Problem 8 - Essential Calculus Early Transcendentals

Question

Find the distance from \((4,-2,6)\) to each of the following. (a) The \(x y\) -plane (b) The \(y z-\) plane (c) The \(x z\) -plane (d) The \(x\) -axis (e) The \(y\) -axis (f) The \(z\) -axis

Step-by-Step Solution

Answer

(a) 6, (b) 4, (c) 2, (d) 2√10, (e) 2√13, (f) 2√5

Step 1: Understanding the Point Coordinate System

The point in question is \(4, -2, 6\). Each of these coordinates represents a distance from the origin along the x-, y-, and z-axes, respectively. To determine distances to planes and axes, we consider which coordinate is relevant for each plane or axis.

Step 2: Finding Distance to the xy-plane

The xy-plane is defined by having constant z-values, specifically z=0. To find the distance from the point \(4, -2, 6\) to the xy-plane, you need to look at the z-coordinate. The distance here is the absolute value of the z-coordinate: \|6| = 6\.

Step 3: Finding Distance to the yz-plane

The yz-plane is defined by having constant x-values, specifically x=0. To find the distance from \(4, -2, 6\) to the yz-plane, the x-coordinate is considered. The distance is the absolute value of the x-coordinate: \|4| = 4\.

Step 4: Finding Distance to the xz-plane

The xz-plane is defined by having constant y-values, specifically y=0. For the point \(4, -2, 6\), consider the y-coordinate. The distance to the xz-plane is the absolute value of the y-coordinate: \|-2| = 2\.

Step 5: Finding Distance to the x-axis

The x-axis is defined by having both y=0 and z=0. The distance from \(4, -2, 6\) to the x-axis is calculated using the Pythagorean theorem in the yz-plane: \sqrt{(-2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10\}.

Step 6: Finding Distance to the y-axis

The y-axis is defined by having both x=0 and z=0. The distance from \(4, -2, 6\) to the y-axis is found using the Pythagorean theorem in the xz-plane: \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13\}.

Step 7: Finding Distance to the z-axis

The z-axis is defined by having both x=0 and y=0. The distance from \(4, -2, 6\) to the z-axis is found using the Pythagorean theorem in the xy-plane: \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5\}.

Key Concepts

3D Cartesian planes

Understanding the 3D Cartesian planes is crucial for dealing with geometric problems in space. In a three-dimensional coordinate system, we have the x, y, and z axes intersecting at a point called the origin. From the origin, planes are formed by setting each axis to zero, one at a time, to define the xy-plane, yz-plane, and xz-plane.

• The **xy-plane** is where the z-coordinate is always zero. It stretches infinitely along the x and y axes.
• The **yz-plane** is where the x-coordinate is consistently zero. It extends along the y and z directions.
• The **xz-plane** sets the y-coordinate to zero and spans across the x and z axes.

These planes help us in visualizing points and determining distances in the 3D space. Whenever calculating distances to one of these planes, the distance is simply the absolute value of the coordinate not included in the plane.

Distance formula

When dealing with points in a 3D Cartesian plane, determining the distance between points or to specific planes and axes is a frequent necessity. The distance formula in three dimensions extends the Pythagorean theorem into 3D space. Given two points \(A(x_1, y_1, z_1)\) and \(B(x_2, y_2, z_2)\), the distance between them is calculated as:\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]For calculating distances from a point to a plane or an axis, we adjust the formula to consider only the relevant dimensions to compute the shortest path in space.

Point-to-plane distance

Calculating the distance from a point to a plane involves determining the perpendicular distance from the point to the plane, which is the shortest possible distance.

When dealing with the standard planes within the 3D Cartesian system:

• **To the xy-plane**: Only the z-coordinate matters, so the distance is simply the absolute value of that coordinate.
• **To the yz-plane**: Similarly, the distance involves the x-coordinate, ignoring y and z coordinates.
• **To the xz-plane**: Here, the y-coordinate is of interest, with z and x being irrelevant.

These scenarios highlight why only one coordinate matters in point-to-plane distance calculations when dealing with the standard Cartesian planes.

Point-to-axis distance

Distances from a point to an axis in a 3D coordinate system require a more involved process because the shortest route may have components spanning multiple dimensions. We rely on the Pythagorean theorem to calculate these.

• **To the x-axis**: Consider the yz-plane. The distance uses both the y and z coordinates: \(\sqrt{y^2 + z^2}\).
• **To the y-axis**: Here, the distance comes from the xz-plane using x and z coordinates: \(\sqrt{x^2 + z^2}\).
• **To the z-axis**: In the xy-plane, x and y coordinates determine the distance: \(\sqrt{x^2 + y^2}\).

This method ensures that the computed path from the point to the axis is always the shortest possible, taking into account the orthogonal structure of the Cartesian system.