Problem 21 - Calculus: Early Transcendentals

Question

Find the distance from (2,-1,1) to \(\langle 2,2,0\rangle+t\langle 1,2,3\rangle .\)

Step-by-Step Solution

Answer

The distance is approximately 3.58 units.

Step 1: Understand the line equation

The line is expressed in parametric form with a point and a direction vector. Here, the line passes through the point \((2,2,0)\) and moves in the direction of the vector \((1,2,3)\). This can be represented as \((2 + t, 2 + 2t, 0 + 3t)\), where \(t\) is a parameter.

Step 2: Determine the vector from the point to the line

Define the vector from point \((2,-1,1)\) to a point on the line \((2 + t, 2 + 2t, 3t)\). This is \(( (2 + t) - 2, (2 + 2t) - (-1), (3t) - 1 )\) or \( ( t, 3 + 2t, 3t - 1 )\).

Step 3: Find vector perpendicular to the line

For the shortest distance, find the projection of the vector \( (t, 3 + 2t, 3t - 1) \) onto the direction vector \( \langle 1, 2, 3 \rangle \). Use the formula for the projection: \((\text{Proj}_\mathbf{b} \mathbf{a}) = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b}\).

Step 4: Calculate dot product and projection length

Compute the dot product \( (t, 3 + 2t, 3t - 1) \cdot (1, 2, 3) = t + 2(3 + 2t) + 3(3t - 1) = 13t + 5 \). The dot product of \( \langle 1, 2, 3 \rangle \) with itself is \( 14 \). Thus, the projection length is \( \frac{13t + 5}{14} \).

Step 5: Solve for t to find perpendicular vector

Set the dot product of the vector \( (t, 3 + 2t, 3t - 1) \) with the direction vector \( \langle 1, 2, 3 \rangle \) to zero to find \( t \). Solving \( 13t + 5 = 0 \) gives \( t = -\frac{5}{13}\).

Step 6: Compute point on line using t

Substitute \( t = -\frac{5}{13}\) back into the line equation to find the point on the line closest to \((2, -1, 1)\). This gives the point \( \left(2 - \frac{5}{13}, 2 - \frac{10}{13}, 0 - \frac{15}{13}\right) = \left(\frac{21}{13}, \frac{16}{13}, -\frac{15}{13}\right)\).

Step 7: Calculate the distance from the point to the line

Find the distance between \((2, -1, 1)\) and the point \( \left(\frac{21}{13}, \frac{16}{13}, -\frac{15}{13}\right)\). Use the distance formula: \[\sqrt{\left(2 - \frac{21}{13}\right)^2 + \left(-1 - \frac{16}{13}\right)^2 + \left(1 + \frac{15}{13}\right)^2}\].

Step 8: Final computation of distance

Compute each component: \( \left(\frac{5}{13}\right)^2 = \frac{25}{169} \), \( \left(-\frac{29}{13}\right)^2 = \frac{841}{169} \), \( \left(\frac{28}{13}\right)^2 = \frac{784}{169} \). Sum them up to get \( \frac{1650}{169} \). The distance is \( \sqrt{\frac{1650}{169}} = \frac{\sqrt{1650}}{13} \), which is approximately \ 3.58 \.

Key Concepts

Parametric Equations

A parametric equation represents a path or a curve by expressing the coordinates of the points as functions of a parameter. This can simplify the calculation of positions along a line or curve, as you only need to change the parameter value.

For instance, take the description of our line: it is given by the equations \( x = 2 + t \), \( y = 2 + 2t \), and \( z = 3t \), where \( t \) is our parameter. Each parameter value \( t \) specifies a unique position on this line.

• The point \((2,2,0)\) is where \(t=0\).
• The line's direction vector \(\langle 1, 2, 3 \rangle \) guides us on how the line extends as \(t\) increases.

By adjusting the parameter \( t \), you can find any point along this line, which is invaluable for tasks like finding the shortest distance to another point.

Vector Projections

Vector projections help us find how much of one vector goes in the direction of another. This is really handy when determining the shortest distance from a point to a line.

Picture projecting \( \mathbf{a} \) onto \( \mathbf{b} \). The formula is: \[ \text{Proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b} \]

• The dot product \( \mathbf{a} \cdot \mathbf{b} \) shows how much \( \mathbf{a} \) aligns with \( \mathbf{b} \).
• Dividing by \( \mathbf{b} \cdot \mathbf{b} \), a constant, adjusts this to the right scale.

This gives us a vector that points in the same direction as \( \mathbf{b} \) but its length is modified to reflect how much of \( \mathbf{a} \) points in that direction.

In our exercise, we projected the vector \( (t, 3+2t, 3t-1) \) from a point on the line to the line's direction vector.

Distance Formula in 3D

The distance formula in 3D extends the classic 2D distance formula by adding a third dimension. It lets us calculate how far apart two points are in space.

The formula is: \[\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\]

• Each squared term accounts for the difference in each respective coordinate.
• These differences are summed to reflect the total squared distance.

By taking the square root, the formula converts the total from squared units back to the original units that match the distance measure.

In the given problem, we used this formula to find the distance from the point \((2, -1, 1)\) to the nearest point on the line, determined in the previous steps.

Perpendicular Vectors

Perpendicular vectors have a dot product of zero. This fact is critical for finding the shortest distance from a point to a line.

To find a perpendicular vector, set the dot product of the vector from the point to the line and the line’s direction vector equal to zero and solve for the unknown parameter.

• If \( \mathbf{a} \) and \( \mathbf{b} \) are perpendicular, \( \mathbf{a} \cdot \mathbf{b} = 0 \).
• This means \( \mathbf{a} \) neither pulls towards nor away from \( \mathbf{b} \).

In our problem, by solving \( 13t + 5 = 0 \), we find the value of \( t \) where the vector from the point is perpendicular to the line. This reveals the exact spot where the line is closest to the point.