Problem 46 - University Calculus: Early Transcendentals

Question

Find the distance from the line \(x=2+t, y=1+t\) \(z=-(1 / 2)-(1 / 2) t\) to the plane \(x+2 y+6 z=10\)

Step-by-Step Solution

Answer

The distance from the line to the plane is \(\sqrt{\frac{3321}{1681}}\).

Step 1: Identify the direction vector

The given line equations can be expressed in vector form as r(t) = [2, 1, -1/2] + t[1, 1, -1/2]. The direction vector of the line is \([1, 1, -1/2]\).

Step 2: Identify a point on the line

A specific point can be found on the line by substituting a value for t, such as t = 0. For t = 0, the point becomes \((2, 1, -1/2)\).

Step 3: Determine the normal vector of the plane

The plane equation is given as \(x + 2y + 6z = 10\). The normal vector to the plane is \([1, 2, 6]\).

Step 4: Find the vector from the point to the plane

Choose a point on the plane. Solving \(x + 2y + 6z = 10\), if we let \(y = 0, z = 0\), then \(x = 10\), giving point \((10, 0, 0)\). The vector from point on the line \((2, 1, -1/2)\) to the point on the plane \((10, 0, 0)\) is \([8, -1, 1/2]\).

Step 5: Project onto the normal vector

We need to calculate the projection of the vector \([8, -1, 1/2]\) onto the normal vector \([1, 2, 6]\). The formula for projection is:\[\text{Proj}_n a = \frac{a \cdot n}{n \cdot n} n\]Calculate \(a \cdot n = 8 \times 1 + (-1) \times 2 + (1/2) \times 6 = 8 - 2 + 3 = 9\).Calculate \(n \cdot n = 1^2 + 2^2 + 6^2 = 1 + 4 + 36 = 41\).The projection is then \(\frac{9}{41} [1, 2, 6]\).

Step 6: Calculate the length of the projection

The distance from the line to the plane is the length of the projection vector. The vector is \(\frac{9}{41} [1, 2, 6] = \left[ \frac{9}{41}, \frac{18}{41}, \frac{54}{41} \right]\). Calculate the magnitude:\[\text{Distance} = \sqrt{\left(\frac{9}{41}\right)^2 + \left(\frac{18}{41}\right)^2 + \left(\frac{54}{41}\right)^2}\]Computing each term:- \(\left(\frac{9}{41}\right)^2 = \frac{81}{1681}\)- \(\left(\frac{18}{41}\right)^2 = \frac{324}{1681}\)- \(\left(\frac{54}{41}\right)^2 = \frac{2916}{1681}\)Adding these terms gives: \(\frac{81 + 324 + 2916}{1681} = \frac{3321}{1681}\).Thus, the distance is \(\sqrt{\frac{3321}{1681}}\).

Key Concepts

Vector Projection

The concept of vector projection is essential when determining the distance between a line and a plane. In simple terms, vector projection helps us project one vector onto another. This is particularly useful when we want to find the shortest or perpendicular distance from a point on a line to a plane.
To compute the projection of a vector \(\mathbf{a}\) onto another vector \(\mathbf{n}\), we use the formula:\[\text{Proj}_n \, \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{n}}{\mathbf{n} \cdot \mathbf{n}} \cdot \mathbf{n}\]

• \(\mathbf{a} \cdot \mathbf{n}\) is the dot product of the two vectors.
• \(\mathbf{n} \cdot \mathbf{n}\) is the dot product of the normal vector with itself.

In our example, we project the vector from a point on the line to a point on the plane onto the plane's normal vector. This results in another vector that points along the normal direction but of adjusted magnitude, helping us find the shortest distance.

Direction Vector

A direction vector gives direction to a line in a spatial context. It's often derived from the parametrization of the line. In our exercise, the line's equations are given as \(x = 2 + t\), \(y = 1 + t\), and \(z = -\frac{1}{2} - \frac{1}{2}t\).
From these, we see that:

• The coefficient of \(t\) in each equation forms the components of the direction vector \([1, 1, -1/2]\).

Direction vectors are crucial for defining lines in vector form and help in various calculations, especially when comparing multiple spatial elements like lines or planes. They don't usually change the length (or magnitude) but represent the flow or direction from one point to another along the line.

Normal Vector

A normal vector is perpendicular to a surface, like a plane. It plays a significant role in geometry and physics by determining the orientation of the surface.
From the equation of the plane \(x + 2y + 6z = 10\), the coefficients of \(x\), \(y\), and \(z\) provide the components of the normal vector \([1, 2, 6]\).

• This vector is essential when computing the perpendicular (shortest) distance from a given point to a plane.
• Being perpendicular, it effectively helps in finding how much of the point-line distance "penetrates" or "lies above" the plane.

The normal vector is thus central when projecting the line-point vector onto it to find the shortest distance.

Magnitude of a Vector

The magnitude of a vector gives its length and is calculated through the Pythagorean theorem applied in three-dimensional spaces. For any vector \(\mathbf{v} = [v_x, v_y, v_z]\), the magnitude \(|\mathbf{v}|\) is given by:\[|\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}\]

• This formula helps us identify how "long" the vector is, which is crucial when we're calculating distances.

In our exercise, the magnitude of the projection vector represents the distance from the line to the plane. So, after determining the projection vector's components, we find its magnitude to derive the definitive distance. Understanding how to calculate and interpret vector magnitudes assists in transitioning from spatial directionality to measurable lengths.