» Required Reading » Table of Contents » Chapter 5 5.3 Representations of a line in two and three dimensions
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Two points P₁ and P₂  on L determine L.

L can be described parametrically as the set of points with coordinates those of P₁ + s * (P₂ - P₁).

(P₂ - P₁) is a vector which points in the direction of L.

In two dimensions so that the vectors here are 2-vectors, there is only one direction perpendicular to L, and that direction can be obtained by switching the coordinates of (P₂ - P₁)  and changing one sign, (thus (7, -4) is perpendicular to (4, 7))
With N the perpendicular vector, the equation of the line becomes Nr = NP₁.

We do this out explicitly  L consists of the points obeying

x  = P_1x  + s*(P_2x -P_1x )

y  = P_1y  + s*(P_2y -P_1y )

and the equation for L is

(P_2y – P_1y ) x - (P_2x -P_1x )y = (P_2y -P_1y )P_1x - (P_2x -P_1x )P_1y

which when solved for y is

   y = ((P_2y - P_1y ) / (P_2x - P_1x ))x + C for some constant C.

The ratio ((P_2y - P_1y ) / (P_2x - P_1x )), the coefficient of x in the equation for the line, is the difference of y coordinates of the two points divided by the difference in x coordinates. It is called the slope of the line L.

In three dimensions, a line is determined by two equations. You can describe it as above parametrically, but you must find two vectors normal to (P₂ - P₁) to find the equations. You have infinite choice but convenient choices  are

i( P₂ - P₁)  and j(P₂ - P₁),

and you can require that the dot product of each of these with r is what it is with P₁. This gives you two equations which together determine the line.

To find a point on a line from equations you fix one coordinate arbitrarily and solve the two equations for the other two coordinates.

If you drop a perpendicular from a point to a line or plane, the point you reach on that line or plane is called the projection of the point onto the line or plane.

Exercises:

5.3 Draw an appropriate picture and find a formula for calculating each of the following:

a. The projection of a point onto a line. (Find a vector from a point Q on the line to the given point P, and project that vector onto the vector representing the line. The resulting vector will point from Q to the desired projection)

b. The projection of a point onto a plane. (Find a vector from a point Q in the plane to the given point P, find its projection normal to the plane, and subtract that projection from it. That will point from Q to the desired projection)

c. The distance between a point and a plane, the plane defined by 3 pts in it. (Find a vector from a point Q in the plane to the given point P. The desired distance is the length of the projection of this vector normal to the plane.)

d. The distance between a point and a line the line described by 2 pts on it. (Find a vector from a point Q on the line to the given point P. The desired distance is the length of the vector obtained by subtracting its projection along the line direction from it.)

e. The distance between two skew lines in 3 space. (Find a vector from a point on one line to one on the other and the desired distance is the length of the projection of this line on the normal to both lines.)

f. The distance between two parallel lines in 3 space.(Find a vector from a point on one line to one on the other and subtract from it its projection in the direction of the lines. The length of the result is the desired distance.)

5.4 Examine the line slope applet. What can you say about lines with slope > 1? > 0? < -1?