Using Vector Projections for Shortest Distances

Finding the shortest distance from a point to a line is a classic geometry problem, but in HSC Mathematics Extension 1, vector methods offer an elegant and insightful solution. This topic connects directly to the syllabus's treatment of vectors and their geometric applications, giving you a technique that bypasses coordinate geometry or calculus. By projecting one vector onto another, you'll learn how to decompose displacement into parallel and perpendicular components—a skill that appears in many applied contexts.

Problem Statement

A line $L$ passes through the points $A(-1, 2)$ and $B(3, 5)$. A point $P(4, 1)$ does not lie on the line.

1. Find the vector $\vec{AB}$ and the vector $\vec{AP}$.
2. Using vector projection, find the shortest distance from the point $P$ to the line $L$.

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Hints

• Recall that the shortest distance from a point to a line is the perpendicular distance.
• The scalar projection of $\vec{AP}$ onto $\vec{AB}$ gives the length of the adjacent side of the right-angled triangle formed by $A$, $P$, and the point on the line closest to $P$.
• You can use Pythagoras' theorem with the length of $\vec{AP}$ and the scalar projection of $\vec{AP}$ onto $\vec{AB}$ to find the perpendicular distance.
• Alternatively, the vector projection $\text{proj}_{\vec{AB}} \vec{AP}$ can be used. The perpendicular vector is $\vec{AP} - \text{proj}_{\vec{AB}} \vec{AP}$, and its magnitude is the shortest distance.

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Solutions

1. We begin by computing the two key displacement vectors. The vector $\vec{AB}$ gives the direction of the line, while $\vec{AP}$ connects a known point on the line to the external point.

   $$\vec{AB} = \begin{pmatrix} 3 - (-1) \\ 5 - 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$$

   $$\vec{AP} = \begin{pmatrix} 4 - (-1) \\ 1 - 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \end{pmatrix}$$

2. The shortest distance is the perpendicular length, so we can decompose $\vec{AP}$ into components along and perpendicular to $\vec{AB}$. Let $N$ be the point on the line $L$ that is closest to $P$. The line segment $PN$ is perpendicular to $L$. The length $AN$ is the scalar projection of $\vec{AP}$ onto $\vec{AB}$. Let's calculate this scalar projection.

   The scalar projection formula gives the signed length of the component along $\vec{AB}$:

   $$\text{Scalar Projection} = \frac{\vec{AP} \cdot \vec{AB}}{|\vec{AB}|}$$

   First, find the dot product $\vec{AP} \cdot \vec{AB}$:

   $$\vec{AP} \cdot \vec{AB} = (5)(4) + (-1)(3) = 20 - 3 = 17$$

   Next, find the magnitude of $\vec{AB}$:

   $$|\vec{AB}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$

   So, the scalar projection $AN = \frac{17}{5}$.

   Now, we find the magnitude of $\vec{AP}$ (the hypotenuse of the right-angled triangle $\triangle APN$):

   $$|\vec{AP}| = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26}$$

   With the hypotenuse and one leg known, the perpendicular distance is simply the other leg, obtained via Pythagoras' theorem in $\triangle APN$:

   $$|\vec{AP}|^2 = AN^2 + PN^2$$

   $$26 = \left(\frac{17}{5}\right)^2 + d^2$$

   $$d^2 = 26 - \frac{289}{25} = \frac{650 - 289}{25} = \frac{361}{25}$$

   Taking the positive square root:

   $$d = \frac{19}{5} = 3.8$$

   The shortest distance from $P$ to the line $L$ is exactly $3.8$ units.

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Takeaways

• Vectors provide a powerful, geometric way to solve shortest distance problems without needing to find the equation of a line or use calculus.
• The scalar projection breaks a vector into orthogonal components, making it ideal for perpendicularity problems.
• Always draw a quick diagram mentally or on paper to visualize how $\vec{AP}$, its projection along the line, and the perpendicular distance form a right-angled triangle.

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Further Readings

HSC Vectors, HSC Mechanics