Find parametric equation for the line
asked Jun 1, 2012 in Calculus Answers by anonymous

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:

To avoid this verification in future, please log in or register.

1 Answer

Given two parametric lines defined by

p = p1 + t(A)

q = q1 + s(B)

these lines will be parallel if the unit direction vectors A and B are the same.

Let us define p as

p = p1 + t(p2 - p1) where

p1 = (1, 2, 0) and p2 - p1 = (2, -1, 3) taken from x = 1 + 2t, y = 2 - t, z = 3t

Let A = (p2-p1)/norm(p2-p1) be the normalized direction vector of p2-p1 where

norm(p2-p1) = sqrt(4 + 1 + 9) = sqrt(14)

Then A = (2/sqrt(14), -1/sqrt(14), 3/sqrt(14)) which is the normalized direction vector for line p so that

p = p1 + t(A)

If we set B = A, then the parametric line q passing through the point q1 = (3, -2, 1) is given by

q = q1 + s(B)

Because B = A, line q is parallel to the line p given above.

Note that when the direction vector is normalized, the parameters "s" and "t" for lines q and p represent the Euclidian distance from the points q1 and p1 respectively.
answered Mar 18, 2013 by anonymous
Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
80,288 questions
84,162 answers
67,172 users