Perpendicular from a point to a line

Perpendicular from a Point to a Line

In geometry, drawing a perpendicular from a point to a line is a fundamental concept. It involves creating a line segment from a given point that intersects a given line at a right angle (90 degrees). This concept is crucial in various geometric constructions and proofs.

Understanding the Concept

When we say that we are drawing a perpendicular from a point to a line, we mean that we are finding the shortest distance from that point to the line. This shortest path will always be along a line that intersects the original line at a right angle.

Formula

In 3D geometry, the distance ( d ) from a point ( P(x_1, y_1, z_1) ) to a line defined by two points ( A(x_2, y_2, z_2) ) and ( B(x_3, y_3, z_3) ) or by its vector equation ( \vec{r} = \vec{a} + \lambda \vec{b} ) can be found using the following formula:

[ d = \frac{|\vec{PA} \times \vec{PB}|}{|\vec{AB}|} ]

where ( \vec{PA} ) and ( \vec{PB} ) are the position vectors of points ( A ) and ( B ) with respect to point ( P ), and ( \vec{AB} ) is the direction vector of the line.

Steps to Draw Perpendicular

  1. Identify the given point ( P ) and the given line ( L ).
  2. If the line is given in the form of an equation, convert it into a vector form if necessary.
  3. Use the formula to calculate the perpendicular distance.
  4. Determine the point of intersection, which is the foot of the perpendicular.

Differences and Important Points

Aspect Perpendicular from a Point to a Line
Definition A line segment from a point that intersects a given line at a right angle.
Importance Used to find the shortest distance from a point to a line.
Formula ( d = \frac{
Application Used in geometric constructions, proofs, and various fields such as engineering and physics.
Geometric Construction Requires a compass and straightedge or equivalent tools.
Analytical Geometry Requires knowledge of vectors and their operations.

Examples

Example 1: Geometric Construction

Given a point ( P ) and a line ( L ), construct a perpendicular from ( P ) to ( L ) using a compass and straightedge.

  1. Place the compass at point ( P ) and draw an arc that intersects line ( L ) at two points, say ( A ) and ( B ).
  2. Without changing the compass width, place the compass at point ( A ) and draw an arc above line ( L ).
  3. Repeat the same process with the compass at point ( B ).
  4. The intersection of the two arcs above line ( L ) determines point ( C ).
  5. Draw a straight line from point ( P ) to point ( C ). This line is the perpendicular from ( P ) to ( L ).

Example 2: Analytical Geometry

Find the perpendicular distance from point ( P(1, 2, 3) ) to the line passing through points ( A(4, 0, -1) ) and ( B(2, -2, 3) ).

  1. Calculate the direction vector ( \vec{AB} ) as ( \vec{B} - \vec{A} ).
  2. Find the vectors ( \vec{PA} ) and ( \vec{PB} ).
  3. Use the cross product to find ( \vec{PA} \times \vec{PB} ).
  4. Calculate the magnitude of ( \vec{AB} ).
  5. Use the formula to find the perpendicular distance ( d ).

Let's calculate it step by step:

  1. ( \vec{AB} = \vec{B} - \vec{A} = (2 - 4) \hat{i} + (-2 - 0) \hat{j} + (3 - (-1)) \hat{k} = -2 \hat{i} - 2 \hat{j} + 4 \hat{k} )
  2. ( \vec{PA} = \vec{A} - \vec{P} = (4 - 1) \hat{i} + (0 - 2) \hat{j} + (-1 - 3) \hat{k} = 3 \hat{i} - 2 \hat{j} - 4 \hat{k} )
  3. ( \vec{PB} = \vec{B} - \vec{P} = (2 - 1) \hat{i} + (-2 - 2) \hat{j} + (3 - 3) \hat{k} = \hat{i} - 4 \hat{j} )
  4. ( \vec{PA} \times \vec{PB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -2 & -4 \ 1 & -4 & 0 \end{vmatrix} = (8 \hat{i} + 4 \hat{j} + (12 - 2) \hat{k}) = 8 \hat{i} + 4 \hat{j} + 10 \hat{k} )
  5. ( |\vec{AB}| = \sqrt{(-2)^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} )
  6. ( d = \frac{|\vec{PA} \times \vec{PB}|}{|\vec{AB}|} = \frac{\sqrt{8^2 + 4^2 + 10^2}}{\sqrt{24}} = \frac{\sqrt{64 + 16 + 100}}{\sqrt{24}} = \frac{\sqrt{180}}{\sqrt{24}} = \frac{6\sqrt{5}}{2\sqrt{6}} = \frac{3\sqrt{5}}{\sqrt{6}} = \frac{3\sqrt{30}}{6} = \frac{\sqrt{30}}{2} )

Therefore, the perpendicular distance from point ( P ) to the line is ( \frac{\sqrt{30}}{2} ) units.

Understanding the concept of drawing a perpendicular from a point to a line is essential for solving problems in geometry and related fields. The ability to construct such a perpendicular geometrically or calculate it analytically is a fundamental skill in mathematics.