Problem 25 - APEX Calculus

Question

Find the distance from the point to the line. \(Q=(0,3), \quad \vec{\ell}(t)=\langle 2,0\rangle+t\langle 1,1\rangle\)

Step-by-Step Solution

Answer

The distance is \( \frac{3}{\sqrt{2}} \).

Step 1: Identify Line Equation

The line is given in vector form as \( \vec{\ell}(t) = \langle 2, 0 \rangle + t \langle 1, 1 \rangle \). This represents a line passing through the point (2, 0) with direction vector \( \langle 1, 1 \rangle \).

Step 2: Define Point P on the Line

A general point \( P \) on the line \( \vec{\ell}(t) = \langle 2 + t, t \rangle \) can be expressed. Thus, for any \( t \), the point on the line is \( (2 + t, t) \).

Step 3: Find Vector from Q to P

We need a vector from \( Q = (0, 3) \) to \( P = (2 + t, t) \). This vector is given by \( \langle (2 + t) - 0, t - 3 \rangle = \langle 2 + t, t - 3 \rangle \).

Step 4: Find Distance Formula

The perpendicular distance from a point to a line can be found using the formula for the magnitude of the cross product of vector \( \overrightarrow{QP} = \langle 2 + t, t - 3 \rangle \) and the unit vector in the direction of the line \( \langle 1, 1 \rangle \).

Step 5: Calculate Cross Product Magnitude

Compute the cross product and find its magnitude as \( |(2 + t) - (t - 3)| = |5| = 5 \).

Step 6: Calculate Distance

The distance from point \( Q \, to \, line \, \ell \) is \( | \overrightarrow{QP} \cdot \mathbf{n} | \), where \( \mathbf{n} \) is a unit vector perpendicular to \( \langle 1, 1 \rangle \), thus the distance is \( \frac{|(2+0) - (3+0)|}{\sqrt{1^2+1^2}} = \frac{3}{\sqrt{2}} \).

Step 7: Conclusion

The perpendicular distance from point \( Q \, to \, line \, \ell \) is \( \frac{3}{\sqrt{2}} \).

Key Concepts

Vector Equation

Understanding vector equations is essential when working with lines in the context of geometry and physics. A vector equation of a line determines every point on that line through a parameter, often denoted as \( t \). This can be thought of as a sophisticated way of representing a line instead of simple slope-intercept form.
Imagine you define a line by a given point on the line and a direction vector. In the problem, the line is described with the equation \( \vec{\ell}(t) = \langle 2, 0 \rangle + t \langle 1, 1 \rangle \).
Here:

• \( \langle 2, 0 \rangle \) is a specific point on the line, often referred to as the position vector.
• \( \langle 1, 1 \rangle \) is the direction vector, representing the line's orientation.
• \( t \) is a scalar, that is freely varying, allowing the equation to cover all points on the line when adjusted.

Using vector equations allows for a more dynamic representation, which can handle multi-dimensional problems with ease.

Distance Formula

When you need to find the distance from a point to a line, the distance formula becomes your primary tool. The distance from a point \( Q \) to a line can be thought of as the shortest path connecting \( Q \) straight to the line—this is always a line segment at right angles (perpendicular) to the line.
To calculate this distance using vectors, first recognize:

• You construct a vector from the point \( Q \) to any point \( P \) on the line.
• The formula involves using a cross product to find a vector orthogonal to the line of interest.

Mathematically, the distance \( d \) can be expressed as:\[d = \frac{|\overrightarrow{QP} \times \mathbf{d}|}{|\mathbf{d}|}\]Where \( \overrightarrow{QP} \) is the vector from \( Q \) to \( P \) and \( \mathbf{d} \) is the direction vector of the line. For our problem, substituting into the formula gave the correct perpendicular distance.
Proper use of the distance formula reveals the elegant symmetry in geometry and is a vital skill for solving vector problems.

Perpendicular Distance

The concept of perpendicular distance is crucial, as it signifies the shortest distance from a point to a line. Think of it as the length of the shadow a perpendicular line from the point would cast. In geometry, this measures the minimum separation between a line and a point, providing an exact measure that doesn't change with orientation or position changes along the line.
In the exercise, after identifying point \( Q \), we seek the perpendicular distance to the line expressed in vector form. With vectors \( \langle 1, 1 \rangle \) describing the line direction, we find a perpendicular using vector operations.
The perpendicular is achieved with vector cross products or projections. The process:

• Compute a perpendicular vector to the direction vector.
• Use this perpendicular vector to calculate magnitude.

Through these steps, you ensure that the derived distance is strictly perpendicular, reflecting the physical notion of shortest separation.

Cross Product

Among the useful vector operations, the cross product stands out when dealing with perpendicular matters. The cross product of two vectors results in a third vector that is orthogonal to the initial pair. This perpendicular aspect is fundamental when calculating distances and is particularly harnessed in geometry problems like the exercise.
For the point-to-line distance, the cross product helps find a vector orthogonal to the line's direction vector \( \langle 1, 1 \rangle \). This involves calculating:

• The vector from the point \( Q \) to any point on the line, noted as \( \overrightarrow{QP} \).
• The cross product \( \overrightarrow{QP} \times \mathbf{d} \), where \( \mathbf{d} = \langle 1, 1 \rangle \).

Even though the exercise didn't explicitly require plotting the full cross product (since the operation is 2D here), its importance remains. The magnitude of this resultant vector reflects the area of a parallelogram composed by the vectors, aligning directly with perpendicular concepts.
Thus, the cross product isn't just a computational trick—it provides structural insights into how vectors interact in space.