/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Find the shortest distance from ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the shortest distance from the point \(P(2,1,-1)\) to the plane \(4 x-3 y+z=5\).

Short Answer

Expert verified
The shortest distance is \(\frac{1}{\sqrt{26}}\).

Step by step solution

01

Identify the formula

The shortest distance from a point \(x_1, y_1, z_1\) to a plane \ ax + by + cz + d = 0 \ is given by the formula: \[ D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]In this problem, the plane is \ 4x - 3y + z = 5 \ and can be rewritten in the form \( ax + by + cz + d = 0 \) with \( a = 4, b = -3, c = 1, d = -5 \). The point \( P(2, 1, -1) \) gives us \( x_1 = 2, y_1 = 1, z_1 = -1 \).
02

Substitute into the formula

Substitute the values from Step 1 into the distance formula: \[ D = \frac{|4(2) - 3(1) + 1(-1) - 5|}{\sqrt{4^2 + (-3)^2 + 1^2}} \]
03

Calculate the numerator

Calculate the expression inside the absolute value:\[ 4(2) - 3(1) + 1(-1) - 5 = 8 - 3 - 1 - 5 \ = -1 \]Since the absolute value of \( -1 \) is 1, we have \ |4(2) - 3(1) + 1(-1) - 5| = 1 \.
04

Calculate the denominator

Calculate the square root in the denominator:\[ \sqrt{4^2 + (-3)^2 + 1^2} = \sqrt{16 + 9 + 1} = \sqrt{26} \]
05

Compute the distance

Now divide the numerator by the denominator to find the distance:\[ D = \frac{1}{\sqrt{26}} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distance formula
The distance formula is a powerful tool in geometry that helps us find the shortest distance between a point and a plane. It may sound complex, but it's rooted in straightforward principles. In this context, the formula is: \[D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\] What does this mean? Each letter in the equation represents a part of the point-to-plane connection. The letters \(x_1, y_1, z_1\) show the coordinates of the point you're measuring from. The letters \(a, b, c\) are the coefficients from the plane equation, and \(d\) is a constant term derived when the equation is rearranged in the form \(ax + by + cz + d = 0\). The numerator \(|ax_1 + by_1 + cz_1 + d|\) calculates the weighted distance between the point and the plane along its normal vector. The denominator \(\sqrt{a^2 + b^2 + c^2}\) scales this by the length of the normal vector to ensure the distance isn't distorted by vector length. This formula is derived from fundamental concepts of geometry and vector calculus, making it a vital part of 3D geometry.
3D geometry
3D geometry is the mathematical study of shapes and figures in three dimensions. Unlike 2D shapes flat on the plane, 3D shapes exist in space and boast attributes like surface area and volume. Understanding these shapes helps us model real-world scenarios. In exercise examples like calculating distances from points to planes, three key coordinates \((x, y, z)\) guide us through the three-dimensional space. This spatial representation allows us to visualize and comprehend concepts like perpendicular distances, angles, and intersections. When examining a plane, it's crucial to identify its equation, frequently expressed as \(ax + by + cz = d\). The magnitude \(\sqrt{a^2 + b^2 + c^2}\) represents the length of the normal vector to the plane in space, a core concept of vector calculus. Such geometry supports applications in fields like architecture, physics, and computer graphics.
Vector calculus
Vector calculus is a branch of mathematics focused on vector fields and differentiation and integration of vector functions. Understanding vector calculus is essential for analyzing scenarios in physics and engineering, where direction and magnitude are critical. Vectors have a magnitude and direction, functioning as arrows in space. In exercises like finding distances or working with planes, vectors help us navigate and solve spatial problems. The plane's normal vector, composed of coefficients \((a, b, c)\) in its equation, is vital in determining perpendicular distances, like from point to plane, using tools from vector calculus. In the given problem, calculating the shortest distance involves projecting the point directly onto the plane along this normal vector. The vector form of the plane's equation helps us balance graphical intuition with numerical precision. This creates a seamless fusion of algebra and geometry, empowering us to solve complex spatial challenges with ease.

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