Scalar Projection: Find Distance Point to Line

In summary, the scalar projection can be used to find the distance between a point (x1, y1) and a line ax + by + c = 0 by taking the scalar product of a unit normal to the line with a difference vector between the point and a point on the line. The unit normal can be found by taking the derivative of the line and the tangent vector will be (1, -a/b). The normal vector can then be found by taking the perpendicular vector to the tangent vector, which can be represented as (a, b). After normalizing it, the dot product can be taken with the difference vector to find the distance.
  • #1
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Homework Statement



Use the scalar projection to show that a distance from a point P(x1, y1) to the line ax + by + c = 0 is
[tex]\frac{ax1 + by1 + c}{\sqrt{a^2 + b^2}}[/tex]

Homework Equations



scalar projection = [tex]\frac{a . b}{|a|}[/tex]

The Attempt at a Solution


The first thing that I did was to say that b = (x1,y1). Unfortunately I'm having a difficult time coming up with a vector for the line. Honestly,, I am having a hard time visualizing how the scalar projection would yield the distance between a point and the line. Does anyone know what I should do to start?
 
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  • #2
First, points aren't vectors. Vectors are differences between points. Second, to use scalar projection the distance between a point and a line is the scalar product of a unit normal to the line with a difference vector between the point and a point on the line. Can you find a unit normal to the line?
 
  • #3
Ok got it thanks.
 
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  • #4
I'm not sure what you are up to in getting that 'normal' vector, P1 isn't even on the line (0,-c/b) is. Start by finding a tangent vector to the line. Since y=(-1/b)x-c/b and point on the line is (x,y), if I take d/dx, I get a tangent vector of (1,-a/b), right? A normal vector is perpendicular to that, like (a,b)? Now remember after you normalize it that you want to dot it with the DIFFERENCE between (x1,y1) and a point on the line.
 

1. What is scalar projection?

Scalar projection is a mathematical concept that involves finding the distance from a given point to a line. It is calculated by taking the dot product of the vector from the point to the line and the unit vector of the line.

2. How is scalar projection used in real-world applications?

Scalar projection has various applications in fields such as physics, engineering, and computer graphics. For example, it can be used to calculate the distance between a moving object and a fixed line, or to determine the height of a projectile based on its initial velocity and angle.

3. Can scalar projection be negative?

Yes, scalar projection can be negative. This occurs when the point is projected onto a line in the opposite direction of the line's unit vector. The negative value indicates that the point is on the opposite side of the line from where the unit vector is pointing.

4. What is the difference between scalar projection and vector projection?

The main difference between scalar projection and vector projection is that scalar projection only gives the magnitude or distance from the point to the line, while vector projection also takes into account the direction of the projected vector.

5. How do you find the distance point to line using scalar projection?

To find the distance from a point to a line using scalar projection, you will need to know the coordinates of the point and at least two points on the line. Then, you can use the formula: d = |(P-A) · u|, where d is the distance, P is the point, A is a point on the line, and u is the unit vector of the line.

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