How do you find the vector equation and parametric equations for the line through $(0, 14, - 10)$ $(0, 14, -10)$ and parallel to the line: $x = - 1 + 2 t, y = 6 - 3 t, z = 3 + 9 t$ $x = -1+2t, y = 6-3t, z = 3+9t$ ?

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Answer 1 · 2016-08-17T17:53:10

${ (x = 2t), (y=14-3t), (z = -10+9t) :}$

The usual parametric representation is written as

$p=p_0+t vec v$ .

Here we have

$p = {x,y,z}$
$p_0 = {-1,6,3}$
$t in RR$
$vec v = {2,-3,9}$

So, given

$p_1 = {0,14,-10}$

the parallel line is given by

$p = p_1 + t vec v$

which detailed is

${ (x = 2t), (y=14-3t), (z = -10+9t) :}$

How do you find the vector equation and parametric equations for the line through (0, 14, -10)(0,14,−10) (0, 14, -10) and parallel to the line:x = -1+2t, y = 6-3t, z = 3+9tx=−1+2t,y=6−3t,z=3+9tx = -1+2t, y = 6-3t, z = 3+9t?