Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the distance from \((2,3,-1)\) to (a) the \(x y\) -plane, (b) the \(y\) -axis, and (c) the origin.

Step-by-Step Solution

Answer

(a) 1, (b) \( \sqrt{5} \), (c) \( \sqrt{14} \)

Step 1: Distance to xy-plane

The distance from a point \(x, y, z\) to the \(xy\) -plane is given by the absolute value of the \(z\) -coordinate. Thus, for the point \( (2, 3, -1) \), the distance is:\[ |z| = |-1| = 1 \]

Step 2: Distance to y-axis

The distance from a point \(x, y, z\) to the \(y\) -axis is given by the formula \sqrt{x^2 + z^2}\. Thus, for the point \( (2, 3, -1) \), the distance is:\[ \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}\]

Step 3: Distance to the origin

The distance from a point \(x, y, z\) to the origin \(0, 0, 0\) is given by the formula \sqrt{x^2 + y^2 + z^2}\. Thus, for the point \( (2, 3, -1) \), the distance is:\[ \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}\]

Key Concepts

Understanding Distance to a Plane in 3D Space

In three-dimensional space, one of the frequent tasks is to find the distance of a given point from a specific plane. For instance, asking for the distance from the point \((2, 3, -1)\) to the \(xy\)-plane means determining how far the point is from being directly on that plane.
The \(xy\)-plane is characterized by a \(z\)-coordinate of zero, meaning any point on this plane has the form \((x, y, 0)\).
To find how far our point \((2, 3, -1)\) is from the \(xy\)-plane, we simply look at the absolute value of the \(z\)-coordinate of the point. This is because the plane's lack of a \(z\) component means only vertical distance (perpendicular to the plane) matters.
Therefore, the distance is straightforward, calculated as:

• Take the absolute value of the \(z\)-coordinate: \(|z| = |-1| = 1\).

Finding Distance to an Axis

In the three-dimensional realm, locating the distance of a point from an axis captures a unique perspective. For our exercise, we focus on the distance from the point \((2, 3, -1)\) to the \(y\)-axis.
The \(y\)-axis can be identified by fixing the \(x\) and \(z\) coordinates to zero, making every point on it resemble \((0, y, 0)\).
The quest is to determine how far a point deviates horizontally and vertically from this axis. Picture an imaginary line extending from the point to the \(y\)-axis, forming a right triangle with legs parallel to the \(x\) and \(z\) axes.
To compute this distance, utilize the Pythagorean theorem as follows:

• Square both the \(x\) and \(z\) coordinates.
• Add these squared values together.
• Finally, calculate the square root of this sum: \(\sqrt{x^2 + z^2} = \sqrt{2^2 + (-1)^2} = \sqrt{5}\).

Distance to the Origin in 3D Space

Distance measurements in three dimensions often involve calculating how far a point is from the origin. The origin in a 3D coordinate system is located at \((0, 0, 0)\).
To find the distance from the point \((2, 3, -1)\) to the origin, envision it as the hypotenuse of a right triangle formed in space.
Use the formula derived from the Pythagorean theorem, extended into three dimensions, to determine this distance:

• Square each coordinate of the point: \(x\), \(y\), and \(z\).
• Add these squares together.
• Take the square root of the sum to find the distance: \(\sqrt{x^2 + y^2 + z^2} = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{14}\).

By applying this method, you can effectively measure how far any point lies from the origin, offering a clear step towards understanding spatial relationships in mathematics.