Scalar Projection: Find Distance Point to Line

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The discussion focuses on using scalar projection to determine the distance from a point P(x1, y1) to the line defined by ax + by + c = 0. The key formula for this distance is derived as \(\frac{ax1 + by1 + c}{\sqrt{a^2 + b^2}}\). Participants clarify that while points are not vectors, the distance can be calculated using a unit normal vector to the line and a difference vector from the point to a point on the line. A tangent vector to the line is identified, leading to the conclusion that a normal vector can be derived as (a, b). The discussion emphasizes the importance of normalizing the vector before applying the dot product to find the distance.
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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
\frac{ax1 + by1 + c}{\sqrt{a^2 + b^2}}

Homework Equations



scalar projection = \frac{a . b}{|a|}

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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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?
 
Ok got it thanks.
 
Last edited:
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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