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Unformatted text preview: Lines and Planes in 3-space John E. Gilbert, Heather Van Ligten, and Benni Goetz There are many ways of expressing the equations of lines in 2-space (you probably learned the slope-intercept and point-slope formulas among others, for example). What we want now is to do the same for lines and planes in 3-space. Here vectors will be particularly convenient. Lines: Two points determine a line both in 2-space and 3-space . So imagine a laser pointer (or a light saber for Star Wars fans) at one of the two points, say b , and shine it towards the other point, say a . If we extend the laser pointer or light saber in both directions, we get a line. To write this as an equation, we just need to write the light saber and the idea of “extending” it mathematically. Represent the light saber as a displacement vector v shown in dark blue, with tail at b and head at a ; so v = a- b . To extend v in both directions we scale the vector by writing t v , shown in lighter blue, where t is a real number. Doing this for all such t gives us the complete line. So in vector-form each point on the line is given by r ( t ) = t v + b . x y z v b a t v + b Isn’t this like the slope-intercept form for a line in the plane? Sometimes it’s useful to write the equation for a line in coordinate-form : if we write r ( t ) = x ( t ) , y ( t ) , z ( t ) , v = k, m, n , b = x 1 , y 1 , z 1 , then the vector equation becomes r ( t ) = tk + x 1 , tm + x 2 , tn + b 3 , giving a second equation for a line: x ( t ) = tk + x 1 , y ( t ) = tm + y 1 , z ( t ) = tn + z 1 . Solving for t in these equations (and writing x instead of x ( t ) and so on), we then get t = x- x 1 k , t = y- y 1 m , t = z- z 1 n , giving a third equation x- x 1 k = y- y 1 m = z- z 1 n ....
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