4(0) + 2(1) = 2 \). The vector \( \overrightarrow{PQ} \) is then \( (4, -2, 3) - (0, 0, 1) = (4, -2, 2) \).

Step 3: Find the normal vector

The normal vector \( \mathbf{n} \) is extracted from the plane equation \( 4x - 4y + 2z = 2 \). It is \( \mathbf{n} = (4, -4, 2) \).

Step 4: Calculate the dot product

Calculate the dot product \( \overrightarrow{PQ} \cdot \mathbf{n} = (4, -2, 2) \cdot (4, -4, 2) = 4 \times 4 + (-2) \times (-4) + 2 \times 2 = 16 + 8 + 4 = 28 \).

Step 5: Calculate the magnitude of n

Find the magnitude of the normal vector \( \mathbf{n} \): \( \|\mathbf{n}\| = \sqrt{4^2 + (-4)^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 \).

Step 6: Apply the distance formula

Use the formula to find the distance: \( d = \frac{|28|}{6} = \frac{28}{6} = \frac{14}{3} \).

Step 7: Verify your calculations

Review each step to ensure accuracy, confirming calculations and logical flow. Check choice of point, vector operations, and ensure the correct application of the formula.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is an essential operation in vector arithmetic. It results in a scalar rather than a vector. The dot product is calculated by multiplying corresponding components of two vectors and then summing these products. For example, if you have two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the dot product is given by:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]In our case, the vectors involved are \( \overrightarrow{PQ} = (4, -2, 2) \) and \( \mathbf{n} = (4, -4, 2) \). Their dot product calculates to **28**.

• Helps determine the angle between vectors.
• Zero dot product indicates orthogonal vectors.

The dot product is crucial in determining the distance from a point to a plane, as it reflects how much of a vector goes in the direction of another.

Normal Vector

The normal vector, often denoted as \( \mathbf{n} \), plays a key role in defining a plane's orientation in three-dimensional space. It is perpendicular to every vector that lies on the plane. Given the plane equation, the normal vector can be easily extracted from the coefficients of \( x, y, \) and \( z \).For instance, in the plane equation \( 4x - 4y + 2z = 2 \), the normal vector is \( \mathbf{n} = (4, -4, 2) \).

• Gives direction to the plane.
• Helps in determining distances and angles related to the plane.

Understanding the normal vector is vital for calculating distances to the plane, since it serves as a reference direction for all perpendicular projections onto the plane.

Plane Equation

The plane equation in three-dimensional space usually takes the form \( ax + by + cz = d \), depicting a plane with normal vector \( \mathbf{n} = (a, b, c) \). This linear equation holds for every point that lies on the plane. Each solution to this equation represents a point on the surface of the plane.In our exercise, the given plane equation is \( 4x - 4y + 2z = 2 \). To find a point \( P \) that satisfies this equation, you can set some variables to zero, and solve for the third. For instance, by setting \( x = 0, y = 0 \), we solve for \( z \), resulting in the point \( P = (0, 0, 1) \).

• Defines the plane position and orientation.
• Integral for finding vectors and calculating distances.

The equation is foundational for understanding how the plane exists in space and interacts with other geometric entities.

Magnitude of Vector

The magnitude of a vector, often symbolized as \( \| \mathbf{v} \| \), represents the length or size of the vector. In a three-dimensional space, for a vector \( \mathbf{v} = (v_1, v_2, v_3) \), the magnitude is computed by:\[ \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]For our particular normal vector \( \mathbf{n} = (4, -4, 2) \), the magnitude is calculated as:\[ \| \mathbf{n} \| = \sqrt{4^2 + (-4)^2 + 2^2} = 6 \]

• Essential for normalizing vectors.
• Used in various computations like distance and force.

Understanding the magnitude gives insights into the vector's scale, and it is a critical component for normalizing vectors when deriving distances, as shown in the formula \( d = \frac{|\overrightarrow{PQ} \cdot \mathbf{n}|}{\|\mathbf{n}\|} \).