|January 21st, 2017, 11:07 PM||#1|
Joined: Jan 2017
How to prove this?
show that in symmetric form of the line through point P'=(x',y',z') and parallel to non zero vector u = (a,b,c) is (x-x')/a = (y-y')/b = (z-z')/c.
|January 22nd, 2017, 12:42 AM||#2|
Joined: Feb 2012
This is just a way to state that P-P' = ku, k real, which is equivalent to saying that the line (P,P') is parallel to u. If you want to impress your teacher, you can say that this is a modern version of Thales theorem.
|January 28th, 2017, 05:26 AM||#3|
Joined: Jan 2015
Vector <p, q, r> is parallel to vector <u, v, w> if and only if there exist a non-zero number k such that p= ku, q= kv, and r= kw. Taking (x, y. z) to be any point on the line, with (x', y', z') the given point, then <x- x', y- y', z- z'> is a vector in the direction of the vector (a, b, c) so that there exist a non-zero number k such that x- x'= ka, y- y'= kb, and z- z'= kc.
from that , , and . Since those are all equal to k they are equal to each other:
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