Problem 36 - University Calculus: Early Transcendentals

Question

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

Answer

The distance from the point to the line is \(\sqrt{3}\).

Step 1: Understand the Point and the Line's Parametric Equations

The point given is \((2, 1, -1)\), and the line is defined by the parametric equations: \(x = 2t\), \(y = 1 + 2t\), and \(z = 2t\). This means a point on the line can be represented as \((2t, 1+2t, 2t)\) for any parameter \(t\).

Step 2: Identify Direction Vector of the Line

From the line's parametric equations, we can extract the direction vector \(\mathbf{v} = (2, 2, 2)\). This vector indicates the direction in which the line extends.

Step 3: Use Vector Projection to Find Shortest Distance Formula

The shortest distance \(d\) from a point \(\mathbf{P} = (x_0, y_0, z_0)\) to a line defined by a point \(\mathbf{A} = (x_1, y_1, z_1)\) and direction vector \(\mathbf{v} = (a, b, c)\) is given by:\[d = \frac{\|\mathbf{AP} \times \mathbf{v}\|}{\|\mathbf{v}\|}\]where \(\mathbf{AP} = (x_0 - x_1, y_0 - y_1, z_0 - z_1)\).

Step 4: Find Vector AP

Let's take the point \((x_1, y_1, z_1)\) on the line when \(t=0\), so \(\mathbf{A} = (0, 1, 0)\). The vector \(\mathbf{AP}\) is given by the following components:\[\mathbf{AP} = (2 - 0, 1 - 1, -1 - 0) = (2, 0, -1)\]

Step 5: Compute Cross Product AP × v

The cross product \(\mathbf{AP} \times \mathbf{v}\) is calculated as follows:\[\mathbf{AP} \times \mathbf{v} = \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 0 & -1 \ 2 & 2 & 2 \end{pmatrix} = \hat{i}(0 + 2) - \hat{j}(-2 - 2) + \hat{k}(4 - 0) = 2\hat{i} + 4\hat{j} + 4\hat{k}\]Thus, \(\mathbf{AP} \times \mathbf{v} = (2, 4, 4)\).

Step 6: Calculate Magnitude of AP × v

The magnitude \(\|\mathbf{AP} \times \mathbf{v}\|\) is:\[\|\mathbf{AP} \times \mathbf{v}\| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6\]

Step 7: Calculate Magnitude of Direction Vector v

The magnitude \(\|\mathbf{v}\|\) is:\[\|\mathbf{v}\| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}\]

Step 8: Calculate Shortest Distance

Using the distance formula:\[d = \frac{\|\mathbf{AP} \times \mathbf{v}\|}{\|\mathbf{v}\|} = \frac{6}{2\sqrt{3}} = \sqrt{3}\]Thus, the distance from the point to the line is \(\sqrt{3}\).

Key Concepts

Parametric Equations

Parametric equations are a way to express a line or curve using one or more parameters, usually denoted by a variable like \(t\). Instead of representing a point on the line with a set of coordinates like \((x, y, z)\), parametric equations describe coordinates as functions of the parameter. In this exercise, the line is described by the equations \(x = 2t\), \(y = 1 + 2t\), and \(z = 2t\).
These equations articulate any point on the line as \((2t, 1+2t, 2t)\). Here, \(t\) can be any real number, and by altering \(t\), you can move along the line. This representation is incredibly handy for mathematical computations and visualizations because it simplifies complex geometric problems to functions that depend on just a single parameter.

Vector Cross Product

The vector cross product is a fundamental operation between two vectors in three-dimensional space. When we have two vectors, such as \(\mathbf{AP}\) (from the point on the line to the given point), and the direction vector \(\mathbf{v}\) of the line, the cross product \(\mathbf{AP} \times \mathbf{v}\) results in a third vector that is perpendicular to both original vectors. The formula used is:

• The cross product is computed as a determinant of a matrix containing the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) in the first row, components of \(\mathbf{AP}\) in the second, and components of \(\mathbf{v}\) in the third row.
• This operation provides us a vector whose length is vital for calculating distances or areas.

The vector we've acquired from this operation in the problem \((2, 4, 4)\) presents an "area" aspect of the triangle where two sides are defined by the point-lie vector \(\mathbf{AP}\) and the direction vector \(\mathbf{v}\).

Direction Vector

A direction vector gives the direction in which a line extends in space. In our problem, the parametric equations \(x = 2t\), \(y = 1 + 2t\), \(z = 2t\) provide a direction vector \(\mathbf{v} = (2, 2, 2)\).
This vector simplifies the line's characteristics to a simple format, making it much easier to understand and perform calculations. For almost any line in 3D space defined by parametric equations, the coefficients of \(t\) represent the line's direction vector:

• The direction vector can tell us if lines are parallel (same direction) or intersecting (different directions).
• It helps in determining angles between two lines or computing projections and cross products.

Bearing in mind its importance, direction vectors can efficiently transform geometric problems into algebraic tasks.

Vector Magnitude

The magnitude of a vector, often thought of as the "length" of the vector, is a measure of how far the vector extends in space. It's computed using the formula \(\|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2}\) for a vector \(\mathbf{v} = (a, b, c)\).
In our case:

• The magnitude of the direction vector \(\mathbf{v} = (2, 2, 2)\) was found to be \(\sqrt{12} = 2\sqrt{3}\).
• The magnitude of the cross product vector \((2, 4, 4)\) turned out to be 6.

Understanding the magnitude is essential because it standardizes vectors, allowing for consistent measurements of "distance" or "length." In the given problem of finding the shortest distance from a point to the line, the magnitude was necessary for the distance formula \(d = \frac{\|\mathbf{AP} \times \mathbf{v}\|}{\|\mathbf{v}\|}\), culminating in an elegant result of \(\sqrt{3}\).