Problem 36
Question
In Exercises \(33-38,\) find the distance from the point to the line. $$ (2,1,-1) ; \quad x=2 t, \quad y=1+2 t, \quad z=2 t $$
Step-by-Step Solution
Verified Answer
The distance from the point to the line is \( \sqrt{5} \).
1Step 1: Understand the Line Representation
The line is given in its parametric form with equations \( x = 2t \), \( y = 1 + 2t \), and \( z = 2t \). This corresponds to a point on the line \((0, 1, 0)\) and a direction vector \(\mathbf{d} = (2, 2, 2)\).
2Step 2: Determine the Vector from the Point to the Line
The given point is \( (2, 1, -1) \). We find the vector from the point \((0, 1, 0)\) on the line to the point \((2, 1, -1)\) which is \( \mathbf{v} = (2, 1, -1) - (0, 1, 0) = (2, 0, -1) \).
3Step 3: Find the Projection of the Vector onto the Line's Direction
To project \( \mathbf{v} \) onto \( \mathbf{d} \), use the formula: \[ \text{proj}_{\mathbf{d}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d} \] Compute \( \mathbf{v} \cdot \mathbf{d} = (2)(2) + (0)(2) + (-1)(2) = 4 \). Compute \( \mathbf{d} \cdot \mathbf{d} = (2)^2 + (2)^2 + (2)^2 = 12 \). Thus, \( \text{proj}_{\mathbf{d}} \mathbf{v} = \frac{4}{12} \cdot (2, 2, 2) = \left( \frac{2}{3}, \frac{2}{3}, \frac{2}{3} \right) \).
4Step 4: Calculate the Perpendicular Vector
The perpendicular vector from the point to the line is the difference between \( \mathbf{v} \) and its projection onto \( \mathbf{d} \). So, \[ \mathbf{p} = \mathbf{v} - \text{proj}_{\mathbf{d}} \mathbf{v} = (2, 0, -1) - \left( \frac{2}{3}, \frac{2}{3}, \frac{2}{3} \right) = \left( \frac{4}{3}, -\frac{2}{3}, -\frac{5}{3} \right) \].
5Step 5: Find the Distance as the Magnitude of the Perpendicular Vector
The distance from the point to the line is the magnitude of \( \mathbf{p} \): \[ \| \mathbf{p} \| = \sqrt{ \left( \frac{4}{3} \right)^2 + \left( -\frac{2}{3} \right)^2 + \left( -\frac{5}{3} \right)^2 } = \sqrt{ \frac{16}{9} + \frac{4}{9} + \frac{25}{9} } = \sqrt{ \frac{45}{9} } = \sqrt{5} \].
Key Concepts
Parametric EquationsVector ProjectionMagnitude of Vector
Parametric Equations
Parametric equations are an effective way to represent lines, especially in 3D space. They express coordinates as functions of a parameter, typically denoted as \( t \). For the line given in the exercise, the parametric equations are \( x = 2t \), \( y = 1 + 2t \), and \( z = 2t \). These equations define each point along the line as \( t \) varies. This convenient form helps us easily identify the direction vector of the line, which is crucial for vectors and distance calculations.
The equations can be rearranged to provide one specific point on the line and a direction vector. In our case, setting \( t = 0 \), the point is \( (0, 1, 0) \). The direction vector, which outlines the line's path, is \( \mathbf{d} = (2, 2, 2) \). This vector indicates how far and in what direction we move in \( x \), \( y \), and \( z \) for each unit of \( t \).
Understanding parametric equations allows us to analyze the line's behavior in space and to use these details in further calculations, like finding distances or intersections.
The equations can be rearranged to provide one specific point on the line and a direction vector. In our case, setting \( t = 0 \), the point is \( (0, 1, 0) \). The direction vector, which outlines the line's path, is \( \mathbf{d} = (2, 2, 2) \). This vector indicates how far and in what direction we move in \( x \), \( y \), and \( z \) for each unit of \( t \).
Understanding parametric equations allows us to analyze the line's behavior in space and to use these details in further calculations, like finding distances or intersections.
