Orthogonal Projections
Given a line and a vector emanating from a point on , it is sometimes convenient to express as the sum of a vector , parallel to , and a vector , perpendicular to . If you have taken a physics course, you may have seen a force vector decomposed into the sum of two components: one parallel and one perpendicular to the direction of motion.
Suppose is a direction vector for . Then for some scalar . Our goal is to find . \begin{align*} \vec{v}\dotp \vec{d}&=(\vec{v}_{\parallel }+\vec{v}_{\perp })\dotp \vec{d}\\ &=(k\vec{d}+\vec{v}_{\perp })\dotp \vec{d}\\ &=k\vec{d}\dotp \vec{d}+\vec{v}_{\perp }\dotp \vec{d}\\ &=k\norm{\vec{d}}^2+0\\ &=k\norm{\vec{d}}^2 \end{align*}
We conclude that and
The vector is called the projection of onto . In our discussion, is a direction vector for line . So, we can also say that is the projection of onto .
To find , observe that .
Distance from a Point to a Line
The shortest distance from a point to a line is the length of the perpendicular line segment dropped from the point to the line. Vector projection formula will help us find the length of such a perpendicular.
Find the distance from to .
The line has a direction vector
We will now find the projection of onto
Next, we find .
Finally, to find the distance between point and line , we find the magnitude of .
Practice Problems
Problems prob:proj1a-prob:proj1b
Find .
Answer:
Answer: The graph below shows the line together with a circle of radius . Change the value of to the radius you have found to visualize the correct answer.