Question

find the distance between the given point $\mathbf{S}$ and the line having the given direction vector $v$ or slope $m$ and passing through the given point T. $\mathbf{S}(5,1) ; m=-3, \mathbf{T}(2,-1)$

   find the distance between the given point $\mathbf{S}$ and the line having the given direction vector $v$ or slope $m$ and passing through the given point T.
$\mathbf{S}(5,1) ; m=-3, \mathbf{T}(2,-1)$
Modern Analytic Geometry
Modern Analytic Geometry
William Wooton,… 1st Edition
Chapter 3, Problem 8 ↓

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We have \(\mathbf{S}(5,1)\) and \(\mathbf{T}(2,-1)\).  Show more…

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find the distance between the given point $\mathbf{S}$ and the line having the given direction vector $v$ or slope $m$ and passing through the given point T. $\mathbf{S}(5,1) ; m=-3, \mathbf{T}(2,-1)$
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Key Concepts

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Vector Projection
Vector projection is the process of projecting one vector onto another to determine the component of the first vector in the direction of the second. When finding the distance from a point to a line, projecting the vector formed between the point and a point on the line onto the line’s normal vector isolates the perpendicular component. The magnitude of this projection is the point-to-line distance.
Point-to-Line Distance
This concept involves finding the shortest (perpendicular) distance from a given point to a straight line. In analytic geometry, this distance is computed by using a formula that leverages the coefficients of the line’s standard form, or equivalently by projecting the vector from any point on the line to the given point onto the normal vector of the line. This method minimizes distance along the perpendicular direction, which is the shortest path from the point to the line.
Line Equation
A line in the plane can be represented in various forms, such as the slope-intercept form (y = mx + b) or the point-slope form (y - y? = m(x - x?)). When given a point on the line and its slope, the point-slope form is particularly useful. It provides a practical way to derive the entire line’s equation, which is crucial for further geometric calculations, including distance measurement.
Direction and Normal Vectors
A direction vector defines the orientation of a line through its components indicating the line’s direction. Correspondingly, a normal vector is perpendicular to the direction vector and thus to the line. In the context of computing distances, the normal vector is essential as it is used to project the vector connecting the given point to a point on the line, directly yielding the shortest distance.

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$69-70$ Use the formula in Exercise 12.4 .45 to find the distance from the point to the given line. $$(0,1,3) ; \quad x=2 t, \quad y=6-2 t, \quad z=3+t$$

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