Vector Projection
Vector projection is an essential concept in understanding how one vector can be expressed in relation to another. When we project one vector onto another, we're finding the component of one vector that lies in the direction of the other vector. In mathematical terms, the projection of vector \( \mathbf{v} \) onto vector \( \mathbf{d} \) is calculated as:
\[ \text{proj}_{\mathbf{d}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d} \]
This formula involves the dot product, which measures how much one vector extends in the direction of another. For the vectors in our problem, \( \mathbf{v} = (2, 0, -1) \) and \( \mathbf{d} = (2, 2, 2) \), the dot product \( \mathbf{v} \cdot \mathbf{d} \) equals \( 4 \), while \( \mathbf{d} \cdot \mathbf{d} \) is \( 12 \). With these values, the projection vector comes out to \( \left( \frac{2}{3}, \frac{2}{3}, \frac{2}{3} \right) \).
Understanding the projection helps us decompose vectors into components parallel and perpendicular to a given direction. This is particularly useful for calculating the shortest distance from a point to a line, as is required in this exercise.
\[ \text{proj}_{\mathbf{d}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d} \]
This formula involves the dot product, which measures how much one vector extends in the direction of another. For the vectors in our problem, \( \mathbf{v} = (2, 0, -1) \) and \( \mathbf{d} = (2, 2, 2) \), the dot product \( \mathbf{v} \cdot \mathbf{d} \) equals \( 4 \), while \( \mathbf{d} \cdot \mathbf{d} \) is \( 12 \). With these values, the projection vector comes out to \( \left( \frac{2}{3}, \frac{2}{3}, \frac{2}{3} \right) \).
Understanding the projection helps us decompose vectors into components parallel and perpendicular to a given direction. This is particularly useful for calculating the shortest distance from a point to a line, as is required in this exercise.
Magnitude of Vector
The magnitude of a vector, often considered its 'length', is another crucial concept. It's calculated using the Pythagorean theorem extended into 3D space. For a vector \( \mathbf{p} = (x, y, z) \), its magnitude is given by:
\[ \| \mathbf{p} \| = \sqrt{x^2 + y^2 + z^2} \]
In the context of the exercise, finding the magnitude of the perpendicular vector \( \mathbf{p} = \left( \frac{4}{3}, -\frac{2}{3}, -\frac{5}{3} \right) \) gives us the shortest distance from the point to the line. Calculating this, we find:
\[ \| \mathbf{p} \| = \sqrt{ \left( \frac{4}{3} \right)^2 + \left( -\frac{2}{3} \right)^2 + \left( -\frac{5}{3} \right)^2 } \]
This simplifies to \( \sqrt{5} \), providing the distance. Understanding how to calculate and interpret vector magnitudes is indispensable in navigating space and solving geometric problems like this.
\[ \| \mathbf{p} \| = \sqrt{x^2 + y^2 + z^2} \]
In the context of the exercise, finding the magnitude of the perpendicular vector \( \mathbf{p} = \left( \frac{4}{3}, -\frac{2}{3}, -\frac{5}{3} \right) \) gives us the shortest distance from the point to the line. Calculating this, we find:
\[ \| \mathbf{p} \| = \sqrt{ \left( \frac{4}{3} \right)^2 + \left( -\frac{2}{3} \right)^2 + \left( -\frac{5}{3} \right)^2 } \]
This simplifies to \( \sqrt{5} \), providing the distance. Understanding how to calculate and interpret vector magnitudes is indispensable in navigating space and solving geometric problems like this.
Other exercises in this chapter
Problem 35
Sketch the surfaces in Exercises \(13-44.\) ASSORTED $$y=-\left(x^{2}+z^{2}\right)$$
View solution Problem 36
Find the areas of the parallelograms whose vertices are given in Exercises \(35-40 .\) $$ A(0,0), \quad B(7,3), \quad C(9,8), \quad D(2,5) $$
View solution Problem 36
Find a. the direction of \(\overrightarrow{P_{1} P}_{2}\) and b. the midpoint of line segment \(P_{1} P_{2}\). \(P_{1}(1,4,5) \quad P_{2}(4,-2,7)\)
View solution Problem 36
Write inequalities to describe the sets in Exercises \(35-40\) The solid cube in the first octant bounded by the coordinate planes and the planes \(x=2, y=2,\)
View solution