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 January 21st, 2017, 11:07 PM #1 Newbie   Joined: Jan 2017 From: Pakistan Posts: 4 Thanks: 0 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 Senior Member   Joined: Feb 2012 Posts: 144 Thanks: 16 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 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,338 Thanks: 588 Vector is parallel to vector 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 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 $\frac{x- x'}{a}= k$, $\frac{y- y'}{b}= k$, and $\frac{z- z'}{c}= k$. Since those are all equal to k they are equal to each other: $\frac{x- x'}{a}= \frac{y- y'}{b}= \frac{z- z'}{c}$

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