Problem 34 - University Calculus: Early Transcendentals

Question

find the distance from the point to the line. $$(0,0,0) ; \quad x=5+3 t, \quad y=5+4 t, \quad z=-3-5 t$$

Step-by-Step Solution

Answer

The distance from the point to the line is 5 units.

Step 1: Identify point and line components

The point given is \(P(0,0,0)\). The parametric equations for the line are \(x = 5 + 3t\), \(y = 5 + 4t\), and \(z = -3 - 5t\). We will denote a point on the line as \(Q(x, y, z) = (5 + 3t, 5 + 4t, -3 - 5t)\).

Step 2: Find a vector from the point to a point on the line

Take the vector PQ by subtracting the point \(P\) coordinates from the line coordinates: \( \vec{PQ} = Q - P = (5 + 3t - 0, 5 + 4t - 0, -3 - 5t - 0) = (5 + 3t, 5 + 4t, -3 - 5t) \).

Step 3: Find the direction vector of the line

The direction vector of the line, \( \vec{d} \), is derived from the coefficients of \(t\) in the parametric equations: \( \vec{d} = (3, 4, -5) \).

Step 4: Project vector \(\vec{PQ}\) onto \(\vec{d}\)

The projection of \(\vec{PQ} \) onto \( \vec{d} \) is given by the formula: \[ \text{Proj}_{\vec{d}} \vec{PQ} = \frac{\vec{PQ} \cdot \vec{d}}{\|\vec{d}\|^2} \vec{d} \]. First, compute the dot product \(\vec{PQ} \cdot \vec{d} = (5 + 3t) \times 3 + (5 + 4t) \times 4 + (-3 - 5t) \times (-5) \).

Step 5: Calculate the magnitude of the direction vector

Calculate \(\|\vec{d}\| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \). Then, \(\|\vec{d}\|^2 = 50\).

Step 6: Calculate the distance

Use the formula for the distance from point to line: \[ \text{Distance} = \|\vec{PQ} - \text{Proj}_{\vec{d}} \vec{PQ}\| \]. First, find \( \vec{PQ} \cdot \vec{d} = 0 \,\) because the line passes through \(P(0,0,0)\). Therefore, the shortest distance from the point to the line is the magnitude of the part of \(\vec{PQ}\) not parallel to \(\vec{d}\), which boils down to the calculation above: \- \sqrt{{(5 + 3t)^2 + (5 + 4t)^2 + (-3 - 5t)^2}} \text{ when t=0, equals 5 }\.

Key Concepts

Parametric equations

Parametric equations are a mathematical way to express geometric objects like lines and curves using parameters. A line is often described with parametric equations in three-dimensional space, just like in the exercise where the equations are:

• \(x = 5 + 3t\)
• \(y = 5 + 4t\)
• \(z = -3 - 5t\)

In these equations, \(t\) is a parameter representing scalar multiples along the line's direction vector.
It helps to determine any point on the line by substituting \(t\) with real numbers. Parametric equations are incredibly useful as they allow us to easily handle programming tasks and mathematical modeling. Understanding the role of \(t\) can help you visualize how a point travels along the line.

Vector projection

Vector projection is a technique used to project one vector onto another. In this exercise, the vector from the point \(P\) to a point \(Q\) on the line (\(\vec{PQ}\)) was projected onto the direction vector \(\vec{d}\). The formula for vector projection is:
\[\text{Proj}_{\vec{d}} \vec{PQ} = \frac{\vec{PQ} \cdot \vec{d}}{\|\vec{d}\|^2} \vec{d}\]This calculation tells us the extent to which \(\vec{PQ}\) aligns with \(\vec{d}\). By visualizing projections, we better understand how vectors relate spatially.
Once you find the projection, you subtract it from \(\vec{PQ}\) to obtain the component of \(\vec{PQ}\) that is perpendicular to \(\vec{d}\). This perpendicular segment is crucial for finding distances and analyzing vector behavior.

Direction vector

A direction vector indicates the direction in which a line travels. It is derived by inspecting the coefficients of the parameter \(t\) in the parametric equations of a line.
For the given exercise, the direction vector \(\vec{d}\) is:

• \(\vec{d} = (3, 4, -5)\)

This vector shows how the line changes in the x, y, and z directions.
In essence, the direction vector possesses all the essential directional attributes of a line, allowing for calculation of parallel lines and comprehension of spatial orientations. Working with direction vectors is crucial for understanding the geometric alignment and displacement in multiple dimensions.

Dot product

The dot product, also known as the scalar product, is a fundamental mathematical operation involving two vectors. It provides a single number (scalar) and measures how much one vector extends in the direction of another.
The formula for the dot product of vectors \(\vec{a} = (a_1, a_2, a_3)\) and \(\vec{b} = (b_1, b_2, b_3)\) is:
\[\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\]In the exercise, the dot product was utilized to find the projection of \(\vec{PQ}\) onto \(\vec{d}\).
This calculation reveals the component of \(\vec{PQ}\) that aligns with \(\vec{d}\). Mastering the dot product helps not only in vector projections but also in understanding angles between vectors, as a dot product of zero signifies perpendicular vectors. This concept expands into both math and physics for tasks requiring directional analysis and vector manipulation.